holt physics appendix

Selected Answers 942

Glossary 952

Index 958

831831

Appendix A Mathematical Review . . . . . . . . . . . . . . . . . . . . . . . . 832

Appendix B Downloading Graphing Calculator Programs . . . . 847

Appendix C Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 848

Appendix D Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854

Appendix E SI Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866

Appendix F Useful Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 868

Appendix G Periodic Table of the Elements . . . . . . . . . . . . . . . . 872

Appendix H Abbreviated Table of Isotopes and Atomic Masses . . . . . . . . . . . . . . . . . . . . . . . . . . 874

Appendix I Additional Problems . . . . . . . . . . . . . . . . . . . . . . . . . 880

Appendix J Advanced Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 897

Angular Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 898

Tangential Speed and Acceleration . . . . . . . . . . . . . . . . . . . 902

Rotation and Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 904

Rotational Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 906

Properties of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 908

Fluid Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 910

The Doppler Effect and the Big Bang . . . . . . . . . . . . . . . . . 912

Special Relativity and Time Dilation . . . . . . . . . . . . . . . . . 914

Special Relativity and Velocities . . . . . . . . . . . . . . . . . . . . . 916

The Equivalence of Mass and Energy . . . . . . . . . . . . . . . . . 918

General Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 920

De Broglie Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 922

Electron Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924

Semiconductor Doping . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926

Superconductors and BCS Theory . . . . . . . . . . . . . . . . . . . 928

Antimatter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 930

Appendix K Selected CBL™ Procedures . . . . . . . . . . . . . . . . . . 932

Free-Fall Acceleration (Chapter 2) . . . . . . . . . . . . . . . . . . . 932

Force and Acceleration (Chapter 4) . . . . . . . . . . . . . . . . . . 934

Specific Heat Capacity (Chapter 9) . . . . . . . . . . . . . . . . . . . 936

Speed of Sound (Chapter 12) . . . . . . . . . . . . . . . . . . . . . . . . 938

Magnetic Field of a Conducting Wire (Chapter 19) . . . . . 940

ContentsReference Section

AP

PE

ND

IX A

Mat

hem

atic

al R

evie

w

832 832 Appendix A: Mathematical Review

Mathematical Review

Positive exponents Many quantities that scientists deal with often have

very large or very small values. For example, the speed of light is about

300 000 000 m/s, and the ink required to make the dot over an i in this text-

book has a mass of about 0.000 000 001 kg. Obviously, it is cumbersome to

work with numbers such as these. We avoid this problem by using a method

based on powers of the number 10.

100 = 1

101 = 10

102 = 10 × 10 = 100

103 = 10 × 10 × 10 = 1000

104 = 10 × 10 × 10 × 10 = 10 000

105 = 10 × 10 × 10 × 10 × 10 = 100 000

The number of zeros determines the power to which 10 is raised, or the expo-

nent of 10. For example, the speed of light, 300 000 000 m/s, can be expressed

as 3 × 108 m/s. In this case, the exponent of 10 is 8.

Negative exponents For numbers less than one, we note the following:

10−1 = ⎯1

1

0⎯ = 0.1

10−2 = ⎯10 ×

1

10⎯ = 0.01

10−3 = ⎯10 × 1

1

0 × 10⎯ = 0.001

10−4 = ⎯10 × 10 ×

1

10 × 10⎯ = 0.0001

10−5 = = 0.000 01

The value of the negative exponent equals the number of places the decimal

point must be moved to be to the right of the first nonzero digit (in these

cases, the digit 1). Numbers that are expressed as a number between 1 and

10 multiplied by a power of 10 are said to be in scientific notation. For exam-

ple, 5 943 000 000 is 5.943 × 109 when expressed in scientific notation, and

0.000 083 2 is 8.32 × 10−5 when expressed in scientific notation.

1⎯⎯⎯10 × 10 × 10 × 10 × 10

1. Word

Scientific Notation

833Appendix A: Mathematical Review 833

Multiplication and division in scientific notation When numbers

expressed in scientific notation are being multiplied, the following general

rule is very useful:

10n × 10m = 10(n +m)

Note that n and m can be any numbers; they are not necessarily integers. For

example, 102 × 105 = 107, and 101/4 × 101/2 = 103/4. The rule also applies to

negative exponents. For example, 103 × 10−8 = 10−5. When dividing numbers

expressed in scientific notation, note the following:

⎯1

1

0

0m

n

⎯ = 10n × 10−m = 10(n −m)

For example, ⎯1

1

0

0

3

2⎯ = 10(3−2) = 101.

The rules for multiplying, dividing, adding, and subtracting fractions are

summarized in Table 1, where a, b, c, and d are four numbers.

1. Triangles

Operation Rule Example

Multiplication �⎯a

b⎯��⎯

d

c⎯� = ⎯

b

a

d

c⎯ �⎯

2

3⎯��⎯

4

5⎯� = ⎯

(

(

2

3

)

)

(

(

4

5

)

)⎯ = ⎯

1

8

5⎯

Division

= ⎯a

b

d

c⎯ = ⎯

(

(

2

3

)

)

(

(

5

4

)

)⎯ = ⎯

1

1

0

2⎯ = ⎯

5

6⎯

Addition and subtraction⎯a

b⎯ ± ⎯

d

c⎯ = ⎯

ad

b

±

d

bc⎯ ⎯

2

3⎯ − ⎯

4

5⎯ = ⎯

(2)(5

(

)

3)

−(5

(

)

3)(4)⎯ = − ⎯

1

2

5⎯

�⎯2

3⎯�

⎯

�⎯4

5⎯�

�⎯a

b⎯�

⎯

�⎯d

c⎯�

Fractions

Table 1 Basic Operations for Fractions

Powers

Rules of exponents When powers of a given quantity, x, are multiplied, the

rule used for scientific notation applies:

(xn)(xm) = x(n+m)

For example, (x2)(x4) = x(2+4) = x6.

When dividing the powers of a given quantity, note the following:

⎯x

xm

n

⎯ = x(n−m)

For example, ⎯x

x8

2⎯ = x(8−2) = x6.

1. Triangles


A power that is a fraction, such as ⎯13

⎯, corresponds to a root as follows:

x1/n = n√

x�

For example, 41/3 = 3√

4� = 1.5874. (A scientific calculator is useful for such

calculations.)

Finally, any quantity, xn, that is raised to the mth power is as follows:

(xn)m = xnm

For example, (x2)3 = x(2)(3) = x6.

The basic rules of exponents are summarized in Table 2.

Algebra

Solving for unknowns When algebraic operations are performed, the laws

of arithmetic apply. Symbols such as x, y, and z are usually used to represent

quantities that are not specified. Such unspecified quantities are called

unknowns.

First, consider the following equation:

8x = 32

If we wish to solve for x, we can divide each side of the equation by the same

factor without disturbing the equality. In this case, if we divide both sides by

8, we have the following:

⎯8

8

x⎯ = ⎯

3

8

2⎯

x = 4

Next, consider the following equation:

x + 2 = 8

In this type of expression, we can add or subtract the same quantity from

each side. If we subtract 2 from each side, we get the following:

x + 2 − 2 = 8 − 2

x = 6

In general, if x + a = b, then x = b − a.

1. Triangles

x0 = 1 x1 = x (xn)(xm) = x(n+m)

⎯x

xm

n

⎯ = x(n −m) x(1/n) = n√

x� (xn)m = x(nm)

Table 2 Rules of Exponents


Now, consider the following equation:

⎯5

x⎯ = 9

If we multiply each side by 5, we are left with x isolated on the left and a value

of 45 on the right.

(5)�⎯5

x⎯� = (9)(5)

x = 45

In all cases, whatever operation is performed on the left side of the equation

must also be performed on the right side.

Factoring Some useful formulas for factoring an equation are given in Table 3. As an

example of a common factor, consider the equation 5x + 5y + 5z = 0. This

equation can be expressed as 5(x + y + z) = 0. The expression a2 + 2ab + b2,

which is an example of a perfect square, is equivalent to the expression

(a + b)2. For example, if a = 2 and b = 3, then 22 + (2)(2)(3) + 32 = (2 + 3)2,

or (4 + 12 + 9) = 52 = 25. Finally, for an example of the difference of two

squares, let a = 6 and b = 3. In this case, (62 − 32) = (6 + 3)(6 − 3), or

(36 − 9) = (9)(3) = 27.

Table 3 Factoring Equations

ax + ay + az = a(x + y + z) common factor

a2 + 2ab + b2 = (a + b)2 perfect square

a2 − b2 = (a + b)(a − b) difference of two squares

Quadratic Equations The general form of a quadratic equation is as follows:

ax2 + bx + c = 0

In this equation, x is the unknown quantity and a, b, and c are numerical fac-

tors known as coefficients. This equation has two roots, given by the following:

x = ⎯−b ±√

2

b�a

2�−� 4�ac�⎯

If b2 ≥ 4ac, the value inside the square-root symbol will be positive or zero

and the roots will be real. If b2 < 4ac, the value inside the square-root symbol

will be negative and the roots will be imaginary numbers. In problems in this

physics book, imaginary roots should not occur.


ExampleFind the solutions for the equation x2 + 5x + 4 = 0.

SolutionThe given equation can be expressed as (1)x2 + (5)x + (4) = 0. In other

words, a = 1, b = 5, and c = 4. The two roots of this equation can be

found by substituting these values into the quadratic equation, as follows:

x = = = = ⎯−5

2

± 3⎯

The two roots are x = ⎯−5

2

+ 3⎯ = −1 and x = ⎯

−5

2

− 3⎯ = −4.

We can evaluate these answers by substituting them into the given

equation and verifying that the result is zero.

x2 + 5x + 4 = 0

For x = −1, (−1)2 + 5(−1) + 4 = 1 − 5 + 4 = 0.

For x = −4, (−4)2 + 5(−4) + 4 = 16 − 20 + 4 = 0.

x = −1 and x = −4

−5 ±√

9�⎯

2

−5 ±√

52� −� (�4)�(1�)(�4)�⎯⎯⎯

(2)(1)

−b ±√

b�2�−� 4�ac�⎯⎯

2a

ExampleFactor the equation 2x2 − 3x − 4 = 0.

SolutionThe given equation can be expressed as (2)x2 + (−3)x + (−4) = 0. Thus,

a = 2, b = −3, and c = −4. Substitute these values into the quadratic

equation to factor the given equation.

x = = = = ⎯3 ± 6

4

.403⎯

The two roots are x = ⎯3 + 6

4

.403⎯ = 2.351 and x = ⎯

3 − 6

4

.403⎯ = −0.851.

Again, evaluate these answers by substituting them into the given equation.

2x2 − 3x − 4 = 0

For x = 2.351, 2(2.351)2 − 3(2.351) − 4 = 11.054 − 7.053 − 4 ≈ 0.

For x = −0.851, 2(−0.851)2 − 3(−0.851) − 4 = 1.448 + 2.553 − 4 ≈ 0.

x = 2.351 and x = −0.851

3 ±√

41�⎯

4

3 ±√

(−�3)�2�−� (�4)�(2�)(�−�4)�⎯⎯⎯

(2)(2)

−b ±√

b�2�−� 4�ac�⎯⎯

2a


Linear Equations A linear equation has the following general form:

y = ax + b

In this equation, a and b are constants. This equation is called linear because

the graph of y versus x is a straight line, as shown in Figure 1. The constant b,

called the intercept, represents the value of y where the straight line intersects

the y-axis. The constant a is equal to the slope of the straight line and is also

equal to the tangent of the angle that the line makes with the x-axis (q). If

any two points on the straight line are specified by the coordinates (x1, y1)

and (x2, y2), as in Figure 1, then the slope of the straight line can be

expressed as follows:

slope = ⎯x

y2

2

−−

y

x1

1⎯ = ⎯

ΔΔ

x

y⎯

For example, if the two points shown in Figure 1 are (2, 4) and (6, 9), then

the slope of the line is as follows:

slope = ⎯(

(

9

6

−−

4

2

)

)⎯ = ⎯

5

4⎯

Note that a and b can have either positive or negative values. If a > 0, the

straight line has a positive slope, as in Figure 1. If a < 0, the straight line has

a negative slope. Furthermore, if b > 0, the y intercept is positive (above the

x-axis), while if b < 0, the y intercept is negative (below the x-axis). Figure 2gives an example of each of these four possible cases, which are summarized

in Table 4.

x

Δx

Δy

y (x2, y2)(x1, y1)

(0, 0)

(0, b)θ

θ

ya > 0b > 0

a > 0b < 0

a < 0b > 0

a < 0b < 0

x

(1) (2)

(3)

(4)

Solving SimultaneousLinear Equations

Consider the following equation:

3x + 5y = 15

This equation has two unknowns, x and y. Such an equation does not have a

unique solution. That is, (x = 0, y = 3), (x = 5, y = 0), and �x = 2, y = ⎯95

⎯� are all

solutions to this equation.

If a problem has two unknowns, a unique solution is possible only if there

are two independent equations. In general, if a problem has n unknowns, its

solution requires n independent equations. There are three basic methods

that can be used to solve simultaneous equations. Each of these methods is

discussed below, and an example is given for each.

a > 0, b > 0 positive slope positive y intercept

a > 0, b < 0 positive slope negative y intercept

a < 0, b > 0 negative slope positive y intercept

a < 0, b < 0 negative slope negative y intercept

Table 4 Linear Equations

Figure 1

Figure 2

Constants Slope y intercept


ExampleSolve the following two simultaneous equations:

1. 5x + y = −8

2. 2x − 2y = 4

SolutionFirst solve for either x or y in one of the equations. We’ll begin by solv-

ing equation 2 for x.

2. 2x − 2y = 4

2x = 4 + 2y

x = ⎯4 +

2

2y⎯ = 2 + y

Next, we substitute this equation for x into equation 1 and solve for y.

1. 5x + y = −8

5(2 + y) + y = −8

10 + 5y + y = −8

6y = −18

To find x, substitute this value for y into the equation for x derived from

equation 2.

x = 2 + y = 2 + −3

x = −1

y = −3

First method: substitution One way to solve two simultaneous equations

involving two unknowns, x and y, is to solve one of the equations for one of

the unknown values in terms of the other unknown value. In other words,

either solve one equation for x in terms of y or solve one equation for y in

terms of x. Once you have an expression for either x or y, substitute this

expression into the other original equation. At this point, the equation has

only one unknown quantity. This unknown can be found through algebraic

manipulations and then can be used to determine the other unknown.

There is always more than one way to solve simultaneous equations by sub-

stitution. In this example, we first solved equation 2 for x. However, we could

have begun by solving equation 2 for y or equation 1 for x or y. Any of these

processes would result in the same answer.


Second method: canceling one term Simultaneous equations can also be

solved by multiplying both sides of one of the equations by a value that will

make either the x value or the y value in that equation equal to and opposite

the corresponding value in the second equation. When the two equations are

added together, that unknown value drops out and only one of the unknown

values remains. This unknown can be found through algebraic manipula-

tions and then can be used to determine the other unknown.

In this example, we multiplied both sides of equation 1 by 2 so that the y

terms would cancel when the two equations were added together. As with

substitution, this is only one of many possible ways to solve the equations.

For example, we could have multiplied both sides of equation 2 by ⎯34

⎯ so that

the x terms would cancel when the two equations were added together.


1. 3x + y = −6

2. −4x − 2y = 6

SolutionFirst, multiply each term of one of the equations by a factor that will

make either the x or the y values cancel when the two equations are

added together. In this case, we can multiply each term in equation 1 by

the factor 2. The positive 2y in equation 1 will then cancel the negative

2y in equation 2.

1. 3x + y = −6

(2)(3x) + (2)(y) = −(2)(6)

6x + 2y = −12

Next, add the two equations together and solve for x.

2. −4x − 2y = 6

1. 6x + 2y = −12

2x = −6

Then, substitute this value of x into either equation to find y.

1. 3x + y = −6

y = −6 − 3x = −6 − (3)(−3) = −6 + 9

y = 3

x = −3


Third method: graphing the equations Two linear equations with two

unknowns can also be solved by a graphical method. If the straight lines cor-

responding to the two equations are plotted in a conventional coordinate sys-

tem, the intersection of the two lines represents the solution.

Suppose that a quantity, x, is expressed as a power of another quantity, a.

x = ay

The number a is called the base number. The logarithm of x with respect to

the base, a, is equal to the exponent to which a must be raised in order to sat-

isfy the expression x = ay.

y = logax

Conversely, the antilogarithm of y is the number x.

x = antilogay


1. x − y = 2

2. x − 2y = −1

SolutionThese two equations are plotted in Figure 3. To plot an equation, rewrite

the equation in the form y = ax + b, where a is the slope and b is the y

intercept. In this example, the equations can be rewritten as follows:

y = x − 2

y = ⎯12

⎯x + ⎯12

⎯

Once one point of a line is known, any other point on that line can be

found with the slope of the line. For example, the slope of the first line is 1,

and we know that (0, −2) is a point on this line. If we choose the point x =2, we have (2, y2). The coordinate y2 can be found as follows:

slope = ⎯x

y2

2

−−

y

x1

1⎯ = ⎯

y2

2

−−(−

0

2)⎯ = 1

y2 = 0

Connecting the two known coordinates, (0, −2) and (2, 0), results in a

graph of the line. The second line can be plotted with the same method.

As shown in Figure 3, the intersection of the two lines has the coor-

dinates x = 5, y = 3. This intersection represents the solution to the

equations. You should check this solution using either of the analytical

techniques discussed above.

x − 2y = −1

x − y = 2

y

x123456

2 3 4 5 6

(5, 3)

1

Figure 3

Logarithms


Common and natural bases In practice, the two bases most often used are

base 10, called the common logarithm base, and base e = 2.718…, called the

natural logarithm base. When common logarithms are used, y and x are

related as follows:

y = log10 x, or x = 10y

When natural logarithms are used, the symbol ln is used to signify that the

logarithm has a base of e; in other words, log ex = ln x.

y = ln x, or x = ey

For example, log10 52 = 1.716, so antilog10 1.716 = 101.716 = 52. Likewise,

ln 52 = 3.951, so antiln 3.951 = e3.951 = 52.

Note that you can convert between base 10 and base e with the equality

ln x = (2.302 585)log10 x.

Some useful properties of logarithms are summarized in Table 5.

Table 5 Properties of Logarithms

Rule Example

log (ab) = log a + log b log (2)(5) = log 2 + log 5

log �⎯a

b⎯� = log a − log b log ⎯3

4⎯ = log 3 − log 4

log (an) = n log a log 73 = 3 log 7

ln e = 1

ln ea = a ln e5 = 5

ln �⎯1

a⎯� = −ln a ln ⎯1

8⎯ = −ln 8

The rules for converting numbers from fractions to decimals and percentages

and from percentages to decimals are summarized in Table 6.

Conversions Between Fractions, Decimals, and Percentages

Conversion Rule Example

Fraction to divide numerator⎯3

4

1

5⎯ = 0.69decimal by denominator

Fraction to convert to decimal, then⎯3

4

1

5⎯ = (0.69)(100%) = 69%percentage multiply by 100%

Percentage to move decimal point two 69% = 0.69decimal places to the left, and

remove the percent sign

Table 6 Conversions


Table 7 provides equations for the area and volume of several geometrical

shapes used throughout this text.

1. Triangles

Table 7 Geometrical Areas and Volumes

Geometrical shape Useful equations

area = lw

area = pr2

circumference = 2pr

area = ⎯1

2⎯bh

surface area = 4pr2

volume = ⎯4

3⎯pr3

surface area = 2pr2 + 2prl

surface area = 2(lh + lw + hw)

volume = lwh

rectangular box

h wl

cylinder

r

l

sphere

r

triangle

h

b

circle

r

rectangle

w

l

Geometry

perimeter = 2(l + w)

volume = pr2l


The portion of mathematics that is based on the relationships between the

sides and angles of triangles is called trigonometry. Many of the concepts of this

branch of mathematics are of great importance in the study of physics. To

review some of the basic concepts of trigonometry, consider the right triangle

shown in Figure 4, where side a is opposite the angle q, side b is adjacent to the

angle q, and side c is the hypotenuse of the triangle (the side opposite the right

angle). The most common trigonometry functions are summarized in Table 8,using this figure as an example.

1. Triangles

ac

90°θb

sin q = ac

cos q = bc

tan q = ab

Trigonometry and the Pythagorean Theorem

Figure 4

Table 8 Trigonometry Functions

sine (sin)sin q = ⎯

sid

h

e

yp

o

o

p

t

p

e

o

n

s

u

it

s

e

e

q⎯ = ⎯

a

c⎯

cosine (cos)cos q = ⎯

side

hy

a

p

d

o

ja

t

c

e

e

n

n

u

t

se

to q⎯ = ⎯

b

c⎯

tangent (tan)tan q = ⎯

si

s

d

id

e

e

ad

o

j

p

a

p

c

o

en

si

t

te

to

qq

⎯ = ⎯a

b⎯

inverse sine (sin−1)q = sin−1�⎯sid

h

e

yp

o

o

p

t

p

e

o

n

s

u

it

s

e

e

q⎯ � = sin−1�⎯

a

c⎯�

inverse cosine (cos−1)q = cos−1�⎯side

hy

a

p

d

o

ja

t

c

e

e

n

n

u

t

se

to q⎯ � = cos−1�⎯

b

c⎯�

inverse tangent (tan−1)q = tan−1�⎯si

s

d

i

e

de

ad

o

j

p

a

p

ce

o

n

si

t

te

to

qq

⎯ � = tan−1�⎯a

b⎯�

When q = 30°, for example, the ratio of a to c is always 0.50. In other words,

sin 30° = 0.50. Sine, cosine, and tangent are quantities without units because

each represents the ratio of two lengths. Furthermore, note the following

trigonometry identity:

⎯c

si

o

n

s

qq

⎯⎯ = = ⎯si

s

d

i

e

de

ad

o

j

p

a

p

ce

o

n

si

t

te

to

qq

⎯ = tan q

Some additional trigonometry identities are as follows:

sin2q + cos2q = 1

sin q = cos(90° − q)

cos q = sin(90° − q)

⎯sid

h

e

yp

o

o

p

t

p

e

o

n

s

u

it

s

e

e

q⎯

⎯⎯⎯side

hy

a

p

d

o

ja

t

c

e

e

n

n

u

t

s

t

e

o q⎯


Determining an unknown side The first three functions given in Table 8can be used to determine any unknown side of a right triangle when one side

and one of the non-right angles are known. For example, if q = 30° and

a = 1.0 m, the other two sides of the triangle can be found as follows:

sin q = ⎯a

c⎯

c = ⎯sin

a

q⎯ = ⎯

s

1

in

.0

3

m

0°⎯

tan q = ⎯a

b⎯

b = ⎯tan

a

q⎯ = ⎯

ta

1

n

.0

3

m

0°⎯

Determining an unknown angle In some cases, you might know the value

of the sine, cosine, or tangent of an angle and need to know the value of the

angle itself. The inverse sine, cosine, and tangent functions given in Table 8can be used for this purpose. For example, in Figure 4, suppose you know

that side a = 1.0 m and side c = 2.0 m. To find the angle q, you could use the

inverse sine function, sin−1, as follows:

q = sin−1�⎯a

c⎯� = sin−1�⎯12.

.

0

0

m

m⎯� = sin−1(0.50)

Converting from degrees to radians The two most common units used

to measure angles are degrees and radians. A full circle is represented by

360 degrees (360°) or by 2p radians (2p rad). As such, the following conver-

sions can be used:

[angle (°)] = ⎯1

p80⎯ [angle (rad)]

[angle (rad)] = ⎯1

p80⎯ [angle (°)]

Pythagorean theorem Another useful equation when working with right

triangles is the Pythagorean theorem. If a and b are the two legs of a right

triangle and c is the hypotenuse, as in Figure 5, the Pythagorean theorem can

be expressed as follows:

c2 = a2 + b2

In other words, the square of the hypotenuse of a right triangle equals the

sum of the squares of the other two legs of the triangle. The Pythagorean

q = 30°

b = 1.7 m

c = 2.0 m

ac

90°

b

Figure 5


Absolute error Some of the laboratory experiments in this book involve finding

a value that is already known, such as free-fall acceleration. In this type of experi-

ment, the accuracy of your measurements can be determined by comparing

your results with the accepted value. The absolute value of the difference

between your experimental or calculated result and the accepted value is called

the absolute error. Thus, absolute error can be found with the following equation:

absolute error = ⏐experimental − accepted⏐

Be sure not to confuse accuracy with precision. The accuracy of a measurement

refers to how close that measurement is to the accepted value for the quantity

being measured. Precision depends on the instruments used to measure a quan-

tity. A meterstick that includes millimeters, for example, will give a more pre-

cise result than a meterstick whose smallest unit of measure is a centimeter.

Thus, a measurement of 9.61 m/s2 for free-fall acceleration is more precise than

a measurement of 9.8 m/s2, but 9.8 m/s2 is more accurate than 9.61 m/s2.

Relative error Note that a measurement that has a relatively large absolute

error may be more accurate than a measurement that has a smaller absolute

error if the first measurement involved much larger quantities. For this rea-

son, the percentage error, or relative error, is often more meaningful than the

absolute error. The relative error of a measured value can be found with the

following equation:

relative error =(experimental − accepted)⎯⎯⎯

accepted

theorem is useful when two sides of a right triangle are known but the third

side is not. For example, if c = 2.0 m and a = 1.0 m, you could find b using the

Pythagorean theorem as follows:

b =√

c2� −� a�2� =√

(2�.0� m�)2� −� (�1.�0�m�)2�b =

√4.�0�m�2�−� 1�.0� m�2� =

√3.�0�m�2�

Law of sines and law of cosines The law of sines may be used to find angles

of any general triangle. The law of cosines is used for calculating one side of a

triangle when the angle opposite and the other two sides are known. If a, b, and

c are the three sides of the triangle and qa, qb, and q c are the three angles oppo-

site those sides, as shown in Figure 6, the following relationships hold true:

⎯sin

a

qa⎯ = ⎯

sin

b

qb⎯ = ⎯

sin

c

q c⎯

c2 = a2 + b2 − 2ab cos q c

b = 1.7 m

1. Triangles

Accuracy in Laboratory Calculations

Figure 6

a

c

θ

b

bθc

θa


Note that the absolute error is less in the first experiment, while the relative

error is less in the second experiment. The absolute error is less in the first

experiment because typical values for free-fall acceleration are much smaller

than typical values for the speed of sound in air. The relative errors take

this difference into account. Thus, comparing the relative errors shows that

the speed of sound is measured with greater accuracy than is the free-fall

acceleration.

For the second experiment, the absolute and relative errors can be cal-

culated as follows:

absolute error = ⏐experimental − accepted⏐ = ⏐355 m/s − 346 m/s⏐

relative error = =

relative error = 0.026 = 2.6%

(355 m/s − 346 m/s)⎯⎯⎯

346 m/s

(experimental − accepted)⎯⎯⎯

accepted

absolute error = 9 m/s

For the first experiment, the absolute and relative errors can be calcu-

lated as follows:

absolute error = ⏐experimental − accepted⏐ = ⏐10.31 m/s2 − 9.81 m/s2⏐

relative error = =

relative error = 0.051 = 5.1%

(10.31 m/s2 − 9.81 m/s2)⎯⎯⎯

9.81 m/s2

(experimental − accepted)⎯⎯⎯

accepted

absolute error = 0.50 m/s2

In other words, the relative error is the difference between the experimental

value and the accepted value divided by the accepted value. Because relative

error takes the size of the quantity being measured into account, the accuracy of

two different measurements can be compared by comparing their relative errors.

For example, consider two laboratory experiments in which you are determin-

ing values that are fairly well known. In the first, you determine that free-fall

acceleration at Earth’s surface is 10.31 m/s2. In the second, you find that the

speed of sound in air at 25°C is 355 m/s. The accepted values for these quanti-

ties are 9.81 m/s2 and 346 m/s, respectively. Now we’ll find the absolute and

relative errors for each experiment.

847Appendix B: Downloading Graphing Calculator Programs 847

If you will be doing the Technology and Learning

exercises, click Download Graphing Calculator

Programs. The file GRAPHING.ZIP will be loaded

onto your computer. Once the file is downloaded,

double-click the icon and the file will be extracted

into a file called Graphing.8xg. Follow the instruc-

Graphing Calculator Exercises

AP

PE

ND

IX B

Calculator P

rograms

Downloading GraphingCalculator Programs

The following steps describe how to transfer the

DataMate data collection program from the CBL2

or LabPro interface to the calculator.

1. Prepare the calculator to receive the program.

• For the TI-73, TI-73 Explorer, TI-82, TI-83, TI-

83 Plus, TI-83 Plus Silver Edition, TI-84 Plus,

and TI-84 Plus Silver Edition, turn on your cal-

culator and press @ Ò. (On the TI-73,

press Ï and then select “Link…”.) Press ®

to RECEIVE and then press e.

[“Waiting…” appears on your screen.]

• For the TI-86, turn on your calculator, press

@ Ò, and then press Ó. [“Waiting…”

appears on your screen.]

• For the TI-89, TI-89 Titanium, TI-92, TI-92

Plus, and Voyage 200, turn on your calculator

and make sure that the calculator is on the

Home screen. (You do not need to put the

calculator into the Receiving mode as is

required with the other calculators.)

CBL2 Labs

Topic: Graphing CalculatorGo To: go.hrw.comKeyword: HF6 CALC

Visit the HRW Web site for links fordownloading graphing calculatorprograms.

2. Press the Transfer button on CBL2 or LabPro

interface. [“Receiving…” followed by a list of

the loaded programs (or application) is

displayed on the calculator.]

3. A “Done” message on the calculator and two

beeps from the interface will indicate that the

transfer is complete. Press @ q.

4. Verify that the programs have been successfully

loaded.

• On the TI-82 and TI-83, press p to see

DATAMATE and its subprograms.

• On the TI-73, TI-73 Explorer, TI-83 Plus,

TI-83 Plus Silver Edition, TI-84 Plus, and

TI-84 Plus Silver Edition, press Ï to see

DATAMATE.

• On the TI-86, press p and then ˝ to

see DATAMATE and its subprograms.

• On the TI-89, TI-89 Titanium, TI-92, TI-92

Plus, and Voyage 200, press @ Ù to

see DATAMATE and its subprograms.

tions for your TI-Graph Link to load Graphing.8xg

onto your TI calculator. When the file is sent to the

calculator, it should expand into 22 programs (one

for each chapter). These programs should appear in

the PRGM menu.

• Calculator and CBL instructions in the Holt

Physics program are written for the TI graphing

calculators, for the CBL2 interface from TI or the

LabPro interface from Vernier Software &

Technology, and for probes from Vernier. You may

use other hardware, but some of the programs and

Troubleshooting

instructions may not work exactly as described.

• If you have problems loading programs onto your

calculator, you may need to clear programs or other

data from your calculator’s memory.

• If you need additional help, both TI and Vernier

can provide technical support.

1. Triangles

AP

PE

ND

IX C

Sym

bols

848 848 Appendix C: Symbols

Symbols

displacement vector,displacement component

velocity vector,velocity component

acceleration vector

force vector,force component

momentum vector

gravitational field vector

angle marking

rotational motion

Mechanics

energy transferred as heat

energy transferred as work

cycle or process

Thermodynamics

ray (light or sound)

positive charge

negative charge

electric field lines

electric field vector

electric current

magnetic field lines

magnetic field vector (into page,out of page)

–

+

Waves and ElectromagnetismSymbol Meaning Symbol Meaning

Symbol Meaning

Diagram Symbols

849Appendix C: Symbols 849

Symbol Meaning Symbol Meaning

Δ (Greek delta) change in some quantity ≤ less than or equal to

Σ (Greek sigma) sum of quantities ∝ is proportional to

q (Greek theta) any angle ≈ is approximately equal to

= equal to ⏐n⏐ absolute value or magnitude of

> greater than sin sine

≥ greater than or equal to cos cosine

< less than tan tangent

Symbol Quantity

A area

D diameter

F, F force

m mass

M total mass

R radius (of a spherical body, a curved mirror, or a curved lens)

r radius (of sphere, shell, or disk)

t time

V volume

Quantity Symbols Used ThroughoutSymbols that are boldfaced refer to vector quantities that have both a magni-

tude and a direction. Symbols that are italicized refer to quantities with only a

magnitude. Symbols that are neither are usually units.

Mathematical Symbols


Symbol Quantity

a, a acceleration

ag free-fall acceleration (acceleration due to gravity)

d, d displacement

FΔt impulse

Fg, Fg gravitational force (weight)

Fk, Fk force of kinetic friction

Fn, Fn normal force

Fnet, Fnet net force

FR, FR force of air resistance

Fs, Fs force of static friction

Fs,max, Fs,max maximum force of static friction

h height

k spring constant

KE kinetic energy

KEtrans translational kinetic energy

MA mechanical advantage

ME mechanical energy (sum of all kinetic and potential energies)

mk (Greek mu) coefficient of kinetic friction

ms (Greek mu) coefficient of static friction

P power

p, p momentum

PE potential energy

PEelastic elastic potential energy

PEg gravitational potential energy

r separation between point masses

v, v velocity or speed

W work

Wfriction work done by a frictional force (or work required to overcome a frictional force)

Wnet net work done

Δx, Δx displacement in the x direction

Δy, Δy displacement in the y direction

Translational Mechanics Symbols Used in This BookSymbols that are boldfaced refer to vector quantities that have both a magni-

tude and a direction. Symbols that are italicized refer to quantities with only a

magnitude. Symbols that are neither are usually units.


Rotational Mechanics Symbols Used in This Book Symbols that are boldfaced refer to vector quantities

that have both a magnitude and a direction. Symbols

that are italicized refer to quantities with only a mag-

nitude. Symbols that are neither are usually units.

Symbol Quantity

at tangential acceleration

ac centripetal acceleration

a (Greek alpha) angular acceleration

d sin q lever arm (for torque calculations)

Fc, Fc centripetal force

I moment of inertia

KErot rotational kinetic energy

L angular momentum

l length of a rotating rod

s arc length

t (Greek tau) torque

tnet (Greek tau) net torque

q (Greek theta) angle of rotation

Δq (Greek delta and theta) angular displacement (in radians)

vt tangential speed

w (Greek omega) angular speed

Fluid Dynamics andThermodynamics Symbols Used in This Book Symbols that are boldfaced refer to vector quantities




Symbol Quantity

cp specific heat capacity

eff efficiency of a simple machine, thermalefficiency of a heat engine

FB, FB buoyant force

L latent heat

Lf latent heat of fusion

Lv latent heat of vaporization

N number of gas particles or nuclei

P pressure

P0 initial pressure, atmospheric pressure

Pnet net pressure

r (Greek rho) mass density

Q heat

Qc energy transferred as heat to or from alow-temperature (cold) substance

Qh energy transferred as heat to or from ahigh-temperature (hot) substance

Qnet net amount of energy transferred asheat to or from a system

T temperature (absolute)

TC temperature in degrees Celsius

Tc temperature of a low-temperature(cool) substance

TF temperature in degrees Fahrenheit

Th temperature of a high-temperature(hot) substance

U internal energy


Vibrations, Waves, and OpticsSymbols Used in This Book Symbols that are boldfaced refer to vector quantities




Symbol Quantity

C center of curvature for sphericalmirror

d slit separation in double-slit interference of light

d sin q path difference for interfering lightwaves

Felastic, Felastic spring force

F focal point

f focal length

f frequency

fn nth harmonic frequency

h object height

h� image height

k spring constant

L length of a pendulum, vibratingstring, or vibrating column of air

l path length of light wave

l (Greek lambda) wavelength

m order number for interference fringes

M magnification of image

n harmonic number (sound)

n index of refraction

p object distance

q image distance

T period of a pendulum (simple har-monic motion)

q (Greek theta) angle of incidence ofa beam of light (reflection)

q (Greek theta) angle of fringe separationfrom center of interference pattern

q � (Greek theta) angle of reflection

qc (Greek theta) critical angle ofrefraction

qi (Greek theta) angle of incidence ofa beam of light (refraction)

qr (Greek theta) angle of refraction

Electromagnetism Symbols Used in This BookSymbols that are boldfaced refer to vector quantities




Symbol Quantity

B, B magnetic field

C capacitance

d separation of plates in a capacitor

E, E electric field

emf emf (potential difference) producedby a battery or electromagneticinduction

Felectric, Felectric electric force

Fmagnetic, Fmagnetic magnetic force

I electric current

i instantaneous current (ac circuit)

Imax maximum current (ac circuit)

Irms root-mean-square current (ac circuit)

L self-inductance

l length of an electrical conductor ina magnetic field

M mutual inductance

N number of turns in a current-carrying loop or a transformer coil

PEelectric electrical potential energy

Q large charge or charge on a fullycharged capacitor

q charge

R resistance

r separation between charges

Req equivalent resistance

V electric potential

ΔV potential difference

Δv instantaneous potential difference(ac circuit)

ΔVmax maximum potential difference (ac circuit)

ΔVrms root-mean-square potential difference (ac circuit)

w (Greek omega) angular frequency


Symbol Quantity

A mass number

b (Greek beta) current or potential difference gain of an amplifier

E photon energy

ER rest energy

ft threshold frequency (photoelectric effect)

hft work function (photoelectric effect)

KEmax maximum energy of ejected photoelectron

l (Greek lambda) decay constant

lN decay rate (activity)

N neutron number, number of decayed particles

n energy quantum number

T1/2 half-life

Z atomic number

Symbol Particle

a alpha particle

b, b–

bottom quark, antiquark

b + (Greek beta) positron (beta particle)

b − (Greek beta) electron (beta particle)

c, c– charmed quark, antiquark

d, d–

down quark, antiquark

e +, 0+1e positron

e −, 0−1e electron

g (Greek gamma) photon (gamma rays)

42He alpha particle (helium-4 nucleus)

m (Greek mu) muon

10n neutron

11p proton

s, s– strange quark, antiquark

t, t– top quark, antiquark

u, u– up quark, antiquark

t (Greek tau) tauon

v, v– (Greek nu) neutrino, antineutrino

W+, W− boson (weak force)

Z boson (weak force)

Particle and Electronic Symbols Used in This Book For this part of the book, two tables are given because some symbols refer to

quantities and others refer to specific particles. The symbol’s context should

make clear which table should be consulted.

AP

PE

ND

IX D

Equa

tion

s

854 854 Appendix D: Equations

Chapter 2 Motion in One Dimension

DISPLACEMENT

AVERAGE VELOCITY

AVERAGE SPEED

AVERAGE ACCELERATION

DISPLACEMENT These equations are valid only for constantly

accelerated, straight-line motion.

FINAL VELOCITYThese equations are valid only for constantly

accelerated, straight-line motion.

Equations

Δx = xf − xi

vavg = =

average speed =

aavg = =

Δx = ⎯12

⎯(vi + vf)Δt

Δx = viΔt + ⎯12

⎯a(Δt)2

vf = vi + aΔt

vf2 = vi

2 + 2aΔx

vf − vi⎯tf − ti

Δv⎯Δt

distance traveled⎯⎯

time of travel

xf − xi⎯tf − ti

Δx⎯Δt

Chapter 3 Two-Dimensional Motion and Vectors

PYTHAGOREAN THEOREMThis equation is valid only for right triangles.

TANGENT, SINE, AND COSINE FUNCTIONSThese equations are valid only

for right triangles.

VERTICAL MOTION OF A PROJECTILETHAT FALLS FROM RESTThese equations assume that air resistance is

negligible, and apply only when the initial

vertical velocity is zero. On Earth’s surface,

ay = −g = −9.81 m/s2.

HORIZONTAL MOTION OF APROJECTILEThese equations assume that air resistance

is negligible.

c2 = a2 + b2

vy,f = ayΔt

vy,f2 = 2ayΔy

Δy = ⎯12

⎯ay(Δt)2

vx = vx,i = constant

Δx = vx Δt

tan q = sin q = cos q =adj⎯hyp

opp⎯hyp

opp⎯adj

855Appendix D: Equations 855

Chapter 4 Forces and the Laws of Motion

NEWTON’S FIRST LAW

NEWTON’S SECOND LAW∑F is the vector sum of all external forces

acting on the object.

NEWTON’S THIRD LAW

WEIGHTOn Earth’s surface, ag = g = 9.81 m/s2.

COEFFICIENT OF STATIC FRICTION

COEFFICIENT OF KINETIC FRICTIONThe coefficient of kinetic friction varies with

speed, but we neglect any such variations here.

FORCE OF FRICTION

An object at rest remains at rest,and an object in motion continuesin motion with constant velocity(that is, constant speed in a straightline) unless the object experiences anet external force.

∑F = ma

If two objects interact, themagnitude of the force exerted onobject 1 by object 2 is equal to themagnitude of the force exerted onobject 2 by object 1, and these twoforces are opposite in direction.

Fg = mag

ms =

mk =

Ff = mFn

Fk⎯Fn

Fs,max⎯Fn

PROJECTILES LAUNCHED AT AN ANGLEThese equations assume that air resistance

is negligible. On Earth’s surface,

ay = −g = −9.81 m/s2.

RELATIVE VELOCITY

vx = vi cos q = constant

Δx = (vi cos q)Δt

vy,f = vi sin q + ayΔt

vy,f2 = vi

2(sin q)2 + 2ayΔy

Δy = (vi sin q)Δt + ⎯12

⎯ay(Δt)2

vac = vab + vbc


Chapter 5 Work and Energy

NET WORKThis equation applies only when the force is

constant.

KINETIC ENERGY

WORK-KINETIC ENERGY THEOREM

GRAVITATIONAL POTENTIAL ENERGY

ELASTIC POTENTIAL ENERGY

MECHANICAL ENERGY

CONSERVATION OF MECHANICAL ENERGYThis equation is valid only if nonmechanical

forms of energy (such as friction) are disregarded.

POWER

Wnet = Fnetd cos q

KE = ⎯12

⎯ mv2

Wnet = ΔKE

PEg = mgh

PEelastic = ⎯12

⎯kx2

ME = KE + ΣPE

MEi = MEf

P = ⎯ΔW

t⎯ = Fv

Chapter 6 Momentum and Collisions

MOMENTUM

IMPULSE-MOMENTUM THEOREMThis equation is valid only when the force

is constant.

CONSERVATION OF MOMENTUMThese equations are valid for a closed system,

that is, when no external forces act on the system

during the collision. When such external forces

are either negligibly small or act for too short

a time to make a significant change in the

momentum, these equations represent a good

approximation. The second equation is valid for

two-body collisions.

p = mv

FΔt = Δp = mvf − mvi

pi = pf

m1v1,i + m2v2,i = m1v1,f + m2v2,f


Chapter 7 Circular Motion and Gravitation

CENTRIPETAL ACCELERATION

CENTRIPETAL FORCE

NEWTON’S LAW OF UNIVERSALGRAVITATIONThe constant of universal gravitation (G) equals

6.673 × 10−11 N•m2/kg2.

KEPLER’S LAWS OF PLANETARY MOTION

PERIOD AND SPEED OF AN OBJECTIN CIRCULAR ORBITThe constant of universal gravitation (G) equals

6.673 × 10−11 N•m2/kg2.

TORQUE

ac = ⎯v

rt2

⎯

Fc = ⎯m

r

vt2

⎯

Fg = G ⎯m

r1m

22⎯

First Law: Each planet travels in anelliptical orbit around the sun, andthe sun is at one of the focal points.

Second Law: An imaginary linedrawn from the sun to any planetsweeps out equal areas in equal timeintervals.

Third Law: The square of a planet’sorbital period (T 2) is proportionalto the cube of the average distance(r3) between the planet and the sun,or T 2 ∝ r3.

T = 2p�⎯G

r

m

3

⎯�vt = �G ⎯

m

r⎯�

t = Fdsin q

CONSERVATION OF MOMENTUM FORA PERFECTLY INELASTIC COLLISIONThis is a simplified version of the conservation of

momentum equation valid only for perfectly

inelastic collisions between two bodies.

CONSERVATION OF KINETIC ENERGYFOR AN ELASTIC COLLISIONNo collision is perfectly elastic; some kinetic

energy is always converted to other forms of

energy. But if these losses are minimal, this

equation can provide a good approximation.

m1 v1,i + m2 v2,i = (m1 + m2) vf

⎯12

⎯m1v1,i2 + ⎯1

2⎯m2v2,i

2 =⎯12

⎯m1v1,f2 + ⎯1

2⎯m2v2,f

2


Chapter 8 Fluid Mechanics

MASS DENSITY

BUOYANT FORCEThe first equation is for an object that is

completely or partially submerged. The second

equation is for a floating object.

PRESSURE

PASCAL’S PRINCIPLE

HYDRAULIC LIFT EQUATION

FLUID PRESSURE AS A FUNCTION OF DEPTH

CONTINUITY EQUATION

BERNOULLI’S PRINCIPLE

r = ⎯m

V⎯⎯

FB = Fg (displaced fluid) = mfg

FB = Fg (object) = mg

P = ⎯A

F⎯

Pressure applied to a fluid in a closedcontainer is transmitted equally toevery point of the fluid and to thewalls of the container.

F2 = F1

P = P0 + rgh

A1v1 = A2v2

The pressure in a fluid decreases asthe fluid’s velocity increases.

A2⎯A1

Chapter 9 Heat

TEMPERATURE CONVERSIONS TF = ⎯95

⎯TC + 32.0

T = TC + 273.15

MECHANICAL ADVANTAGEThis equation disregards friction.

EFFICIENCYThis equation accounts for friction.

MA = ⎯F

Fo

i

u

n

t⎯ = ⎯d

d

o

i

u

n

t⎯

eff = ⎯W

Wo

i

u

n

t⎯


CONSERVATION OF ENERGY

SPECIFIC HEAT CAPACITY

CALORIMETRYThese equations assume that the energy transferred

to the surrounding container is negligible.

LATENT HEAT

ΔPE + ΔKE + ΔU = 0

cp = ⎯m

Q

ΔT⎯

Qw = −Qx

cp,wmwΔTw = −cp,xmxΔTx

Q = mL

Chapter 10 Thermodynamics

WORK DONE BY A GASThis equation is valid only when the pressure is

constant. When the work done by the gas (W) is

negative, positive work is done on the gas.

THE FIRST LAW OFTHERMODYNAMICSQ represents the energy added to the system as heat

and W represents the work done by the system.

CYCLIC PROCESSES

EFFICIENCY OF A HEAT ENGINE

W = PAd = PΔV

ΔU = Q − W

ΔUnet = 0 and Qnet = Wnet

eff = ⎯W

Qn

h

et⎯ = ⎯Qh

Q

−

h

Qc⎯ = 1 − ⎯Q

Q

h

c⎯

Chapter 11 Vibrations and Waves

HOOKE’S LAW

PERIOD OF A SIMPLE PENDULUM INSIMPLE HARMONIC MOTIONThis equation is valid only when the amplitude is

small (less than about 15°).

PERIOD OF A MASS-SPRING SYSTEMIN SIMPLE HARMONIC MOTION

SPEED OF A WAVE

Felastic = –kx

T = 2p��

T = 2p��v = fl

m⎯k

L⎯ag


Chapter 12 Sound

INTENSITY OF A SPHERICAL WAVEThis equation assumes that there is no absorption

in the medium.

HARMONIC SERIES OF A VIBRATINGSTRING OR A PIPE OPEN AT BOTHENDS

HARMONIC SERIES OF A PIPECLOSED AT ONE END

BEATS

intensity = ⎯4p

P

r2⎯

fn = n ⎯2

v

L⎯ n = 1, 2, 3, . . .

fn = n ⎯4

v

L⎯ n = 1, 3, 5, . . .

frequency difference = number ofbeats per second

Chapter 13 Light and Reflection

SPEED OF ELECTROMAGNETICWAVESThis book uses the value c = 3.00 × 108 m/s for the

speed of EM waves in a vacuum or in air.

LAW OF REFLECTION

MIRROR EQUATIONThis equation is derived assuming that the rays

incident on the mirror are very close to the

principal axis of the mirror.

MAGNIFICATION OF A CURVEDMIRROR

c = fl

angle of incidence (q) = angle ofreflection (q�)

⎯⎯p

1⎯⎯ + ⎯⎯

1

q⎯⎯ = ⎯⎯

1

f⎯⎯

M = ⎯h

h

�⎯ = − ⎯⎯

p

q⎯⎯

Chapter 14 Refraction

INDEX OF REFRACTIONFor any material other than a vacuum, the index

of refraction varies with the wavelength of light.

SNELL’S LAW

n = ⎯v

c⎯⎯

ni sin qi = nr sin qr


Chapter 15 Interference and Diffraction

CONSTRUCTIVE AND DESTRUCTIVEINTERFERENCEThe grating spacing multiplied by the sine of the

angle of deviation is the path difference between

two waves. To observe interference effects, the

sources must be coherent and have identical

wavelengths.

DIFFRACTION GRATING

LIMITING ANGLE OF RESOLUTIONThis equation gives the angle q in radians and

applies only to circular apertures.

Constructive Interference:

d sin q = ±mlm = 0, 1, 2, 3, . . .

Destructive Interference:

d sin q = ±(m + ⎯12

⎯)lm = 0, 1, 2, 3, . . .

See the equation above forconstructive interference.

q = 1.22 ⎯D

l⎯

Chapter 16 Electric Forces and Fields

COULOMB’S LAWThis equation assumes either point charges or

spherical distributions of charge.

ELECTRIC FIELD STRENGTH DUE TO A POINT CHARGE

Felectric = kC �⎯q

r1q

22⎯�

E = kC ⎯r

q2⎯

THIN-LENS EQUATIONThis equation is derived assuming that the

thickness of the lens is much less than the focal

length of the lens.

MAGNIFICATION OF A LENSThis equation can be used only when the index of

refraction of the first medium (ni) is greater than

the index of refraction of the second medium (nr).

CRITICAL ANGLEThis equation can be used only when the index of

refraction of the first medium (ni) is greater than

the index of refraction of the second medium (nr).

⎯⎯p

1⎯⎯ + ⎯⎯

1

q⎯⎯ = ⎯⎯

1

f⎯⎯

M = ⎯h

h

�⎯ = − ⎯⎯

p

q⎯⎯ (for ni > nr)

sin qc = ⎯n

nr

i⎯ (for ni > nr)


Chapter 17 Electrical Energy and Current

ELECTRICAL POTENTIAL ENERGYThe displacement, d, is from the reference point

and is parallel to the field. This equation is valid

only for a uniform electric field.

POTENTIAL DIFFERENCEThe second half of this equation is valid only for a

uniform electric field, and Δd is parallel to the

field.

POTENTIAL DIFFERENCE BETWEEN APOINT AT INFINITY AND A POINTNEAR A POINT CHARGE

CAPACITANCE

CAPACITANCE FOR A PARALLEL-PLATE CAPACITOR IN A VACUUMThe permittivity in a vacuum (e0 ) equals

8.85 × 10 −12 C 2/(N•m2).

ELECTRICAL POTENTIAL ENERGYSTORED IN A CHARGED CAPACITORThere is a limit to the maximum energy (or

charge) that can be stored in a capacitor because

electrical breakdown ultimately occurs between

the plates of the capacitor for a sufficiently large

potential difference.

ELECTRIC CURRENT

RESISTANCE

OHM’S LAWOhm’s law is not universal, but it does apply to

many materials over a wide range of applied

potential differences.

ELECTRIC POWER

PEelectric = −qEd

ΔV = ⎯ΔPE

qelectric⎯ = −EΔd

ΔV = kC ⎯q

r⎯

C = ⎯ΔQ

V⎯

C = e0 ⎯A

d⎯

PEelectric = ⎯12

⎯QΔV = ⎯12

⎯C (ΔV)2 = ⎯Q

2C

2

⎯

I = ⎯ΔΔQ

t⎯

R = ⎯Δ

I

V⎯

⎯Δ

I

V⎯ = constant

P = IΔV = I2R = ⎯(Δ

R

V )2

⎯


Chapter 18 Circuits and Circuit Elements

RESISTORS IN SERIES: EQUIVALENT RESISTANCE AND CURRENT

RESISTORS IN PARALLEL: EQUIVALENT RESISTANCE AND CURRENT

Req = R1 + R2 + R3 . . .

The current in each resistor is thesame and is equal to the total current.

⎯R

1

eq⎯ = ⎯

R

1

1⎯ + ⎯

R

1

2⎯ + ⎯

R

1

3⎯ . . .

The sum of the current in eachresistor equals the total current.

Chapter 20 Electromagnetic Induction

FARADAY’S LAW OF MAGNETICINDUCTIONN is assumed to be a whole number.

EMF PRODUCED BY A GENERATORN is assumed to be a whole number.

FARADAY’S LAW FOR MUTUALINDUCTANCE

emf = −N ⎯Δ

ΔΦ

tM⎯

emf = NABw sin wt

maximum emf = NABw

emf = −M ⎯ΔΔ

I

t⎯

Chapter 19 Magnetism

MAGNETIC FLUX

MAGNITUDE OF A MAGNETIC FIELDThe direction of Fmagnetic is always perpendicular

to both B and v, and can be found with the right-

hand rule.

FORCE ON A CURRENT-CARRYINGCONDUCTOR PERPENDICULAR TO A MAGNETIC FIELDThis equation can be used only when the current

and the magnetic field are at right angles to each

other.

ΦM = ABcosq

B = ⎯Fma

q

g

v

netic⎯

Fmagnetic = BI l


Chapter 21 Atomic Physics

ENERGY OF A LIGHT QUANTUM

MAXIMUM KINETIC ENERGY OF A PHOTOELECTRON

WAVELENGTH AND FREQUENCY OF MATTER WAVESPlanck’s constant (h) equals 6.63 × 10−34 J•s.

E = hf

KEmax = hf − hft

l = ⎯h

p⎯⎯ = ⎯

m

h

v⎯

f = ⎯E

h⎯⎯

Chapter 22 Subatomic Physics

RELATIONSHIP BETWEEN RESTENERGY AND MASS

BINDING ENERGY OF A NUCLEUS

MASS DEFECT

ACTIVITY (DECAY RATE)

HALF-LIFE

ER = mc2

Ebind = Δmc2

Δm = Z(atomic mass of H) + Nmn − atomic mass

activity = − ⎯ΔΔN

t⎯ = lN

T1/2 = ⎯0.6

l93⎯

RMS CURRENT AND POTENTIALDIFFERENCE

TRANSFORMERSN is assumed to be a whole number.

Irms = ⎯�Ima

2�x⎯ = 0.707 Imax

ΔVrms = ⎯Δ

�Vm

2�ax⎯ = 0.707 ΔV

ΔV2 = ⎯N

N2

1⎯ ΔV1


Appendix J Advanced Topics

CONVERSION BETWEEN RADIANSAND DEGREES

ANGULAR DISPLACEMENTThis equation gives Δq in radians.

AVERAGE ANGULAR VELOCITY

AVERAGE ANGULAR ACCELERATION

ROTATIONAL KINEMATICSThese equations apply only when the angular

acceleration is constant. The symbol w represents

instantaneous rather than average angular velocity.

TANGENTIAL SPEEDFor this equation to be valid, w must be in rad/s.

TANGENTIAL ACCELERATIONFor this equation to be valid, a must be in rad/s2.

NEWTON’S SECOND LAW FORROTATING OBJECTS

ANGULAR MOMENTUM

ROTATIONAL KINETIC ENERGY

IDEAL GAS LAWBoltzmann’s constant (kB) equals 1.38 × 10−23 J/K.

BERNOULLI’S EQUATION

q(rad) = ⎯18

p0°⎯q(deg)

Δq = ⎯Δr

s⎯

wavg = ⎯⎯ΔΔ

qt

⎯

aavg = ⎯ΔΔwt⎯

wf = w i + aΔt

Δq = w iΔt + ⎯12

⎯a(Δt)2

wf2 = w i

2 + 2a(Δq)

Δq = ⎯12

⎯(w i + wf)Δt

vt = rw

at = ra

t = Ia

L = Iw

KErot = ⎯12

⎯Iw2

PV = NkBT

P + ⎯12

⎯ rv2 + rgh = constant

AP

PE

ND

IX E

SI U

nits

866 866 Appendix E: SI Units

SI Units

Symbol Name Quantity

A ampere current

K kelvin absolute temperature

kg kilogram mass

m meter length

s second time

Symbol Name Numerical equivalent

a atto 10−18

f femto 10−15

p pico 10−12

n nano 10−9

μ micro 10− 6

m milli 10−3

c centi 10−2

d deci 10−1

k kilo 103

M mega 106

G giga 109

T tera 1012

P peta 1015

E exa 1018

SI Base Units Used in This Book SI Prefixes

Symbol Name Quantity Conversions

atm standard atmosphere pressure 1.013 250 × 105 Pa

Btu British thermal unit energy 1.055 × 103 J

Cal food calorie energy = 1 kcal = 4.186 × 103 J

cal calorie energy 4.186 J

Ci curie decay rate or activity 3.7 × 1010 s−1

°F degree Fahrenheit temperature 0.5556°C

ft foot length 0.3048 m

ft •lb foot-pound work and energy 1.356 J

g gram mass 0.001 kg

gal gallon volume 3.785 × 10−3 m3

hp horsepower power 746 W

in inch length 2.54 × 10−2 m

kcal kilocalorie energy 4.186 × 103 J

lb pound force 4.45 N

mi mile length 1.609 × 103 m

rev revolution angular displacement 2p rad

° degrees angular displacement= �⎯

3

2

6

p0

⎯� rad = 1.745 × 10−2 rad

Other Commonly Used Units

867Appendix E: SI Units 867

Symbol Name Quantity Conversion

Bq becquerel decay rate or activity ⎯1

s⎯

C coulomb electric charge 1 A•s

°C degree Celsius temperature 1 K

dB decibel relative intensity (sound) (unitless)

eV electron volt energy 1.60 × 10−19 J

F farad capacitance1 ⎯

k

A

g

2

•

•

m

s4

2⎯ = 1 ⎯V

C⎯

H henry inductance1 ⎯

k

A

g2

•

•

m

s2

2

⎯ = 1 ⎯A

J2⎯

h hour time 3.600 × 103 s

Hz hertz frequency⎯1

s⎯

J joule work and energy1 ⎯

kg

s

•2

m2

⎯ = 1 N •m

kW •h kilowatt-hour energy 3.60 × 106 J

L liter volume 10−3 m3

min minute time 6.0 × 101 s

N newton force1 ⎯

kg

s

•2

m⎯

Pa pascal pressure1 ⎯

m

k

•

g

s2⎯ = 1 ⎯m

N2⎯

rad radian angular displacement (unitless)

T tesla magnetic field strength1 ⎯

A

k

•

g

s2⎯ = 1 ⎯A

N

•m⎯ = 1 ⎯

V

m

•2

s⎯

u unified mass unit mass (atomic masses) 1.660 538 782 × 10−27 kg

V volt electric potential difference1 ⎯

k

A

g •

•

m

s3

2

⎯ = 1 ⎯C

J⎯

W watt power1 ⎯

kg

s

•3

m2

⎯ = 1 ⎯S

J⎯

Ω ohm resistance1 ⎯

k

A

g2

•

•

m

s3

2

⎯ = 1 ⎯V

A⎯

Other Units Acceptable with SI

s

AP

PE

ND

IX F

Use

ful T

able

s

868 868 Appendix F: Useful Tables

Useful Tables

Value used forSymbol Quantity Established value calculations in this book

c speed of light in a vacuum 299 792 458 m/s 3.00 × 108 m/s

e− elementary charge 1.602 176 487 × 10−19 C 1.60 × 10−19 C

e1 base of natural logarithms 2.718 2818 28 2.72

e0 (Greek epsilon) permittivity of 8.854 187 817 × 10−12 C2/(N •m2) 8.85 × 10−12 C2/(N •m2)a vacuum

G constant of universal gravitation 6.672 59 × 10−11 N•m2/kg2 6.673 × 10−11 N•m2/kg2

g free-fall acceleration 9.806 65 m/s2 9.81 m/s2

at Earth’s surface

h Planck’s constant 6.626 068 96 × 10−34 J •s 6.63 × 10−34 J •s

kB Boltzmann’s constant (R/NA) 1.380 6504 × 10−23 J/K 1.38 × 10−23 J/K

kC Coulomb constant 8.987 551 787 × 109 N •m2/C2 8.99 × 109 N •m2/C2

R molar (universal) gas constant 8.314 472 J/(mol •K) 8.31 J/(mol •K)

p (Greek pi) ratio of the circum- 3.141 592 654 calculator valueference to the diameter of a circle

Fundamental Constants

µs µk µs µk

steel on steel 0.74 0.57 waxed wood on wet snow 0.14 0.1

aluminum on steel 0.61 0.47 waxed wood on dry snow — 0.04

rubber on dry concrete 1.0 0.8 metal on metal (lubricated) 0.15 0.06

rubber on wet concrete — 0.5 ice on ice 0.1 0.03

wood on wood 0.4 0.2 Teflon on Teflon 0.04 0.04

glass on glass 0.9 0.4 synovial joints in humans 0.01 0.003

Coefficients of Friction (Approximate Values)

Value used forSymbol Quantity calculations in this book

IE moment of inertia of Earth 8.03 × 1037 kg •m2

ME mass of Earth 5.97 × 1024 kg

RE radius of Earth 6.38 × 106 m

Average Earth–moon distance 3.84 × 108 m

Average Earth–sun distance 1.50 × 1011 m

mass of the moon 7.35 × 1022 kg

mass of the sun 1.99 × 1030 kg

yr period of Earth’s orbit 3.16 × 107 s

Useful Astronomical Data

869Appendix F: Useful Tables 869

Shape Moment of inertia Shape Moment of inertia

thin spherical shell ⎯23

⎯MR2about diameterR

disk or cylinder about ⎯12

⎯MR2symmetry axis

R

solid sphere ⎯25

⎯MR2about diameterRpoint mass about axis MR2

R

thin rod about perpendicular axis ⎯1

3⎯Ml 2

through end

lthin hoop about ⎯12

⎯MR2diameterR

thin rod aboutperpendicular axis ⎯

1

1

2⎯Ml 2

through center

lthin hoop aboutMR2

symmetry axisR

The Moment of Inertia for a Few Shapes

Substance r (kg/m3)

hydrogen 0.0899

helium 0.179

steam (100°C) 0.598

air 1.29

oxygen 1.43

carbon dioxide 1.98

ethanol 0.806 × 103

ice 0.917 × 103

fresh water (4°C) 1.00 × 103

sea water (15°C) 1.025 × 103

glycerine 1.26 × 103

aluminum 2.70 × 103

iron 7.86 × 103

copper 8.92 × 103

silver 10.5 × 103

lead 11.3 × 103

mercury 13.6 × 103

gold 19.3 × 103

*All densities are measured at 0°C and 1 atm unless otherwise noted.

Densities of Some CommonSubstances*

Substance cp (J/kg •°C)

aluminum 8.990 × 102

copper 3.870 × 102

glass 8.370 × 102

gold 1.290 × 102

ice 2.090 × 103

iron 4.480 × 102

lead 1.280 × 102

mercury 1.380 × 102

silver 2.340 × 102

steam 2.010 × 103

water 4.186 × 103

Specific Heat Capacities

870 870 Appendix F: Useful Tables

Medium v (m/s) Medium v (m/s) Medium v (m/s)

Gases Liquids at 25°C Solidsair (0°C) 331 methyl alcohol 1140 aluminum 5100

air (25°C) 346 sea water 1530 copper 3560

air (100°C) 366 water 1490 iron 5130

helium (0°C) 972 lead 1320

hydrogen (0°C) 1290 vulcanized rubber 54

oxygen (0°C) 317

Speed of Sound in Various Media

Intensity (W/m2) Decibel level (dB) Examples

1.0 × 10− 12 0 threshold of hearing

1.0 × 10− 11 10 rustling leaves

1.0 × 10− 10 20 quiet whisper

1.0 × 10−9 30 whisper

1.0 × 10−8 40 mosquito buzzing

1.0 × 10−7 50 normal conversation

1.0 × 10−6 60 air conditioning at 6 m

1.0 × 10−5 70 vacuum cleaner

1.0 × 10−4 80 busy traffic, alarm clock

1.0 × 10−3 90 lawn mower

1.0 × 10−2 100 subway, power motor

1.0 × 10− 1 110 auto horn at 1 m

1.0 × 100 120 threshold of pain

1.0 × 101 130 thunderclap, machine gun

1.0 × 103 150 nearby jet airplane

Conversion of Intensity to Decibel Level

Substance Melting point (°C) Lf (J/kg) Boiling point (°C) Lv (J/kg)

nitrogen −209.97 2.55 × 104 −195.81 2.01 × 105

oxygen −218.79 1.38 × 104 −182.97 2.13 × 105

ethyl alcohol −114 1.04 × 105 78 8.54 × 105

water 0.00 3.33 × 105 100.00 2.26 × 106

lead 327.3 2.45 × 104 1745 8.70 × 105

aluminum 660.4 3.97 × 105 2467 1.14 × 107

Latent Heats of Fusion and Vaporization at Standard Pressure

871Appendix F: Useful Tables 871

Value used for calculationsSymbol Quantity Established value in this book

me mass of electron 9.109 382 15 × 10−31 kg 9.109 × 10−31 kg5.485 799 0943 × 10−4 u 5.49 × 10−4 u0.510 998 910 MeV 5.110 × 10−1 MeV

mn mass of neutron 1.674 927 211 × 10−27 kg 1.675 × 10−27 kg1.008 664 915 97 u 1.008 665 u939.565 346 MeV 9.396 × 102 MeV

mp mass of proton 1.672 621 637 × 10−27 kg 1.673 × 10−27 kg1.007 276 466 77 u 1.007 276 u938.272 013 MeV 9.383 × 102 MeV

Useful Atomic Data

Solids at 20°C n Liquids at 20°C n Gases at 0°C, 1 atm n

cubic zirconia 2.20 benzene 1.501 air 1.000 293

diamond 2.419 carbon disulfide 1.628 carbon dioxide 1.000 450

fluorite 1.434 carbon tetrachloride 1.461

fused quartz 1.458 ethyl alcohol 1.361

glass, crown 1.52 glycerine 1.473

glass, flint 1.66 water 1.333

ice (at 0°C) 1.309

polystyrene 1.49

sodium chloride 1.544

zircon 1.923

*measured with light of vacuum wavelength = 589 nm

Indices of Refraction for Various Substances*

1

2

3

4

5

6

7

Perio

d

* The systematic names and symbolsfor elements greater than 111 willbe used until the approval of trivialnames by IUPAC.

Key:6

CCarbon12.0107

Atomic number

Symbol

Name

Average atomic mass

Group 3 Group 4 Group 5 Group 6 Group 7 Group 8 Group 9

Group 1 Group 2

[He]2s 1

[Ne]3s1

[Ar]4s1

[He]2s2

[Ne]3s2

[Ar]4s2 [Ar]3d 14s 2

1s1

[Ar]3d 24s 2 [Ar]3d 34s 2 [Ar]3d 54s1 [Ar]3d 54 s 2 [Ar]3d 64s 2 [Ar]3d 74s 2

[Kr]5s1

[Xe]6s1

[Rn]7s1

[Kr]5s2

[Xe]6s 2

[Rn]7s2

[Kr]4d 15 s 2

[Xe]5d 16s 2

[Rn]6d 17s 2

[Kr]4d 25s 2

[Xe]4 f14 5d 26s2

[Rn]5 f 146d 27s2

[Kr]4d45s1

[Xe]4 f 145d 36s2

[Rn]5 f 146d 37s2

[Kr]4d 55s 1

[Xe]4 f 145d 46s2

[Rn]5 f 146d 47s2

[Kr]4d 65s1

[Xe]4 f 145d 56s2

[Rn]5 f 146d 57s2

[Kr]4d 75s1

[Xe]4 f 145d 66s2

[Rn]5f 146d 67s2

[Kr]4d 85s1

[Xe]4 f 145d 76s2

[Rn]5 f 146d 77s2

[Xe]4 f15d16s2

[Rn]6d 27s 2

[Xe]4 f 36s 2

[Rn]5 f 26d 17s 2

[Xe]4 f 46s 2

[Rn]5 f 36d 17s2

[Xe]4 f 56s 2

[Rn]5 f 46d 17s2

[Xe]4 f 66s 2

[Rn]5f 67s 2

Electron configuration [He]2s22p26.941

22.989 770

39.0983

85.4678

132.905 43

(223)

140.116

232.0381

140.907 65

231.035 88

144.24

238.028 91

(145)

(237)

150.36

(244)

9.012 182

24.3050

40.078

87.62

137.327

(226)

44.955 910

88.905 85

138.9055

(227)

47.867

91.224

178.49

(261)

50.9415

92.906 38

180.9479

(262)

51.9961

95.94

183.84

(266)

54.938 049

(98)

186.207

(264)

55.845

101.07

190.23

(277)

58.933 200

102.905 50

192.217

(268)

1.007 94

Lithium

Sodium

Potassium

Rubidium

Cesium

Francium

Cerium

Thorium

Praseodymium

Protactinium

Neodymium

Uranium

Promethium

Neptunium

Samarium

Plutonium

Beryllium

Magnesium

Calcium

Strontium

Barium

Radium

Scandium

Yttrium

Lanthanum

Actinium

Titanium

Zirconium

Hafnium

Rutherfordium

Vanadium

Niobium

Tantalum

Dubnium

Chromium

Molybdenum

Tungsten

Seaborgium

Manganese

Technetium

Rhenium

Bohrium

Iron

Ruthenium

Osmium

Hassium

Cobalt

Rhodium

Iridium

Meitnerium

Hydrogen

Li

V

Na

K

Rb

Cs

Fr

Be

Mg

Ca

Sr

Ba

Ra

Sc

Y

La

Ac

Ti

Zr

Hf

Rf

Nb

Ta

Db

Cr

Mo

W

Sg

Mn

Tc

Re

Bh

Ir Os

Ce

Th

Pr

Pa

Nd

U

Pm

Np

Sm

Pu

Fe

Ru

Hs

Co

Rh

Mt

H

3

11

19

37

55

87

4

12

20

38

56

88

21

39

57

89

22

40

72

104

23

41

73

105

24

42

74

106

25

43

75

107

26

44

76 77

108

27

45

109

1

58

90

59

91

60

92

61

93

62

94

AP

PE

ND

IX G

Per

iodi

c Ta

ble

872 872 Appendix G: Periodic Table of the Elements

Periodic Table of the Elements

Topic: Periodic TableGo To: go.hrw.comKeyword: HOLT PERIODIC

Visit the HRW Web site for updateson the periodic table.

The atomic masses listed in this table reflect the precision of current measurements. (Each value listed in parentheses is the mass number of that radioactive element’s most stable or most common isotope.)

The discoveries of elements with atomic numbers 112, 114, and 116 have been reported but not fully confirmed.

HydrogenSemiconductors(also known as metalloids)

MetalsAlkali metalsAlkaline-earth metalsTransition metalsOther metals

NonmetalsHalogensNoble gasesOther nonmetals

Group 14 Group 15 Group 16 Group 17

Group 18

Group 10 Group 11 Group 12

Group 13

[Xe]4f 145d16s2[Xe]4f 146s2[Xe]4f 136s2[Xe]4f 126s2[Xe]4f 116s2[Xe]4f 106s2

[Rn]5f 146d17s2[Rn]5f 147s2[Rn]5f 137s2[Rn]5f 127s2[Rn]5f 117s2[Rn]5f 107s2

[Xe]4f 145d106s26p6

[Kr]4d 105s25p6

[Ar]3d 104s24p6

[Ne]3s 23p6

1s 2

[He]2s 22p5[He]2s 22p4[He]2s 22p3[He]2s 22p2[He]2s 22p1

[Ne]3s 23p5[Ne]3s 23p4[Ne]3s 23p3[Ne]3s 23p2[Ne]3s 23p1

[Ar]3d 104s24p5[Ar]3d 104s24p4[Ar]3d 104s24p3[Ar]3d 104s24p2[Ar]3d 104s24p1

[Kr]4d 105s25p5[Kr]4d 105s25p4[Kr]4d 105s25p3[Kr]4d 105s25p2[Kr]4d 105s25p1

[Xe]4f 145d106s26p5[Xe]4f 145d106s26p4[Xe]4f 145d106s26p3[Xe]4f 145d106s26p2[Xe]4f 145d106s26p1

[He]2s 22p6

[Rn]5f 146d 107s27p2

[Xe]4f 96s2[Xe]4f 75d16s2[Xe]4f 76s2

[Rn]5f 97s2[Rn]5f 76d17s2[Rn]5f 77s2

[Ar]3d 104s2[Ar]3d 104s1[Ar]3d 84s2

[Kr]4d 105s2[Kr]4d 105s1[Kr]4d 105s0

[Xe]4f 145d106s2[Xe]4f 145d96s1

[Rn]5f 146d 97s1 [Rn]5f 146d 107s2[Rn]5f 146d 107s1

[Xe]4f 145d106s1

[Rn]5f 146d 107s27p4

151.964

(243)

157.25

(247)

158.925 34

(247)

162.500

(251)

164.930 32

(252)

167.259

(257)

168.934 21

(258)

173.04

(259)

174.967

(262)

58.6934 63.546 65.409 69.723 72.64 74.921 60 78.96 79.904 83.798

26.981 538 28.0855 30.973 761 32.065 35.453 39.948

12.0107 14.0067 15.9994 18.998 4032 20.1797

4.002 602

106.42 107.8682 112.411 114.818 118.710 121.760 127.60 126.904 47 131.293

195.078

(271) (272) (285) (289)

196.966 55 200.59 204.3833 207.2 208.980 38 (209) (210) (222)

10.811

Europium

Americium

Gadolinium

Curium

Terbium

Berkelium

Dysprosium

Californium

Holmium

Einsteinium

Erbium

Fermium

Thulium

Mendelevium

Ytterbium

Nobelium

Lutetium

Lawrencium

Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton

Aluminum Silicon Phosphorus Sulfur Chlorine Argon

Carbon Nitrogen Oxygen Fluorine Neon

Helium

Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon

Platinum

Darmstadtium Roentgenium Ununbium Ununquadium

Gold Mercury Thallium Lead Bismuth Polonium Astatine Radon

Boron

Eu

Am

Gd

Cm

Tb

Bk

Dy

Cf

Ni Cu Zn Ga Ge As Se Br Kr

Al Si P S Cl Ar

C N O F Ne

He

Pd Ag Cd In Sn Sb Te I Xe

Pt

Ds Rg Uub* Uuq*

Au Hg Tl Pb Bi Po At Rn

Ho

Es

Er

Fm

Tm

Md

Yb

No

Lu

Lr

B

63

95

64

96

65

97

66

98

67

99

68

100

69

101

70

102

71

103

28 29 30 31 32 33 34 35 36

13 14 15 16 17 18

6 7 8 9 10

2

46 47 48 49 50 51 52 53 54

78

110 111 112 114

79 80 81 82 83 84 85 86

5

(292)UnunhexiumUUh*

116

873Appendix G: Periodic Table of the Elements 873

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874 874 Appendix H: Abbreviated Table of Isotopes and Atomic Masses

Abbreviated Table of Isotopes and Atomic Masses

Z Element Symbol Average Mass number Atomic Percent Half-lifeatomic (* indicates mass (u) abundance (if radioactive)mass (u) radioactive) A T1/2

0 (Neutron) n 1* 1.008 665 10.4 m

1 Hydrogen H 1.0079 1* 1.007 825 99.985Deuterium D 2* 2.014 102 0.015Tritium T 3* 3.016 049 12.33 y

2 Helium He 4.002 60 3* 3.016 029 0.000 144* 4.002 602 99.999 866* 6.018 886 0.81 s

3 Lithium Li 6.941 6* 6.015 121 7.57* 7.016 003 92.5

4 Beryllium Be 9.0122 7* 7.016 928 53.3 d*8* 8.005 305 6.7 × 10–17 s9* 9.012 174 100

10* 10.013 584 1.5 × 106 y

5 Boron B 10.81 10* 10.012 936 19.911* 11.009 305 80.1

6 Carbon C 12.011 10* 10.016 854 19.3 s11* 11.011 433 20.4 m12* 12.000 000 98.913* 13.003 355 1.1014* 14. 003 242 5715 y

7 Nitrogen N 14.0067 13* 13.005 738 996 m14* 14.003 074 99.6315* 15.000 108 0.3716* 16.006 100 7.13 s

8 Oxygen O 15.9994 15* 15.003 065 122 s16* 15.994 915 99.76117* 16.999 132 0.03918* 17.999 160 0.20019* 19.003 577 26.9 s

9 Fluorine F 18.998 40 18* 18.000 937 109.8 m19* 18.998 404 10020* 19.999 982 11.0 s

10 Neon Ne 20.180 19* 19.001 880 17.2 s20* 19.992 435 90.4821* 20.993 841 0.2722* 21.991 383 9.25

11 Sodium Na 22.989 87 22* 21.994 434 2.61 y23* 22.989 767 10024* 23.990 961 14.96 h

12 Magnesium Mg 24.305 23* 22.994 124 11.3 s24* 23.985 042 78.9925* 24.985 838 10.0026* 25.982 594 11.01

13 Aluminum Al 26.981 54 26* 25.986 892 7.4 × 105 y27* 26.981 534 100

875Appendix H: Abbreviated Table of Isotopes and Atomic Masses 875

Z Element Symbol Average Mass number Atomic Percent Half-lifeatomic (* indicates mass (u) abundance (if radioactive)mass (u) radioactive) A T1/2

14 Silicon Si 28.086 28* 27.976 927 92.2329* 28.976 495 4.6730* 29.973 770 3.10

15 Phosphorus P 30.973 76 30* 29.978 307 2.50 m31* 30.973 762 10032* 31.973 907 14.263 d

16 Sulfur S 32.066 32* 31.972 071 95.0233* 32.971 459 0.7534* 33.967 867 4.2135* 34.969 033 87.5 d

17 Chlorine Cl 35.453 35* 34.968 853 75.7736* 35.968 307 3.0 × 105 y37* 36.975 893 24.23

18 Argon Ar 39.948 36* 35.967 547 0.33737* 36.966 776 35.04 d38* 37.962 732 0.06339* 38.964 314 269 y40* 39.962 384 99.600

19 Potassium K 39.0983 39* 38.963 708 93.258140* 39.964 000 0.0117 1.28 × 109 y41* 40.961 827 6.7302

20 Calcium Ca 40.08 40* 39.962 591 96.94141* 40.962 279 1.0 × 105 y42* 41.958 618 0.64743* 42.958 767 0.13544* 43.955 481 2.086

21 Scandium Sc 44.9559 41* 40.969 250 0.596 s45* 44.955 911 100

22 Titanium Ti 47.88 44* 43.959 691 60 y47* 46.951 765 7.348* 47.947 947 73.8

23 Vanadium V 50.9415 50* 49.947 161 0.25 1.5 × 1017 y51* 50.943 962 99.75

24 Chromium Cr 51.996 48* 47.954 033 21.6 h52* 51.940 511 83.7953* 52.940 652 9.50

25 Manganese Mn 54.938 05 54* 53.940 361 312.1 d55* 54.938 048 100

26 Iron Fe 55.847 54* 53.939 613 5.955* 54.938 297 2.7 y56* 55.934 940 91.72

27 Cobalt Co 58.933 20 59* 58.933 198 10060* 59.933 820 5.27 y

28 Nickel Ni 58.793 58* 57.935 345 68.07759* 58.934 350 7.5 × 104 y60* 59.930 789 26.223

29 Copper Cu 63.54 63* 62.929 599 69.1765* 64.927 791 30.83

30 Zinc Zn 65.39 64* 63.929 144 48.666* 65.926 035 27.967* 66.927 129 4.168* 67.924 845 18.8


Mathematical ReviewAverage Mass number Half-lifeatomic (* indicates Atomic mass Percent (if radioactive)

Z Element Symbol mass (u) radioactive) A (u) abundance T1/2

31 Gallium Ga 69.723 69* 68.925 580 60.10871* 70.924 703 39.892

32 Germanium Ge 72.61 70* 69.924 250 21.2372* 71.922 079 27.6673* 72.923 462 7.7374* 73.921 177 35.9476* 75.921 402 7.44

33 Arsenic As 74.9216 75* 74.921 594 100

34 Selenium Se 78.96 76* 75.919 212 9.3677* 76.919 913 7.6378* 77.917 397 23.7880* 79.916 519 49.6182* 81.916 697 8.73 1.4 × 1020 y

35 Bromine Br 79.904 79* 78.918 336 50.6981* 80.916 287 49.31

36 Krypton Kr 83.80 81* 80.916 589 2.1 × 105 y82* 81.913 481 11.683* 82.914 136 11.484* 83.911 508 57.085* 84.912 531 10.76 y86* 85.910 615 17.3

37 Rubidium Rb 85.468 85* 84.911 793 72.1787* 86.909 186 27.83 4.75 × 1010 y

38 Strontium Sr 87.62 86* 85.909 266 9.8687* 86.908 883 7.0088* 87.905 618 82.5890* 89.907 737 29.1 y

39 Yttrium Y 88.9058 89* 88.905 847 100

40 Zirconium Zr 91.224 90* 89.904 702 51.4591* 90.905 643 11.2292* 91.905 038 17.1593* 92.906 473 1.5 × 106 y94* 93.906 314 17.38

41 Niobium Nb 92.9064 93* 92.906 376 10094* 93.907 280 2 × 104 y

42 Molybdenum Mo 95.94 92* 91.906 807 14.8493* 92.906 811 3.5 × 103 y94* 93.905 085 9.2595* 94.905 841 15.9296* 95.904 678 16.6897* 96.906 020 9.5598* 97.905 407 24.13

100* 99.907 476 9.63

43 Technetium Tc 97* 96.906 363 2.6 × 106 y98* 97.907 215 4.2 × 106 y99* 98.906 254 2.1 × 105 y

44 Ruthenium Ru 101.07 99* 98.905 939 12.7100* 99.904 219 12.6101* 100.905 558 17.1102* 101.904 348 31.6104* 103.905 558 18.6

45 Rhodium Rh 102.9055 103* 102.905 502 100


Average Mass number Half-lifeatomic (* indicates Atomic mass Percent (if radioactive)


46 Palladium Pd 106.42 104* 103.904 033 11.14105* 104.905 082 22.33106* 105.903 481 27.33108* 107.903 898 26.46110* 109.905 158 11.72

47 Silver Ag 107.868 107* 106.905 091 51.84109* 108.904 754 48.16

48 Cadmium Cd 112.41 109* 108.904 984 462 d110* 109.903 004 12.49111* 110.904 182 12.80112* 111.902 760 24.13113* 112.904 401 12.22 9.3 × 1015 y114* 113.903 359 28.73

49 Indium In 114.82 113* 112.904 060 4.3115* 114.903 876 95.7 4.4 × 1014 y

50 Tin Sn 118.71 116* 115.901 743 14.53117* 116.902 953 7.58118* 117.901 605 24.22119* 118.903 308 8.58120* 119.902 197 32.59121* 120.904 237 55 y

51 Antimony Sb 121.76 121* 120.903 820 57.36123* 122.904 215 42.64

52 Tellurium Te 127.60 125* 124.904 429 7.12126* 125.903 309 18.93128* 127.904 468 31.79 > 8 × 1024 y130* 129.906 228 33.87 < 1.25 × 1021 y

53 Iodine I 126.9045 127* 126.904 474 100129* 128.904 984 1.6 × 107 y

54 Xenon Xe 131.29 129* 128.904 779 26.4131* 130.905 069 21.2132* 131.904 141 26.9134* 133.905 394 10.4136* 135.907 214 8.9 > 2.36 × 1021 y

55 Cesium Cs 132.9054 133* 132.905 436 100135* 134.905 891 2 × 106 y137* 136.907 078 30 y

56 Barium Ba 137.33 133* 132.905 990 10.5 y137* 136.905 816 11.23138* 137.905 236 71.70

57 Lanthanum La 138.905 138* 137.907 105 0.0902 1.05 × 1011 y139* 138.906 346 99.9098

58 Cerium Ce 140.12 138* 137.905 986 0.25140* 139.905 434 88.43142* 141.909 241 11.13 > 5 × 1016 y

59 Praseodymium Pr 140.9076 141* 140.907 647 100

60 Neodymium Nd 144.24 142* 141.907 718 27.13143* 142.909 809 12.18144* 143.910 082 23.80 2.3 × 1015 y145* 144.912 568 8.30146* 145.913 113 17.19


Mathematical ReviewAverage Mass number Half-lifeatomic (* indicates Atomic mass Percent (if radioactive)


61 Promethium Pm 145* 144.912 745 17.7 y146* 145.914 968 5.5 y

62 Samarium Sm 150.36 147* 146.914 894 15.0 1.06 × 1011 y148* 147.914 819 11.3 7 × 1015 y149* 148.917 180 13.8 > 2 × 1015 y150* 149.917 273 7.4152* 151.919 728 26.7154* 153.922 206 22.7

63 Europium Eu 151.96 151* 150.919 846 47.8152* 151.921 740 13.5 y153* 152.921 226 52.2

64 Gadolinium Gd 157.25 155* 154.922 618 14.80156* 155.922 119 20.47157* 156.923 957 15.65158* 157.924 099 24.84160* 159.927 050 21.86

65 Terbium Tb 158.9253 159* 158.925 345 100

66 Dysprosium Dy 162.5 161* 160.926 930 18.9162* 161.926 796 25.5163* 162.928 729 24.9164* 163.929 172 28.2

67 Holmium Ho 164.9303 165* 164.930 316 100

68 Erbium Er 167.26 166* 165.930 292 33.6167* 166.932 047 22.95168* 167.932 369 27.8170* 169.935 462 14.9

69 Thulium Tm 168.9342 169* 168.934 213 100171* 170.936 428 1.92 y

70 Ytterbium Yb 173.04 171* 170.936 324 14.3172* 171.936 379 21.9173* 172.938 209 16.12174* 173.938 861 31.8176* 175.942 564 12.7

71 Lutetium Lu 174.967 175* 174.940 772 97.41176* 175.942 679 2.59 3.78 × 1010 y

72 Hafnium Hf 178.49 177* 176.943 218 18.606178* 177.943 697 27.297179* 178.945 813 13.029180* 179.946 547 35.100

73 Tantalum Ta 180.9479 181* 180.947 993 99.988

74 Tungsten W 183.85 182* 181.948 202 26.3183* 182.950 221 14.28184* 183.950 929 30.7186* 185.954 358 28.6

75 Rhenium Re 186.207 185* 184.952 951 37.40187* 186.955 746 62.60 4.4 × 1010 y

76 Osmium Os 190.2 188* 187.955832 13.3189* 188.958 139 16.1190* 189.958 439 26.4192* 191.961 468 41.0

77 Iridium Ir 192.2 191* 190.960 585 37.3193* 192.962 916 62.7


Average Mass number Half-lifeatomic (* indicates Atomic mass Percent (if radioactive)


78 Platinum Pt 195.08 194* 193.962 655 32.9195* 194.964 765 33.8196* 195.964 926 25.3

79 Gold Au 196.9665 197* 196.966 543 100

80 Mercury Hg 200.59 198* 197.966 743 9.97199* 198.968 253 16.87200* 199.968 299 23.10201* 200.970 276 13.10202* 201.970 617 29.86

81 Thallium Tl 204.383 203* 202.972 320 29.524204* 203.073 839 3.78 y205* 204.974 400 70.476208* 207.981 992 3.053 m

82 Lead Pb 207.2 206* 205.974 440 24.1207* 206.974 871 22.1208* 207.976 627 52.4212* 211.991 872 10.64 h

83 Bismuth Bi 208.9803 209* 208.980 374 100212* 211.991 259 60.6 m

84 Polonium Po 209* 208.982 405 102 y212* 211.988 842 0.30 µs216* 216.001 889 0.145 s

85 Astatine At 218* 218.008 685 1.6 s219* 219.011 294 0.9 m

86 Radon Rn 220* 220.011 369 55.6 s222* 222.017 571 3.823 d

87 Francium Fr 223* 223.019 733 22 m

88 Radium Ra 224* 224.020 187 3.66 d226* 226.025 402 1.6 × 103 y228* 228.031 064 5.75 y

89 Actinium Ac 227* 227.027 701 18.72 y228* 228.028 716 1.913 y

90 Thorium Th 232* 232.038 051 100 1.40 × 1010 y234* 234.043 593 24.1 d

91 Protactinium Pa 231* 231.035 880 32.760 y234* 234.043 300 6.7 h

92 Uranium U 234* 234.040 946 0.0055 2.46 × 105 y235* 235.043 924 0.720 7.04 × 108 y238* 238.050 784 99.2745 4.47 × 109 y

93 Neptunium Np 236* 236.046 560 1.15 × 105 y237* 237.048 168 2.14 × 106 y

94 Plutonium Pu 239* 239.052 157 2.412 × 105 y244* 244.064 200 8.1 × 107 y

AP

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Answers1. 11.68 m

2. 4.0469 × 10−3 km2

3. 6.4 × 10−2 m3

4. 6.0 × 109 mg

5. 6.7 × 10−5 ps

6. 3.53 × 103 km/h, west

7. 2.80 h = 2 h, 48 min

8. 107 s

9. 4.0 × 101 km/h

10. 0.46 m/s2

11. 48 m/h

12. 1.74 m/s

13. +25.0 m/s = 25.0 m/s, upward

14. −3.31 m/s

15. 44.8 m/s

16. +6.2 m/s2 = 6.2 m/s2, upward

17. −21.5 m/s2 = 21.5 m/s2,backward

18. 15.8 m

19. 38.5 m

20. −221 m = 221 m, downward

21. 126 s

22. 1.26 × 103 cm = 12.6 m

23. 1.27 s

24. 15.8 km/s

25. 11 km/h

26. 5.4 m/s2

27. 2.74 s

28. 6.50 s

29. 10.5 m, forward

30. −8.6 m/s = 8.6 m/s, backward

31. 5.9 s

880 Appendix I: Additional Problems

Additional ProblemsChapter 1 The Science of Physics

1. Mt. Waialeale in Hawaii gets 1.168 × 103 cm ofrainfall per year. Express this quantity in meters.

2. An acre is equal to about 4.0469 × 103 m2.Express this area in square kilometers.

3. A group drinks about 6.4 × 104 cm3 of water perperson per year. Express this in cubic meters.

4. The largest stone jar on the Plain of Jars in Laoshas a mass of 6.0 × 103 kg. Express this mass inmilligrams.

5. Half of a sample of the radioactive isotopeberyllium-8 decays in 6.7 × 10−17 s. Express thistime in picoseconds.

Chapter 2 Motion in One Dimension6. The fastest airplane is the Lockheed SR-71. If an

SR-71 flies 15.0 km west in 15.3 s, what is itsaverage velocity in kilometers per hour?

7. Except for a 22.0 min rest stop, Emily drives witha constant velocity of 89.5 km/h, north. Howlong does the trip take if Emily’s average velocityis 77.8 km/h, north?

8. A spaceship accelerates uniformly for 1220 km.How much time is required for the spaceship toincrease its speed from 11.1 km/s to 11.7 km/s?

9. A polar bear initially running at 4.0 m/s acceler-ates uniformly for 18 s. If the bear travels 135 min this time, what is its maximum speed?

10. A walrus accelerates from 7.0 km/h to 34.5 km/hover a distance of 95 m. What is the magnitudeof the walrus’s acceleration?

11. A snail can move about 4.0 m in 5.0 min. What isthe average speed of the snail?

12. A crate is accelerated at 0.035 m/s2 for 28.0 salong a conveyor belt. If the crate’s initial speedis 0.76 m/s, what is its final speed?

13. A person throws a ball vertically and catches itafter 5.10 s. What is the ball’s initial velocity?

14. A bicyclist accelerates –0.870 m/s2 during a3.80 s interval. What is the change in the velocityof the bicyclist and bicycle?

15. A hockey puck slides 55.0 m in 1.25 s with a uni-form acceleration. If the puck’s final speed is43.2 m/s, what was its initial speed?

16. A small rocket launched from rest travels 12.4 m upward in 2.0 s. What is the rocket’s net acceleration?

17. A jet slows uniformly from 153 km/h to 0 km/hover 42.0 m. What is the jet’s acceleration?

18. A softball thrown straight up at 17.5 m/s iscaught 3.60 s later. How high does the ball rise?

19. A child, starting from rest, sleds down a snow-covered slope in 5.50 s. If the child’s final speedis 14.0 m/s, what the length of the slope?

20. A sky diver opens her parachute and drifts downfor 34.0 s with a constant velocity of 6.50 m/s.What is the sky diver’s displacement?

21. In a race, a tortoise runs at 10.0 cm/s and a hareruns at 200.0 cm/s. Both start at the same time,but the hare stops to rest for 2.00 min. The tor-toise wins by 20.0 cm. At what time does the tortoise cross the finish line?

22. What is the length of the race in problem 21?

23. The cable pulling an elevator upward at 12.5 m/sbreaks. How long does it take for the elevator tocome to rest?

24. A disk is uniformly accelerated from rest for0.910 s over 7.19 km. What is its final speed?

25. A tiger accelerates 3.0 m/s2 for 4.1 s to reach afinal speed of 55.0 km/h. What was its initialspeed in kilometers per hour?

26. A shark accelerates uniformly from 2.8 km/h to32.0 km/h in 1.5 s. How large is its acceleration?

27. The 1903 Wright flyer was accelerated at 4.88 m/s2

along a track that was 18.3 m long. How long didit take to accelerate the flyer from rest?

28. A drag racer starts at rest and reaches a speed of386.0 km/h with an average acceleration of16.5 m/s2. How long does this acceleration take?

29. A hummingbird accelerates at −9.20 m/s2 suchthat its velocity changes from +50.0 km/h to0 km/h. What is its displacement?

30. A train backs up from an initial velocity of−4.0 m/s and an average acceleration of−0.27 m/s2. What is the train’s velocity after 17 s?

31. A cross-country skier skiing with an initial veloc-ity of +4.42 m/s slows uniformly at −0.75 m/s2.How long does it take the skier to stop?

32. 13 m, forward

33. 8.3 s

34. 6.2 m/s

35. 7.4 s

36. −125 m = 125 m, downward

37. −490 m/s2 = 490 m/s2,backward

38. 16.9 m/s

39. 17.3 s

40. −1200 m/s2 = 1200 m/s2,backward

41. 7.0 m

42. 2.0 × 101 m

43. 2.6 m/s

44. 0.76 s

45. −11.4 m/s = 11.4 m/s,downward

46. 592 m, east

47. 8.5° north of east

48. 475 km, west

49. 5.0° south of west

50. 6700 km, west

51. 770 m

52. 14.9 m, south

53. −33 km/h = 33 km/h,downward

54. 115 km, 17° north of east

55. 18.9 km, 76° north of west

56. 5.05 s

57. 17.0 m

58. 15 s

59. 52.0°

60. 0.47 h = 28 min

61. 79 s

62. 171 m, 15° below the horizontal

63. 15.8 m, 55° below the horizontal

881Appendix I: Additional Problems 881

32. What is the skier’s displacement in problem 31?

33. A speedboat uniformly increases its speed from25 m/s west to 35 m/s west. How long does ittake the boat to travel 250 m west?

34. A ship accelerates at −7.6 × 10−2 m/s2 so that itcomes to rest at the dock 255 m away in 82.0 s.What is the ship’s initial speed?

35. A student skates downhill with an average accel-eration of 0.85 m/s2. Her initial speed is 4.5 m/s,and her final speed is 10.8 m/s. How long doesshe take to skate down the hill?

36. A wrench dropped from a tall building is caughtin a safety net when the wrench has a velocity of−49.5 m/s. How far did it fall?

37. A rocket sled comes to a complete stop from aspeed of 320 km/h in 0.18 s. What is the sled’saverage acceleration?

38. A racehorse uniformly accelerates 7.56 m/s2,reaching its final speed after running 19.0 m. Ifthe horse starts at rest, what is its final speed?

39. An arrow is shot upward at a speed of 85.1 m/s.How long does the archer have to move from thelaunching spot before the arrow returns to Earth?

40. A handball strikes a wall with a forward speed of13.7 m/s and bounces back with a speed of11.5 m/s. If the ball changes velocity in 0.021 s,what is the handball’s average acceleration?

41. A ball accelerates at 6.1 m/s2 from 1.8 m/s to9.4 m/s. How far does the ball travel?

42. A small sandbag is dropped from rest from ahovering hot-air balloon. After 2.0 s, what is thesandbag’s displacement below the balloon?

43. A hippopotamus accelerates at 0.678 m/s2 untilit reaches a speed of 8.33 m/s. If the hippopota-mus runs 46.3 m, what was its initial speed?

44. A ball is hit upward with a speed of 7.5 m/s. Howlong does the ball take to reach maximum height?

45. A surface probe on the planet Mercury falls17.6 m downward from a ledge. If free-fall accel-eration near Mercury is −3.70 m/s2, what is theprobe’s velocity when it reaches the ground?

Chapter 3 Two-Dimensional Motionand Vectors46. A plane moves 599 m northeast along a runway.

If the northern component of this displacementis 89 m, how large is the eastern component?

47. Find the displacement direction in problem 46.

48. A train travels 478 km southwest along a straightstretch. If the train is displaced south by 42 km,what is the train’s displacement to the west?

49. Find the displacement direction in problem 48.

50. A ship’s total displacement is 7400 km at 26°south of west. If the ship sails 3200 km south,what is the western component of its journey?

51. The distance from an observer on a plain to thetop of a nearby mountain is 5.3 km at 8.4° abovethe horizontal. How tall is the mountain?

52. A skyrocket travels 113 m at an angle of 82.4°with respect to the ground and toward the south.What is the rocket’s horizontal displacement?

53. A hot-air balloon descends with a velocity of55 km/h at an angle of 37° below the horizontal.What is the vertical velocity of the balloon?

54. A stretch of road extends 55 km at 37° north ofeast, then continues for 66 km due east. What isa driver’s resultant displacement along this road?

55. A driver travels 4.1 km west, 17.3 km north, andfinally 1.2 km at an angle of 24.6° west of north.What is the driver’s displacement?

56. A tornado picks up a car and hurls it horizontal-ly 125 m with a speed of 90.0 m/s. How longdoes it take the car to reach the ground?

57. A squirrel knocks a nut horizontally at a speed of10.0 cm/s. If the nut lands at a horizontal dis-tance of 18.6 cm, how high up is the squirrel?

58. A flare is fired at an angle of 35° to the ground atan initial speed of 250 m/s. How long does ittake for the flare to reach its maximum altitude?

59. A football kicked with an initial speed of 23.1 m/sreaches a maximum height of 16.9 m. At whatangle was the ball kicked?

60. A bird flies north at 58.0 km/h relative to thewind. The wind is blowing at 55.0 km/h southrelative to Earth. How long will it take the bird tofly 1.4 km relative to Earth?

61. A race car moving at 286 km/h is 0.750 kmbehind a car moving at 252 km/h. How long willit take the faster car to catch up to the slower car?

62. A helicopter flies 165 m horizontally and thenmoves downward to land 45 m below. What isthe helicopter’s resultant displacement?

63. A toy parachute floats 13.0 m downward. If theparachute travels 9.0 m horizontally, what is theresultant displacement?

882

64. 2.6 m along table’s length;0.61 m along table’s width

65. 0.290 m/s, east; 1.16 m/s,north

66. 12.4 km/h, upward;53.6 km/h, forward

67. 2.6 km

68. 1.42 × 103 m, 16° to the side ofthe initial displacement

69. 66 km, 46° south of east

70. 404 m

71. 10.7 m

72. 70.0 m/s, 1.23° from the vertical

73. 3.0 s

74. +2.0 × 101 m = 2.0 × 101 m,forward; −16 m = 16 m,downward

75. 76.9 km/h, 60.1° west of north

76. 9.58 km, west; 16.7 km, north

77. 7.0 × 102 m, 3.8° above thehorizontal

78. 1.26 × 103 km, 48° north ofwest

79. 47.2 m

80. 3.6 m/s

81. 6.36 m/s

82. 11 m, west; 18 m, north

83. 13.6 km/h, 73° south of east

84. 47° north of west

85. 58 N

86. 49 N

87. 14.0 N; 2.0 N

88. −6.12 m/s2 = 6.12 m/s2,downward

89. 9.5 × 104 kg

90. +13 N = 13 N, upward

91. 258 N, up the slope

92. 0.087

93. 15.9 N

94. 0.73


64. A billiard ball travels 2.7 m at an angle of 13°with respect to the long side of the table. Whatare the components of the ball’s displacement?

65. A golf ball has a velocity of 1.20 m/s at 14.0° eastof north. What are the velocity components?

66. A tiger leaps with an initial velocity of 55.0 km/hat an angle of 13.0° with respect to the horizontal.What are the components of the tiger’s velocity?

67. A tramway extends 3.88 km up a mountain froma station 0.8 km above sea level. If the horizontaldisplacement is 3.45 km, how far above sea levelis the mountain peak?

68. A bullet travels 850 m, ricochets, and movesanother 640 m at an angle of 36° from its previ-ous forward motion. What is the bullet’s result-ant displacement?

69. A bird flies 46 km at 15° south of east, then 22 kmat 13° east of south, and finally 14 km at 14° westof south. What is the bird’s displacement?

70. A ball is kicked with a horizontal speed of9.37 m/s off the top of a mountain. The ballmoves 85.0 m horizontally before hitting theground. How tall is the mountain?

71. A ball is kicked with a horizontal speed of1.50 m/s from a height of 2.50 × 102 m. What is itshorizontal displacement when it hits the ground?

72. What is the velocity of the ball in problem 71when it reaches the ground?

73. A shingle slides off a roof at a speed of 2.0 m/sand an angle of 30.0° below the horizontal. Howlong does it take the shingle to fall 45 m?

74. A ball is thrown with an initial speed of 10.0 m/sand an angle of 37.0° above the horizontal. Whatare the vertical and horizontal components ofthe ball’s displacement after 2.5 s?

75. A rocket moves north at 55.0 km/h with respectto the air. It encounters a wind from 17.0° northof west at 40.0 km/h with respect to Earth. Whatis the rocket’s velocity with respect to Earth?

76. How far to the north and west does the rocket inproblem 75 travel after 15.0 min?

77. A cable car travels 2.00 × 102 m on level ground,then 3.00 × 102 m at an incline of 3.0°, and then2.00 × 102 m at an incline of 8.8°. What is thefinal displacement of the cable car?

78. A hurricane moves 790 km at 18° north of west,then due west for 150 km, then north for 470 km,and finally 15° east of north for 240 km. What isthe hurricane’s resultant displacement?

79. What is the range of an arrow shot horizontallyat 85.3 m/s from 1.50 m above the ground?

80. A drop of water in a fountain takes 0.50 s totravel 1.5 m horizontally. The water is projectedupward at an angle of 33°. What is the drop’s ini-tial speed?

81. A golf ball is hit up a 41.0° ramp to travel 4.46 mhorizontally and 0.35 m below the edge of theramp. What is the ball’s initial speed?

82. A flare is fired with a velocity of 87 km/h westfrom a car traveling 145 km/h north. Withrespect to Earth, what is the flare’s resultant dis-placement 0.45 s after being launched?

83. A sailboat travels south at 12.0 km/h withrespect to the water against a current 15.0° southof east at 4.0 km/h. What is the boat’s velocity?

Chapter 4 Forces and the Laws of Motion84. A boat exerts a 9.5 × 104 N force 15.0° north of

west on a barge. Another exerts a 7.5 × 104 Nforce north. What direction is the barge moved?

85. A shopper exerts a force on a cart of 76 N at anangle of 40.0° below the horizontal. How muchforce pushes the cart in the forward direction?

86. How much force pushes the cart in problem 85against the floor?

87. What are the magnitudes of the largest andsmallest net forces that can be produced by com-bining a force of 6.0 N and a force of 8.0 N?

88. A buoyant force of 790 N lifts a 214 kg sinkingboat. What is the boat’s net acceleration?

89. A house is lifted by a net force of 2850 N andmoves from rest to an upward speed of 15 cm/sin 5.0 s. What is the mass of the house?

90. An 8.0 kg bag is lifted 20.0 cm in 0.50 s. If it isinitially at rest, what is the net force on the bag?

91. A 90.0 kg skier glides at constant speed down a17.0° slope. Find the frictional force on the skier.

92. A snowboarder slides down a 5.0° slope at a con-stant speed. What is the coefficient of kineticfriction between the snow and the board?

93. A 2.00 kg block is in equilibrium on a 36.0°incline. What is the normal force on the block?

94. A 1.8 × 103 kg car is parked on a hill on a 15.0°incline. A 1.25 × 104 N frictional force holds thecar in place. Find the coefficient of static friction.

95. 2.0 m/s2

96. 4.0 N, to the right

97. Fx = 8.60 N; Fy = 12.3 N


99. −448 m/s2 = 448 m/s2,backward

100. 1.3 N, upward

101. 15 kg

102. 57 N

103. 0.085

104. 0.40

105. 1.7 × 108 N

106. 1.8 × 107 kg

107. 24 N, downhill

108. 22.3° up from west

109. 1.150 × 103 N

110. 3.4 × 103 kg

111. 1.2 × 104 N

112. 1400 kg

113. 0.60

114. 7.10 × 103 N

115. 38.0 m

116. −2.77 × 105 J

117. 2.5 × 104 J

118. 4.25 × 104 J

119. 247 m/s

120. 1.27 × 103 kg

121. −5.46 × 104 J

122. 5.1 × 102 N


95. The coefficient of kinetic friction between a jarslid across a table and the table is 0.20. What isthe magnitude of the jar’s acceleration?

96. A force of 5.0 N to the left causes a 1.35 kg bookto have a net acceleration of 0.76 m/s2 to theleft. What is the frictional force on the book?

97. A child pulls a toy by exerting a force of 15.0 Nat an angle of 55.0° with respect to the floor.What are the components of the force?

98. A car is pulled by three forces: 600.0 N to thenorth, 750.0 N to the east, and 675 N at 30.0°south of east. What direction does the car move?

99. Suppose a catcher exerts a force of −65.0 N tostop a baseball with a mass of 0.145 kg. What isthe ball’s net acceleration as it is being caught?

100. A 2.0 kg fish pulled upward by a fishermanrises 1.9 m in 2.4 s, starting from rest. What isthe net force on the fish during this interval?

101. An 18.0 N force pulls a cart against a 15.0 Nfrictional force. The speed of the cart increases1.0 m/s every 5.0 s. What is the cart’s mass?

102. A 47 kg sled carries a 33 kg load. The coefficientof kinetic friction between the sled and snow is0.075. What is the magnitude of the frictionalforce on the sled as it moves up a hill with a 15°incline?

103. Ice blocks slide with an acceleration of 1.22 m/s2

down a chute at an angle of 12.0° below the hori-zontal. What is the coefficient of kinetic frictionbetween the ice and chute?

104. A 1760 N force pulls a 266 kg load up a 17°incline. What is the coefficient of static frictionbetween the load and the incline?

105. A 4.26 × 107 N force pulls a ship at a constantspeed along a dry dock. The coefficient of ki-netic friction between the ship and dry dock is0.25. Find the normal force exerted on the ship.

106. If the incline of the dry dock in problem 105 is10.0°, what is the ship’s mass?

107. A 65.0 kg skier is pulled up an 18.0° slope by a force of 2.50 × 102 N. If the net accelerationuphill is 0.44 m/s2, what is the frictional forcebetween the skis and the snow?

108. Four forces are acting on a hot-air balloon:F1 = 2280.0 N up, F2 = 2250.0 N down, F3 =85.0 N west, and F4 = 12.0 N east. What is thedirection of the net external force on the balloon?

109. A traffic signal is supported by two cables, eachof which makes an angle of 40.0° with the ver-tical. If each cable can exert a maximum forceof 7.50 × 102 N, what is the largest weight theycan support?

110. A certain cable of an elevator is designed toexert a force of 4.5 × 104 N. If the maximumacceleration that a loaded car can withstand is3.5 m/s2, what is the combined mass of the carand its contents?

111. A frictional force of 2400 N keeps a crate ofmachine parts from sliding down a ramp withan incline of 30.0°. The coefficient of static fric-tion between the box and the ramp is 0.20. Whatis the normal force of the ramp on the box?

112. Find the mass of the crate in problem 111.

113. A 5.1 × 102 kg bundle of bricks is pulled up aramp at an incline of 14° to a construction site.The force needed to move the bricks up theramp is 4.1 × 103 N. What is the coefficient ofstatic friction between the bricks and the ramp?

Chapter 5 Work and Energy114. If 2.13 × 106 J of work must be done on a

roller-coaster car to move it 3.00 × 102 m, howlarge is the net force acting on the car?

115. A force of 715 N is applied to a roller-coastercar to push it horizontally. If 2.72 × 104 J ofwork is done on the car, how far has it beenpushed?

116. In 0.181 s, through a distance of 8.05 m, a testpilot’s speed decreases from 88.9 m/s to 0 m/s.If the pilot’s mass is 70.0 kg, how much work isdone against his body?

117. What is the kinetic energy of a disk with a massof 0.20 g and a speed of 15.8 km/s?

118. A 9.00 × 102 kg walrus is swimming at a speedof 35.0 km/h. What is its kinetic energy?

119. A golf ball with a mass of 47.0 g has a kineticenergy of 1433 J. What is the ball’s speed?

120. A turtle, swimming at 9.78 m/s, has a kineticenergy of 6.08 × 104 J. What is the turtle’s mass?

121. A 50.0 kg parachutist is falling at a speed of47.00 m/s when her parachute opens. Her speedupon landing is 5.00 m/s. How much work isdone by the air to reduce the parachutist’s speed?

122. An 1100 kg car accelerates from 48.0 km/h to59.0 km/h over 100.0 m. What was the magni-tude of the net force acting on it?

884

123. 3.35 × 106 J

124. 4.0 × 10−2 J

125. 1.23 J

126. −2.05 × 103 J

127. 12 s

128. 1.17 × 1010 J

129. 0.600 m

130. 17.2 N

131. 133 J

132. 9.4 × 109 J

133. 53.3 m/s

134. 0.17 kg

135. 72.2 m

136. 9.6 × 102 kg

137. 0.13 m = 13 cm

138. 33.5 m/s

139. 7.7 m/s

140. 2.7 × 1010 W = 27 GW

141. 8.0 s

142. 3.0 × 107 J

143. 230 J

144. 8.72 × 106 N/m

145. 7.96 m

146. 12 m

147. 6.0 × 101 m/s

148. 15 m

149. 1.58 × 103 kg•m/s, north

150. 6.66 m/s, south

151. 3.38 × 1031 kg

152. −0.897 N = 0.879 N, to theleft

153. 18 s


123. What is the gravitational potential energy of a64.0 kg person at 5334 m above sea level?

124. A spring has a force constant of 550 N/m. Whatis the elastic potential energy stored in thespring when the spring is compressed 1.2 cm?

125. What is the kinetic energy of a 0.500 g raindropthat falls 0.250 km? Ignore air resistance.

126. A 50.0 g projectile is fired upward at 3.00 ×102 m/s and lands at 89.0 m/s. How muchmechanical energy is lost to air resistance?

127. How long does it take for 4.5 × 106 J of work tobe done by a 380.3 kW engine?

128. A ship’s engine has a power output of 13.0 MW.How much work can it do in 15.0 min?

129. A catcher picks up a baseball from the groundwith a net upward force of 7.25 × 10−2 N sothat 4.35 × 10−2 J of net work is done. How faris the ball lifted?

130. A crane does 1.31 × 103 J of net work when lift-ing cement 76.2 m. How large is the net forcedoing this work?

131. A girl exerts a force of 35.0 N at an angle of20.0° to the horizontal to move a wagon 15.0 malong a level path. What is the net work doneon it if a frictional force of 24.0 N is present?

132. The Queen Mary had a mass of 7.5 × 107 kgand a top cruising speed of 57 km/h. What wasthe kinetic energy of the ship at that speed?

133. How fast is a 55.0 kg sky diver falling when herkinetic energy is 7.81 × 104 J?

134. A hockey puck with an initial speed of 8.0 m/scoasts 45 m to a stop. If the force of friction onthe puck is 0.12 N, what is the puck’s mass?

135. How far does a 1.30 × 104 kg jet travel if it isslowed from 2.40 × 102 km/h to 0 km/h by anacceleration of −30.8 m/s2?

136. An automobile is raised 7.0 m, resulting in anincrease in gravitational potential energy of6.6 × 104 J. What is the automobile’s mass?

137. A spring in a pogo stick has a force constant of1.5 × 104 N/m. How far is the spring compressedwhen its elastic potential energy is 120 J?

138. A 100.0 g arrow is pulled back 30.0 cm againsta bowstring. The bowstring’s force constant is1250 N/m. What speed will the arrow leave the bow?

139. A ball falls 3.0 m down a vertical pipe, the endof which bends horizontally. How fast does theball leave the pipe if no energy is lost to friction?

140. A spacecraft’s engines do 1.4 × 1013 J of work in8.5 min.What is the power output of these engines?

141. A runner exerts a force of 334 N against theground while using 2100 W of power. How longdoes it take him to run a distance of 50.0 m?

142. A high-speed boat has four 300.0 kW motors.How much work is done in 25 s by the motors?

143. A 92 N force pushes an 18 kg box of books, ini-tially at rest, 7.6 m across a floor. The coeffi-cient of kinetic friction between the floor andthe box is 0.35. What is the final kinetic energyof the box of books?

144. A guardrail can be bent by 5.00 cm and thenrestore its shape. What is its force constant if struck by a car with 1.09 × 104 J of kineticenergy?

145. A 25.0 kg trunk strikes the ground with a speed of12.5 m/s. If no energy is lost from air resistance,what is the height from which the trunk fell?

146. Sliding a 5.0 kg stone up a frictionless ramp witha 25.0° incline increases its gravitational poten-tial energy by 2.4 × 102 J. How long is the ramp?

147. A constant 4.00 × 102 N force moves a 2.00 ×102 kg iceboat 0.90 km. Frictional force is negli-gible, and the boat starts at rest. Find the boat’sfinal speed.

148. A 50.0 kg circus clown jumps from a platforminto a net 1.00 m above the ground. The net isstretched 0.65 m and has a force constant of 3.4 ×104 N/m. What is the height of the platform?

Chapter 6 Momentum and Collisions149. If a 50.0 kg cheetah, initially at rest, runs 274 m

north in 8.65 s, what is its momentum?

150. If a 1.46 × 105 kg whale has a momentum of9.73 × 105 kg•m/s to the south, what is itsvelocity?

151. A star has a momentum of 8.62 × 1036 kg•m/sand a speed of 255 km/s. What is its mass?

152. A 5.00 g projectile has a velocity of 255 m/s right.Find the force to stop this projectile in 1.45 s.

153. How long does it take a 0.17 kg hockey puck todecrease its speed by 9.0 m/s if the coefficientof kinetic friction is 0.050?

154. −6.81 × 103 N = 6.81 × 103 N,to the left

155. 637 m, to the right

156. −6.67 m/s = 6.67 m/s,backward

157. 7.5 g

158. 14 kg

159. 0.0 m/s

160. −1.61 × 105 J

161. −5.0 × 101 percent

162. 3.0 m/s

163. 16.4 m/s, west

164. 6.0 km/h, south

165. 5.33 × 107 kg•m/s

166. 2.00 × 10−2 kg = 20.0 g

167. 1.0 × 101 m/s

168. 6.2 m/s

169. 560 N, east

170. 17 s

171. −3.3 × 108 N = 3.3 × 108 N,backward

172. −3.9 × 103 kg•m/s = 3.9 ×103 kg•m/s, opposite thepolar bear’s motion

173. 52 m

174. 4.06 s

175. 24 kg

176. 2.1 m/s

177. 90.6 km/h, east

178. 1.7 m/s, west

179. 26 km/h, 37° north of east

180. −6.9 J

181. −157 J

182. 4.2 m/s

183. 0.125 kg


154. A 705 kg race car driven by a 65 kg driver moveswith a velocity of 382 km/h right. Find the forceto bring the car and driver to a stop in 12.0 s.

155. Find the stopping distance in problem 154.

156. A 50.0 g shell fired from a 3.00 kg rifle has aspeed of 400.0 m/s. With what velocity does therifle recoil in the opposite direction?

157. A twig at rest in a pond moves with a speed of0.40 cm/s opposite a 2.5 g snail, which has aspeed of 1.2 cm/s. What is the mass of the twig?

158. A 25.0 kg sled holding a 42.0 kg child has aspeed of 3.50 m/s. They collide with and pickup a snowman, initially at rest. The resultingspeed of the snowman, sled, and child is2.90 m/s. What is the snowman’s mass?

159. An 8500 kg railway car moves right at 4.5 m/s,and a 9800 kg railway car moves left at 3.9 m/s.The cars collide and stick together. What is thefinal velocity of the system?

160. What is the change in kinetic energy for thetwo railway cars in problem 159?

161. A 55 g clay ball moving at 1.5 m/s collides with a55 g clay ball at rest. By what percentage does thekinetic energy change after the inelastic collision?

162. A 45 g golf ball collides elastically with an iden-tical ball at rest and stops. If the second ball’sfinal speed is 3.0 m/s, what was the first ball’sinitial speed?

163. A 5.00 × 102 kg racehorse gallops with amomentum of 8.22 × 103 kg•m/s to the west.What is the horse’s velocity?

164. A 3.0 × 107 kg ship collides elastically with a2.5 × 107 kg ship moving north at 4.0 km/h.After the collision, the first ship moves north at3.1 km/h and the second ship moves south at6.9 km/h. Find the unknown velocity.

165. A high-speed train has a mass of 7.10 × 105 kgand moves at a speed of 270.0 km/h. What isthe magnitude of the train’s momentum?

166. A bird with a speed of 50.0 km/h has amomentum of magnitude of 0.278 kg•m/s.What is the bird’s mass?

167. A 75 N force pulls a child and sled initially atrest down a snowy hill. If the combined mass ofthe sled and child is 55 kg, what is their speedafter 7.5 s?

168. A student exerts a net force of −1.5 N over aperiod of 0.25 s to bring a falling 60.0 g egg to astop. What is the egg’s initial speed?

169. A 1.1 × 103 kg walrus starts swimming eastfrom rest and reaches a velocity of 9.7 m/s in19 s. What is the net force acting on the walrus?

170. A 12.0 kg wagon at rest is pulled by a 15.0 Nforce at an angle of 20.0° above the horizontal.If an 11.0 N frictional force resists the forwardforce, how long will the wagon take to reach aspeed of 4.50 m/s?

171. A 42 g meteoroid moving forward at 7.82 ×103 m/s collides with a spacecraft. What force isneeded to stop the meteoroid in 1.0 × 10−6 s?

172. A 455 kg polar bear slides for 12.2 s across theice. If the coefficient of kinetic friction betweenthe bear and the ice is 0.071, what is the changein the bear’s momentum as it comes to a stop?

173. How far does the bear in problem 172 slide?

174. How long will it take a −1.26 × 104 N force tostop a 2.30 × 103 kg truck moving at a speed of22.2 m/s?

175. A 63 kg skater at rest catches a sandbag movingnorth at 5.4 m/s. The skater and bag then movenorth at 1.5 m/s. Find the sandbag’s mass.

176. A 1.36 × 104 kg barge is loaded with 8.4 × 103 kgof coal. What was the unloaded barge’s speed ifthe loaded barge has a speed of 1.3 m/s?

177. A 1292 kg automobile moves east at 88.0 km/h.If all forces remain constant, what is the car’svelocity if its mass is reduced to 1255 kg?

178. A 68 kg student steps into a 68 kg boat at rest, causing both to move west at a speed of0.85 m/s. What was the student’s initial velocity?

179. A 1400 kg automobile, heading north at45 km/h, collides inelastically with a 2500 kgtruck traveling east at 33 km/h. What is thevehicles’ final velocity?

180. An artist throws 1.3 kg of paint onto a 4.5 kgcanvas at rest. The paint-covered canvas slidesbackward at 0.83 m/s. What is the change in thekinetic energy of the paint and canvas?

181. Find the change in kinetic energy if a 0.650 kgfish leaping to the right at 15.0 m/s collidesinelastically with a 0.950 kg fish leaping to theleft at 13.5 m/s.

182. A 10.0 kg cart moving at 6.0 m/s hits a 2.5 kg cartmoving at 3.0 m/s in the opposite direction. Findthe carts’ final speed after an inelastic collision.

183. A ball, thrown right 6.00 m/s, hits a 1.25 kg panelat rest, then bounces back at 4.90 m/s. The panelmoves right at 1.09 m/s. Find the ball’s mass.

886

184. 2.1 m/s, east

185. −4.1 × 104 J

186. 12.8 cm/s, to the right

187. 9.8 kg

188. −6.6 × 1016 J

189. 1.0 m/s, 60° south of east

190. 1.2 kg

191. 4.04 × 103 m/s2

192. 256 m

193. 42 m/s

194. 6350 N

195. 8.9 kg

196. 4.47 × 1015 m

197. 1.04 × 104 m/s = 10.4 km/s

198. 8.34 × 10−7 N

199. 1.48 × 1023 kg

200. 6.96 × 108 m

201. 1.10 × 1012 m

202. 1.2 × 105 s = 34 h

203. 6.6 × 103 m/s = 6.6 km/s

204. 7.49 × 104 N•m

205. 0.87 m

206. 2.35 × 107 m = 2.35 × 104 km

207. 254 N

208. 3.4 × 10−5 m/s2

209. 0.42 m = 42 cm

210. 38 m/s

211. 25 N


184. A 2150 kg car, moving east at 10.0 m/s, collidesand joins with a 3250 kg car. The cars moveeast together at 5.22 m/s. What is the 3250 kgcar’s initial velocity?

185. Find the change in kinetic energy in problem 184.

186. A 15.0 g toy car moving to the right at 20.0 cm/scollides elastically with a 20.0 g toy car movingleft at 30.0 cm/s. The 15.0 g car then moves leftat 37.1 cm/s. Find the 20.0 g car’s final velocity.

187. A remora swimming right at 5.0 m/s attaches to a150.0 kg shark moving left at 7.00 m/s. Bothmove left at 6.25 m/s. Find the remora’s mass.

188. A 6.5 × 1012 kg comet, moving at 420 m/s,catches up to and collides inelastically with a 1.50 × 1013 kg comet moving at 250 m/s.Find the change in the comets’ kinetic energy.

189. A 7.00 kg ball moves east at 2.00 m/s, collideswith a 7.00 kg ball at rest, and then moves 30.0°north of east at 1.73 m/s. What is the secondball’s final velocity?

190. A 2.0 kg block moving at 8.0 m/s on a friction-less surface collides elastically with a block atrest. The first block moves in the same directionat 2.0 m/s. What is the second block’s mass?

Chapter 7 Circular Motion and Gravitation191. A pebble that is 3.81 m from the eye of a tor-

nado has a tangential speed of 124 m/s. What is the magnitude of the pebble’s centripetalacceleration?

192. A race car speeds along a curve with a tangentialspeed of 75.0 m/s. The centripetal acceleration onthe car is 22.0 m/s2. Find the radius of the curve.

193. A subject in a large centrifuge has a radius of 8.9 m and a centripetal acceleration of 20g(g = 9.81 m/s2). What is the tangential speed ofthe subject?

194. A 1250 kg automobile with a tangential speed of48.0 km/h follows a circular road that has a radiusof 35.0 m. How large is the centripetal force?

195. A rock in a sling is 0.40 m from the axis of rotationand has a tangential speed of 6.0 m/s. What is therock’s mass if the centripetal force is 8.00 × 102 N?

196. A 7.55 × 1013 kg comet orbits the sun with aspeed of 0.173 km/s. If the centripetal force onthe comet is 505 N, how far is it from the sun?

197. A 2.05 × 108 kg asteroid has an orbit with a7378 km radius. The centripetal force on theasteroid is 3.00 × 109 N. Find the asteroid’s tan-gential speed.

198. Find the gravitational force between a 0.500 kgmass and a 2.50 × 1012 kg mountain that is10.0 km away.

199. The gravitational force between Ganymede and Jupiter is 1.636 × 1022 N. Jupiter’s mass is1.90 × 1027 kg, and the distance between thetwo bodies is 1.071 × 106 km. What isGanymede’s mass?

200. At the sun’s surface, the gravitational force on1.00 kg is 274 N. The sun’s mass is 1.99 × 1030 kg.If the sun is assumed spherical, what is the sun’sradius?

201. At the surface of a red giant star, the gravitationalforce on 1.00 kg is only 2.19 × 10−3 N. If its massequals 3.98 × 1031 kg, what is the star’s radius?

202. Uranus has a mass of 8.6 × 1025 kg. The meandistance between the centers of the planet andits moon Miranda is 1.3 × 105 km. If the orbitis circular, what is Miranda’s period in hours?

203. What is the tangential speed in problem 202?

204. The rod connected halfway along the 0.660 mradius of a wheel exerts a 2.27 × 105 N force.How large is the maximum torque?

205. A golfer exerts a torque of 0.46 N•m on a golfclub. If the club exerts a force of 0.53 N on a sta-tionary golf ball, what is the length of the club?

206. What is the orbital radius of the Martian moonDeimos if it orbits 6.42 × 1023 kg Mars in 30.3 h?

207. A 4.00 × 102 N•m torque is produced applyinga force 1.60 m from the fulcrum and at an angleof 80.0° to the lever. How large is the force?

208. A customer 11 m from the center of a revolvingrestaurant has a speed of 1.92 × 10−2 m/s. Howlarge a centripetal acceleration acts on the customer?

209. A toy train on a circular track has a tangentialspeed of 0.35 m/s and a centripetal accelerationof 0.29 m/s2. What is the radius of the track?

210. A person against the inner wall of a hollowcylinder with a 150 m radius feels a centripetalacceleration of 9.81 m/s2. Find the cylinder’stangential speed.

211. The tangential speed of 0.20 kg toy carts is 5.6 m/swhen they are 0.25 m from a turning shaft. Howlarge is the centripetal force on the carts?

212. 15.0 m/s = 54.0 km/h

213. 165 kg

214. 0.11 N

215. 5.09 × 105 s = 141 h

216. 4.39 × 103 m/s = 4.39 km/s

217. 5.5 × 109 m = 5.5 × 106 km

218. 4.0 N

219. 1.6 N•m

220. 15.5 m

221. 6.62 × 103 N

222. 9 N

223. 0.574 m

224. 1.56 × 103 N

225. 8.13 × 10−3 m2

226. 1.0 × 105 Pa

227. 2.25 × 104 kg/m3

228. 1.30 × 103 kg/m3

229. 2.0 × 101 m2

230. 2.49 × 106 N

231. 4.30 kg

232. 1.90 × 102°C to −1.80 × 102°C

233. 374°F to −292°F

234. 24 K

235. 6.6 × 10−2°C

236. 65.0 kg

237. 1.29 × 104 J

238. −79°C

239. 4.1 × 10−2 kg

240. 957 J/kg•°C

241. 1.200 × 103°C

242. −2.70 × 102°C


212. A 1250 kg car on a curve with a 35.0 m radius hasa centripetal force from friction and gravity of8.07 × 103 N. What is the car’s tangential speed?

213. Two wrestlers, 2.50 × 10−2 m apart, exert a 2.77 ×10−3 N gravitational force on each other. Onehas a mass of 157 kg. What is the other’s mass?

214. A 1.81 × 105 kg blue whale is 1.5 m from a2.04 × 104 kg whale shark. What is the gravita-tional force between them?

215. Triton’s orbit around Neptune has a radius of3.56 × 105 km. Neptune’s mass is 1.03 × 1026 kg.What is Triton’s period?

216. Find the tangential speed in problem 215.

217. A moon orbits a 1.0 × 1026 kg planet in 365 days.What is the radius of the moon’s orbit?

218. What force is required to produce a 1.4 N •mtorque when applied to a door at a 60.0° angleand 0.40 m from the hinge?

219. What is the maximum torque that the force inproblem 218 can exert?

220. A worker hanging 65.0° from the vane of awindmill exerts an 8.25 × 103 N•m torque. If theworker weighs 587 N, what is the vane’s length?

Chapter 8 Fluid Mechanics221. A cube of volume 1.00 m3 floats in gasoline,

which has a density of 675 kg/m3. How large abuoyant force acts on the cube?

222. A cube 10.0 cm on each side has a density of2.053 × 104 kg/m3. Its apparent weight in freshwater is 192 N. Find the buoyant force.

223. A 1.47 × 106 kg steel hull has a base that is 2.50 ×103 m2 in area. If it is placed in sea water (r =1.025 × 103 kg/m3), how deep does the hull sink?

224. What size force will open a door of area 1.54 m2

if the net pressure on the door is 1.013 × 103 Pa?

225. Gas at a pressure of 1.50 × 106 Pa exerts a forceof 1.22 × 104 N on the upper surface of a pis-ton. What is the piston’s upper surface area?

226. In a barometer, the mercury column’s weightequals the force from air pressure on the mer-cury’s surface. Mercury’s density is 13.6 ×103 kg/m3. What is the air’s pressure if the column is 760 mm high?

227. A cube of osmium with a volume of 166 cm3 isplaced in fresh water. The cube’s apparent weightis 35.0 N. What is the density of osmium?

228. A block of ebony with a volume of 2.5 × 10−3 m3

is placed in fresh water. If the apparent weight ofthe block is 7.4 N, what is the density of ebony?

229. One piston of a hydraulic lift holds 1.40 × 103 kg.The other holds an ice block (r = 917 kg/m3)that is 0.076 m thick. Find the first piston’s area.

230. A hydraulic-lift piston raises a 4.45 × 104 Nweight by 448 m. How large is the force on theother piston if it is pushed 8.00 m downward?

231. A platinum flute with a density of 21.5 g/cm3 issubmerged in fresh water. If its apparent weightis 40.2 N, what is the flute’s mass?

Chapter 9 Heat 232. Surface temperature on Mercury ranges from

463 K during the day to 93 K at night. Expressthis temperature range in degrees Celsius.

233. Solve problem 233 for degrees Fahrenheit.

234. The temperature in Fort Assiniboine, Montana,went from −5°F to +37°F on January 19, 1892.Calculate this change in temperature in kelvins.

235. An acorn falls 9.5 m, absorbing 0.85 of its ini-tial potential energy. If 1200 J/kg will raise theacorn’s temperature 1.0°C, what is its tempera-ture increase?

236. A bicyclist on level ground brakes from 13.4 m/sto 0 m/s. What is the cyclist’s and bicycle’s massif the increase in internal energy is 5836 J?

237. A 61.4 kg roller skater on level ground brakesfrom 20.5 m/s to 0 m/s. What is the totalchange in the internal energy of the system?

238. A 0.225 kg tin can (cp = 2.2 × 103 J/kg•°C) iscooled in water, to which it transfers 3.9 × 104 Jof energy. By how much does the can’s tem-perature change?

239. What mass of bismuth (cp = 121 J/kg•°C)increases temperature by 5.0°C when 25 J areadded by heat?

240. Placing a 0.250 kg pot in 1.00 kg of water raisesthe water’s temperature 1.00°C. The pot’s tem-perature drops 17.5°C. Find the pot’s specificheat capacity.

241. Lavas at Kilauea in Hawaii have temperatures of2192°F. Express this quantity in degrees Celsius.

242. The present temperature of the backgroundradiation in the universe is 2.7 K. What is thistemperature in degrees Celsius?

888

243. 315 K

244. 1.4 × 104 J

245. 1.91 × 10−2 kg = 19.1 g

246. 1.2 × 10−4 kg

247. 530 J/kg•°C

248. 820 kg

249. −930°C

250. 4.70 × 106 J

251. 1.50 × 103 Pa = 1.50 kPa

252. 1.89 × 10−2 m3

253. 873 J

254. −1.0 × 101 J

255. 244 J

256. 2.0 × 102 J

257. 5.3 × 103 J

258. 0.189

259. 2.4 × 103 Pa = 2.4 kPa

260. 6.00 × 10−4 m3

261. 5895 J

262. 978 kJ = 9.78 × 105 J

263. 5.30 × 102 kJ = 5.30 × 105 J

264. 2.6 × 108 J

265. 1.0 × 104 J

266. 0.220

267. −18 N

268. 1.0 × 104 N/m

269. −0.11 m = −11 cm

270. 3.57 m/s2

271. 4.0 × 10−2 m = 4.0 cm

272. 4.993 s

273. 0.2003 Hz

274. 1.6 s

275. 730 N/m


243. The human body cannot survive at a tempera-ture of 42°C for very long. Express this quantityin kelvins.

244. Two sticks rubbed together gain 2.15 × 104 Jfrom kinetic energy and lose 33 percent of it tothe air. How much does the sticks’ internalenergy change?

245. A stone falls 561.7 m. When the stone lands, theinternal energy of the ground and the stoneincreases by 105 J. What is the stone’s mass?

246. A 2.5 kg block of ice at 0.0°C slows on a levelfloor from 5.7 m/s to 0 m/s. If 3.3 × 105 J cause1.0 kg of ice to melt, how much of the ice melts?

247. Placing a 3.0 kg skillet in 5.0 kg of water raisesthe water’s temperature 2.25°C and lowers theskillet’s temperature 29.6°C. Find the skillet’sspecific heat.

248. Air has a specific heat of 1.0 × 103 J/kg•°C. Ifair’s temperature increases 55°C when 45 × 106 Jare added to it by heat, what is the air’s mass?

249. A 0.23 kg tantalum part has a specific heatcapacity of 140 J/kg •°C. By how much does the part’s temperature change if it gives up 3.0 × 104 J as heat?

Chapter 10 Thermodynamics250. A volume of air increases 0.227 m3 at a net

pressure of 2.07 × 107 Pa. How much work isdone on the air?

251. The air in a hot-air balloon does 3.29 × 106 J of work, increasing the balloon’s volume by2190 m3. What is the net pressure in the balloon?

252. Filling a fire extinguisher with nitrogen gas at anet pressure of 25.0 kPa requires 472.5 J of workon the gas. Find the change in the gas’s volume.

253. The internal energy of air in a closed car rises873 J. How much heat energy is transferred tothe air?

254. A system’s initial internal energy increases from39 J to 163 J. If 114 J of heat are added to thesystem, how much work is done on the system?

255. A gas does 623 J of work on its surroundingswhen 867 J are added to the gas as heat. What isthe change in the internal energy of the gas?

256. An engine with an efficiency of 0.29 takes in 693 Jas heat. How much work does the engine do?

257. An engine with an efficiency of 0.19 does 998 Jof work. How much energy is taken in by heat?

258. Find the efficiency of an engine that receives571 J as heat and loses 463 J as heat per cycle.

259. A 5.4 × 10−4 m3 increase in steam’s volume does1.3 J of work on a piston. What is the pressure?

260. A pressure of 655 kPa does 393 J of work inflat-ing a bike tire. Find the change in volume.

261. An engine’s internal energy changes from 8093 Jto 2.0920 × 104 J. If 6932 J are added as heat, howmuch work is done on or by the system?

262. Steam expands from a geyser to do 192 kJ ofwork. If the system’s internal energy increases by786 kJ, how much energy is transferred as heat?

263. If 632 kJ are added to a boiler and 102 kJ ofwork are done as steam escapes from a safetyvalve, what is the net change in the system’sinternal energy?

264. A power plant with an efficiency of 0.35 per-cent requires 7.37 × 108 J of energy as heat.How much work is done by the power plant?

265. An engine with an efficiency of 0.11 does 1150 Jof work. How much energy is taken in as heat?

266. A test engine performs 128 J of work andreceives 581 J of energy as heat. What is theengine’s efficiency?

Chapter 11 Vibrations and Waves267. A scale with a spring constant of 420 N/m is

compressed 4.3 cm. What is the spring force?

268. A 669 N weight attached to a giant springstretches it 6.5 cm. What is the spring constant?

269. An archer applies a force of 52 N on a bow-string with a spring constant of 490 N/m. Whatis the bowstring’s displacement?

270. On Mercury, a pendulum 1.14 m long wouldhave a 3.55 s period. Calculate ag for Mercury.

271. Find the length of a pendulum that oscillateswith a frequency of 2.5 Hz.

272. Calculate the period of a 6.200 m long pendu-lum in Oslo, Norway, where ag = 9.819 m/s2.

273. Find the pendulum’s frequency in problem 272.

274. A 24 kg child jumps on a trampoline with aspring constant of 364 N/m. What is the oscil-lation period?

275. A 32 N weight oscillates with a 0.42 s periodwhen on a spring scale. Find the spring constant.

276. 1.0 × 10−2 kg = 1.0 × 101 g

277. 1.4 × 103 m/s

278. 16.6 m

279. 2.2 × 104 Hz

280. 9.8 N

281. 8.6 × 103 N/m

282. −0.82 m = −82 cm

283. 3.177 s

284. 0.99 m = 99 cm

285. 82 kg

286. 850 N/m

287. 1.2 s

288. 31.2 Hz

289. 1.5 × 103 m/s

290. 1.1 m

291. 1.1 W/m2

292. 1.3 × 10−2 W

293. 294 Hz

294. 0.420 m

295. 408 m/s

296. 3.30 × 102 m/s

297. 0.155 m

298. 5.2 × 10−8 W

299. 0.211 m = 21.1 cm

300. 0.447 m = 44.7 cm

301. 2.9971 × 108 m/s

302. 9.4 × 1016 Hz

303. 3.2 × 10−7 m = 320 nm

304. 65 cm

305. −0.96 cm

306. 24.7 cm

307. −1.9 cm

308. 0.19 m

309. 3.8 m

310. 36 cm

311. 0.25

312. 19 cm


276. Find the mass of a ball that oscillates at a periodof 0.079 s on a spring with a constant of 63 N/m.

277. A dolphin hears a 280 kHz sound with a wave-length of 0.51 cm. What is the wave’s speed?

278. If a sound wave with a frequency of 20.0 Hzhas a speed of 331 m/s, what is its wavelength?

279. A sound wave has a speed of 2.42 × 104 m/s and awavelength of 1.1 m. Find the wave’s frequency.

280. An elastic string with a spring constant of65 N/m is stretched 15 cm and released. Whatis the spring force exerted by the string?

281. The spring in a seat compresses 7.2 cm under a620 N weight. What is the spring constant?

282. A 3.0 kg mass is hung from a spring with a springconstant of 36 N/m. Find the displacement.

283. Calculate the period of a 2.500 m long pendu-lum in Quito, Ecuador, where ag = 9.780 m/s2.

284. How long is a pendulum with a frequency of0.50 Hz?

285. A tractor seat supported by a spring with aspring constant of 2.03 × 103 N/m oscillates at a frequency of 0.79 Hz. What is the mass on the spring?

286. An 87 N tree branch oscillates with a period of0.64 s. What is the branch’s spring constant?

287. What is the oscillation period for an 8.2 kg babyin a seat that has a spring constant of 221 N/m?

288. An organ creates a sound with a speed of 331 m/sand a wavelength of 10.6 m. Find the frequency.

289. What is the speed of an earthquake s-wave with a 2.3 × 104 m wavelength and a 0.065 Hzfrequency?

Chapter 12 Sound290. What is the distance from a sound with 5.88 ×

10−5 W power if its intensity is 3.9 × 10−6 W/m2?

291. Sound waves from a stereo have a power outputof 3.5 W at 0.50 m. What is the sound’s intensity?

292. What is a vacuum cleaner’s power output if thesound’s intensity 1.5 m away is 4.5 × 10−4 W/m2?

293. Waves travel at 499 m/s on a 0.850 m long cellostring. Find the string’s fundamental frequency.

294. A mandolin string’s first harmonic is 392 Hz.How long is the string if the wave speed on it is329 m/s?

295. A 1.53 m long pipe that is closed on one endhas a seventh harmonic frequency of 466.2 Hz.What is the speed of the waves in the pipe?

296. A pipe open at both ends has a fundamentalfrequency of 125 Hz. If the pipe is 1.32 m long,what is the speed of the waves in the pipe?

297. Traffic has a power output of 1.57 × 10−3 W. Atwhat distance is the intensity 5.20 × 10−3 W/m2?

298. If a mosquito’s buzzing has an intensity of 9.3 ×10−8 W/m2 at a distance of 0.21 m, how muchsound power does the mosquito generate?

299. A note from a flute (a pipe with a closed end)has a first harmonic of 392.0 Hz. How long isthe flute if the sound’s speed is 331 m/s?

300. An organ pipe open at both ends has a firstharmonic of 370.0 Hz when the speed of soundis 331 m/s. What is the length of this pipe?

Chapter 13 Light and Reflection301. A 7.6270 × 108 Hz radio wave has a wavelength

of 39.296 cm. What is this wave’s speed?

302. An X ray’s wavelength is 3.2 nm. Using the speedof light in a vacuum, calculate the frequency ofthe X ray.

303. What is the wavelength of ultraviolet light witha frequency of 9.5 × 1014 Hz?

304. A concave mirror has a focal length of 17 cm.Where must a 2.7 cm tall coin be placed for itsimage to appear 23 cm in front of the mirror’ssurface?

305. How tall is the coin’s image in problem 304?

306. A concave mirror’s focal length is 9.50 cm. A3.0 cm tall pin appears to be 15.5 cm in front ofthe mirror. How far from the mirror is the pin?

307. How tall is the pin’s image in problem 306?

308. A convex mirror’s magnification is 0.11. Supposeyou are 1.75 m tall. How tall is your image?

309. How far in front of the mirror in problem 308 areyou if your image is 42 cm behind the mirror?

310. A mirror’s focal length is −12 cm. What is theobject distance if an image forms 9.00 cmbehind the surface of the mirror?

311. What is the magnification in problem 310?

312. A metal bowl is like a concave spherical mirror.You are 35 cm in front of the bowl and see animage at 42 cm. What is the bowl’s focal length?

890

313. 38 cm

314. −84.0 cm

315. 2.40

316. −7.7 cm

317. 0.98 cm

318. 5.67 × 1018 Hz

319. 10.5 cm

320. 17 cm to 15 cm

321. 64.0 cm in front of the mirror

322. 1.0 × 101 cm

323. 8.3 cm

324. −17 cm

325. 0.40

326. −5.6 cm

327. −11 cm

328. 4.1 cm

329. 2.9979 × 108 m/s

330. 1.2 cm

331. 33.3 cm

332. 27 cm

333. 0.19

334. 63°

335. 32.2°

336. 1.52

337. −10.4 cm

338. −0.50

339. 18 cm

340. 9.0 cm

341. 1.63

342. 34.49°

343. 39.38°


313. For problem 312, find the bowl’s radius ofcurvature.

314. A concave spherical mirror on a dressing tablehas a focal length of 60.0 cm. If someone sits35.0 cm in front of it, where is the image?


316. An image appears 5.2 cm behind the surface of aconvex mirror when the object is 17 cm in frontof the mirror. What is the mirror’s focal length?

317. If the object in problem 316 is 3.2 cm tall, howtall is its image?

318. In order for someone to observe an object, thewavelength of the light must be smaller than theobject. The Bohr radius of a hydrogen atom is5.291 770 × 10−11 m. What is the lowest frequen-cy that can be used to locate a hydrogen atom?

319. Meteorologists use Doppler radar to watch themovement of storms. If a weather station useselectromagnetic waves with a frequency of2.85 × 109 Hz, what is the wavelength of theradiation?

320. PCS cellular phones have antennas that use radiofrequencies from 1800–2000 MHz. What rangeof wavelengths corresponds to these frequencies?

321. Suppose you have a mirror with a focal lengthof 32.0 cm. Where would you place your righthand so that you appear to be shaking handswith yourself?

322. A car’s headlamp is made of a light bulb infront of a concave spherical mirror. If the bulbis 5.0 cm in front of the mirror, what is theradius of the mirror?

323. Suppose you are 19 cm in front of the bell ofyour friend’s trumpet and you see your imageat 14 cm. If the trumpet’s bell is a concave mir-ror, what would be its focal length?

324. A soup ladle is like a spherical convex mirrorwith a focal length of 27 cm. If you are 43 cm infront of the ladle, where does the image appear?


326. Just after you dry a spoon, you look into the con-vex part of the spoon. If the spoon has a focallength of −8.2 cm and you are 18 cm in front ofthe spoon, where does the image appear?

327. The base of a lamp is made of a convex spheri-cal mirror with a focal length of −39 cm. Wheredoes the image appear when you are 16 cmfrom the base?

328. Consider the lamp and location in problem 327.If your nose is 6.0 cm long, how long does theimage appear?

329. How fast does microwave radiation that has afrequency of 1.173 06 × 1011 Hz and a wave-length of 2.5556 mm travel?

330. Suppose the microwaves in your microwaveoven have a frequency of 2.5 × 1010 Hz. What isthe wavelength of these microwaves?

331. You place an electric heater 3.00 m in front of aconcave spherical mirror that has a focal length of30.0 cm. Where would your hand feel warmest?

332. You see an image of your hand as you reach fora doorknob with a focal length of 6.3 cm. Howfar from the doorknob is your hand when theimage appears at 5.1 cm behind the doorknob?

333. What is the magnification of the image inproblem 332?

Chapter 14 Refraction334. A ray of light in air enters an amethyst crystal

(n = 1.553). If the angle of refraction is 35°,what is the angle of incidence?

335. Light passes from air at an angle of incidence of 59.2° into a nephrite jade vase (n = 1.61).Determine the angle of refraction in the jade.

336. Light entering a pearl travels at a speed of1.97 × 108 m/s. What is the pearl’s index ofrefraction?

337. An object in front of a diverging lens of focallength 13.0 cm forms an image with a magnifi-cation of +5.00. How far from the lens is theobject placed?

338. An object with a height of 18 cm is placed infront of a converging lens. The image height is−9.0 cm. What is the magnification of the lens?

339. If the focal length of the lens in problem 338 is6.0 cm, how far in front of the lens is the object?

340. Where does the image appear in problem 339?

341. The critical angle for light traveling from agreen tourmaline gemstone into air is 37.8°.What is tourmaline’s index of refraction?

342. Find the critical angle for light traveling fromruby (n = 1.766) into air.

343. Find the critical angle for light traveling fromemerald (n =1.576) into air.

344. 17.6° and 20.3°

345. 58°

346. 1.583

347. −21 cm

348. 4.8 cm

349. ∞350. ∞351. 1.486

352. 1.45

353. 1.54

354. ∞355. 4.8 cm

356. 17 cm

357. 1.73 to 1.83

358. 1.54

359. 5.18 × 10−4 m = 0.518 mm

360. 6.07 × 10−7 m = 607 nm

361. 0.137°

362. 1.775 × 104 lines/cm

363. 9.0 × 10−7 m = 9.0 × 102 nm

364. 5.56°

365. 11.2°

366. 4.44 × 10−7 m = 444 nm

367. 0.227°

368. 2.40 × 10−4 m = 0.240 mm

369. 1.445 × 104 lines/cm

370. 4.27 × 10−7 m = 427 nm

371. 140 N attractive

372. 1.4 × 10−3 N attractive

373. 2.2 × 10−17 C


344. Malachite has two indices of refraction:n1 = 1.91 and n2 = 1.66. A ray of light in airenters malachite at an incident angle of 35.2°.Calculate both of the angles of refraction.

345. A ray of light in air enters a serpentine figurine(n = 1.555). If the angle of refraction is 33°,what is the angle of incidence?

346. The critical angle for light traveling from anaquamarine gemstone into air is 39.18°. What isthe index of refraction for aquamarine?

347. A 15 cm tall object is placed 44 cm in front of adiverging lens. A virtual image appears 14 cm infront of the lens. What is the lens’s focal length?

348. What is the image height in problem 347?

349. A lighthouse converging lens has a focal lengthof 4 m. What is the image distance for an objectplaced 4 m in front of the lens?


351. Light moves from olivine (n = 1.670) into onyx.If the critical angle for olivine is 62.85°, what isthe index of refraction for onyx?

352. When light in air enters an opal mounted on aring, the light travels at a speed of 2.07 × 108

m/s. What is opal’s index of refraction?

353. When light in air enters albite, it travels at avelocity of 1.95 × 108 m/s. What is albite’s indexof refraction?

354. A searchlight is constructed by placing a 500 Wbulb 0.5 m in front of a converging lens. Thefocal length of the lens is 0.5 m. What is theimage distance?

355. A microscope slide is placed in front of a converg-ing lens with a focal length of 3.6 cm. The lensforms a real image of the slide 15.2 cm behind thelens. How far is the lens from the slide?

356. Where must an object be placed to form animage 12 cm in front of a diverging lens with afocal length of 44 cm?

357. The critical angle for light traveling fromalmandine garnet into air ranges from 33.1° to35.3°. Calculate the range of almandine garnet’sindex of refraction.

358. Light moves from a clear andalusite (n = 1.64)crystal into ivory. If the critical angle forandalusite is 69.9°, what is the index ofrefraction for ivory?

Chapter 15 Interference andDiffraction359. Light with a 587.5 nm wavelength passes through

two slits. A second-order bright fringe forms0.130° from the center. Find the slit separation.

360. Light passing through two slits with a separa-tion of 8.04 × 10−6 m forms a third bright fringe13.1° from the center. Find the wavelength.

361. Two slits are separated by 0.0220 cm. Find theangle at which a first-order bright fringe isobserved for light with a wavelength of 527 nm.

362. For 546.1 nm light, the first-order maximumfor a diffraction grating forms at 75.76°. Howmany lines per centimeter are on the grating?

363. Infrared light passes through a diffraction gratingof 3600 lines/cm. The angle of the third-ordermaximum is 76.54°. What is the wavelength?

364. A diffraction grating with 1950 lines/cm is usedto examine light with a wavelength of 497.3 nm.Find the angle of the first-order maximum.

365. At what angle does the second-order maximumin problem 364 appear?

366. Light passes through two slits separated by 3.92 ×10−6 m to form a second-order bright fringe atan angle of 13.1°. What is the light’s wavelength?

367. Light with a wavelength of 430.8 nm shines ontwo slits that are 0.163 mm apart. What is theangle at which a second dark fringe is observed?

368. Light of wavelength 656.3 nm passes through twoslits. The fourth-order dark fringe is 0.548° fromthe central maximum. Find the slit separation.

369. The first-order maximum for light with a wave-length of 447.1 nm is found at 40.25°. Howmany lines per centimeter does the grating have?

370. Light through a diffraction grating of9550 lines/cm forms a second-order maximumat 54.58°. What is the wavelength of the light?

Chapter 16 Electric Forces and Fields371. Charges of −5.3 μC and +5.3 μC are separated

by 4.2 cm. Find the electric force between them.

372. A dog’s fur is combed, and the comb gains acharge of 8.0 nC. Find the electric force betweenthe fur and comb when they are 2.0 cm apart.

373. Two equal charges are separated by 6.5 × 10−11 m.If the magnitude of the electric force between thecharges is 9.92 × 10−4 N, what is the value of q?

892

374. 0.387 m = 38.7 cm

375. 0.00 N

376. 1.91 × 10−10 N, 45.0° abovethe horizontal

377. 4.0 × 10−8 N, 9.3° below thenegative x-axis

378. 0.16 m = 16 cm

379. 260 N from either charge

380. −54 N (54 N along the negative x-axis)

381. 1.6 × 10−12 C

382. 3.15 × 105 N/C, upward

383. 0.585 m = 58.5 cm

384. 1.60 × 10−19 C

385. 3.97 × 10−6 N, upward

386. 1.74 × 10−7 N, 21.2° abovethe negative x-axis

387. 0.073 m = 7.3 cm

388. 9.2 C

389. 7.5 × 10−6 N, along the positive y-axis

390. 5.06 × 10−12 C

391. 4.40 × 105 N/C, 89.1° abovethe negative x-axis

392. 1.66 × 10−10 m

393. −7.4 C

394. 1.6 × 10−19 C

395. 1.6 × 10−19 C

396. 4.82 × 10−19 N, 45° above thepositive x-axis

397. 36 cm

398. −512 C

399. 4.4 × 10−4 J

400. 56 C


374. Two point charges of −13.0 μC and −16.0 μCexert repulsive forces on each other of 12.5 N.What is the distance between the two charges?

375. Three equal point charges of 4.00 nC lie 4.00 mapart on a line. Calculate the magnitude anddirection of the net force on the middle charge.

376. A proton is at each corner of a square with sides1.52 × 10−9 m long. Calculate the resultant forcevector on the proton at the upper right corner.

377. Three 2.0 nC charges are located at coordinates(0 m, 0 m), (1.0 m, 0 m), and (1.0 m, 2.0 m).Find the resultant force on the first charge.

378. Charges of 7.2 nC and 6.7 nC are 32 cm apart.Find the equilibrium position for a −3.0 nCcharge.

379. A −12.0 μC charge is between two 6.0 μCcharges, 5.0 cm away from each. What electricforce keeps the central charge in equilibrium?

380. A 9.0 N/C electric field is directed along the x-axis.Find the electric force vector on a −6.0 C charge.

381. What charge experiences an electric force of 6.43 × 10−9 N in an electric field of4.0 × 103 N/C?

382. A 5.00 μC charge is 0.500 m above a 15.0 μCcharge. Calculate the electric field at a point1.00 m above the 15.0 mC charge.

383. Two static point charges of 99.9 μC and 33.3 μCexert repulsive forces on each other of 87.3 N.What is the distance between the two charges?

384. Two particles are separated by 9.30 × 10−11 m. Ifthe magnitude of the electric force between thecharges is 2.66 × 10−8 N, what is the value of q?

385. A −23.4 nC charge is 0.500 m below a 4.65 nCcharge and 1.00 m below a 0.299 nC charge. Findthe resultant force vector on the −23.4 nC charge.

386. Three point charges are on the corners of a trian-gle: q1 = −9.00 nC is at the origin; q2 = −8.00 nCis at x = 2.00 m; and q3 = 7.00 nC is at y = 3.00 m.Find the magnitude and direction of the resul-tant force on q1.

387. Charges of −2.50 nC and −7.50 nC are 20.0 cmapart. Find a 5.0 nC charge’s equilibrium position.

388. A −4.6 C charge is in equilibrium with a −2.3 Ccharge 2.0 m to the right, and an unknown charge4.0 m to the right. What is the unknown charge?

389. Find the electric force vector on a 5.0 nC chargein a 1500 N/C electric field directed along they-axis.

390. What electric charge experiences an 8.42 × 10−9 Nelectric force in an electric field of 1663 N/C?

391. Two 3.00 μC charges lie 2.00 m apart on thex-axis. Find the resultant electric field vector ata point 0.250 m on the y-axis, above the chargeon the left.

392. Two electrons are 2.00 × 10−10 m and 3.00 ×10−10 m, respectively, from a point. Where withrespect to that point must a proton be placed sothat the resultant electric field strength is zero?

393. A −7.0 C charge is in equilibrium with a 49 Ccharge 18 m to the right and an unknown charge25 m to the right. What is the unknown charge?

394. Suppose two pions are separated by 8.3 × 10−10 m.If the magnitude of the electric force between thecharges is 3.34 × 10−10 N, what is the value of q?

395. Suppose two muons having equal but oppositecharge are separated by 6.4 × 10−8 m. If themagnitude of the electric force between thecharges is 5.62 × 10−14 N, what is the value of q?

396. Consider four electrons at the corners of asquare. Each side of the square is 3.02 ×10−5 m. Find the magnitude and direction ofthe resultant force on q3 if it is at the origin.

397. A charge of 5.5 nC and a charge of 11 nC areseparated by 88 cm. Find the equilibrium posi-tion for a −22 nC charge.

398. Three charges are on the y-axis. At the origin is acharge, q1 = 72 C; an unknown charge, q2, is aty = 15 mm. A third charge, q3 = −8.0 C, is placedat y = −9.0 mm so that it is in electrostatic equi-librium with q1 and q2. What is the charge on q2?

Chapter 17 Electrical Energy and Current399. A helium-filled balloon with a 14.5 nC charge

rises 290 m above Earth’s surface. By how muchdoes the electrical potential energy change ifEarth’s electric field is −105 N/C?

400. A charged airplane rises 7.3 km in a 3.4 ×105 N/C electric field. The electrical potentialenergy changes by −1.39 × 1011 J. What is thecharge on the plane?

401. 7.1 × 10−4 F

402. 7.5 × 10−13 C

403. 4.0 A

404. 7.0 × 101 s

405. 160 Ω406. 9.84 A

407. 1.7 × 106 W = 1.7 MW

408. 8.23 Ω409. 6.4 × 102 N/C

410. 0.96 m

411. 12 V

412. 1 μF

413. 1.2 × 10−5 m

414. 0.21 C

415. 1.4 × 102 C

416. 3.2 × 10−4 A = 0.32 mA

417. 7.2 s

418. 3.6 × 102 C

419. 4.8 V

420. 6.5 A

421. 116 V

422. 220 Ω423. 5.0 × 105 W = 0.50 MW

424. 1.59 A

425. 7.5 × 106 V

426. 2.2 A

427. 3.00 × 102 Ω428. 114 Ω429. 6.0 Ω430. 11 Ω431. 13 Ω432. 24 Ω


401. Earth’s radius is 6.4 × 106 m. What is Earth’scapacitance if it is regarded as a conductingsphere?

402. A 0.50 pF capacitor is connected across a 1.5 Vbattery. How much charge can this capacitor store?

403. A 76 C charge passes through a wire’s cross-sectional area in 19 s. Find the current in thewire.

404. The current in a telephone is 1.4 A. How long does98 C of charge take to pass a point in the wire?

405. What is a television’s total resistance if it isplugged into a 120 V outlet and carries 0.75 Aof current?

406. A motor with a resistance of 12.2 Ω is pluggedinto a 120.0 V outlet. What is the current in themotor?

407. The potential difference across a motor with a0.30 Ω resistance is 720 V. How much power isused?

408. What is a microwave oven’s resistance if it uses1750 W of power at a voltage of 120.0 V?

409. A 64 nC charge moves 0.95 m with an electricalpotential energy change of −3.88 × 10−5 J. Whatis the electric field strength?

410. A −14 nC charge travels through a 156 N/Celectric field with a change of 2.1 × 10−6 J in theelectrical potential energy. How far does thecharge travel?

411. A 5.0 × 10−5 F polyester capacitor stores 6.0 ×10−4 C. Find the potential difference across thecapacitor.

412. Some ceramic capacitors can store 3 × 10−2 Cwith a potential difference of 30 kV across them.What is the capacitance of such a capacitor?

413. The area of the plates in a 4550 pF parallel-plate capacitor is 6.4 × 10−3 m2. Find the plateseparation.

414. A television receiver contains a 14 μF capacitorcharged across a potential difference of 1.5 × 104 V.How much charge does this capacitor store?

415. A photocopier uses 9.3 A in 15 s. How muchcharge passes a point in the copier’s circuit inthis time?

416. A 114 μC charge passes through a gold wire’scross-sectional area in 0.36 s. What is the current?

417. If the current in a blender is 7.8 A, how long do56 C of charge take to pass a point in the circuit?

418. A computer uses 3.0 A in 2.0 min. How muchcharge passes a point in the circuit in this time?

419. A battery-powered lantern has a resistance of6.4 Ω. What potential difference is provided bythe battery if the total current is 0.75 A?

420. The potential difference across an electric eel is650 V. How much current would an electric eeldeliver to a body with a resistance of 1.0 × 102 Ω?

421. If a garbage-disposal motor has a resistance of25.0 Ω and carries a current of 4.66 A, what is thepotential difference across the motor’s terminals?

422. A medium-sized oscillating fan draws 545 mA ofcurrent when the potential difference across itsmotor is 120 V. How large is the fan’s resistance?

423. A generator produces a 2.5 × 104 V potentialdifference across power lines that carry 20.0 Aof current. How much power is generated?

424. A computer with a resistance of 91.0 Ω uses230.0 W of power. Find the current in the computer.

425. A laser uses 6.0 × 1013 W of power. What is thepotential difference across the laser’s circuit ifthe current in the circuit is 8.0 × 106 A?

426. A blender with a 75 Ω resistance uses 350 W ofpower. What is the current in the blender’s circuit?

Chapter 18 Circuits and CircuitElements427. A theater has 25 surround-sound speakers

wired in series. Each speaker has a resistance of12.0 Ω. What is the equivalent resistance?

428. In case of an emergency, a corridor on an air-plane has 57 lights wired in series. Each lightbulb has a resistance of 2.00 Ω. Find the equiva-lent resistance.

429. Four resistors with resistances of 39 Ω, 82 Ω,12 Ω, and 42 Ω are connected in parallel acrossa 3.0 V potential difference. Find the equivalentresistance.

430. Four resistors with resistances of 33 Ω, 39 Ω,47 Ω, and 68 Ω are connected in parallel acrossa 1.5 V potential difference. Find the equivalentresistance.

431. A 16 Ω resistor is connected in series with an-other resistor across a 12 V battery. The current inthe circuit is 0.42 A. Find the unknown resistance.

432. A 24 Ω resistor is connected in series with an-other resistor across a 3.0 V battery. The current inthe circuit is 62 mA. Find the unknown resistance.

894

433. 6.0 Ω434. 4.0 Ω435. 0.056 A = 56 mA

436. 0.665 A = 665 mA

437. 1.6 A (refrigerator);1.3 A (oven)

438. 1.5 A (computer);5.0 A (printer)

439. 12.6 Ω440. 0.952 A

441. 2.6 V

442. 0.43 A

443. 9.4 Ω444. 1.6 A

445. 1.6 A

446. 16.6 Ω447. 1.45 A

448. 0.97 A

449. 4 × 10−12 N

450. 3.0 × 10−10 T

451. 7.6 × 106 m/s

452. 1 × 10−2 N

453. 0.70 A

454. 0.63 m

455. 5.1 × 10−4 T

456. 0.30 T


433. A 3.3 Ω resistor and another resistor are con-nected in parallel across a 3.0 V battery. Thecurrent in the circuit is 1.41 A. Find theunknown resistance.

434. A 56 Ω resistor and another resistor are con-nected in parallel across a 12 V battery. Thecurrent in the circuit is 3.21 A. Find theunknown resistance.

435. Three bulbs with resistances of 56 Ω, 82 Ω, and24 Ω are wired in series. If the voltage across thecircuit is 9.0 V, what is the current in the circuit?

436. Three bulbs with resistances of 96 Ω, 48 Ω, and29 Ω are wired in series. What is the currentthrough the bulbs if the voltage across them is115 V?

437. A refrigerator (R1 = 75 Ω) wired in parallelwith an oven (R2 = 91 Ω) is plugged into a120 V outlet. What is the current in the circuitof each appliance?

438. A computer (R1 = 82 Ω) and printer (R2 = 24 Ω)are wired in parallel across a 120 V potential difference. Find the current in each machine’scircuit.

439. For the figure above, what is the equivalentresistance of the circuit?

440. For the figure above, find the current in the circuit.

441. For the figure above, what is the potential dif-ference across the 6.0 Ω resistor?

442. For the figure above, what is the currentthrough the 6.0 Ω resistor?

443. For the figure above, calculate the equivalentresistance of the circuit.

444. For the figure above, what is the total current inthe circuit?

445. For the figure above, what is the current in the3.0 Ω resistors?

446. For the figure above, calculate the equivalentresistance of the circuit.

447. For the figure above, what is the total current inthe circuit?

448. For the figure above, what is the current ineither of the 8.0 Ω resistors?

Chapter 19 Magnetism449. A proton moves at right angles to a magnetic

field of 0.8 T. If the proton’s speed is 3.0 ×107 m/s, how large is the magnetic force exerted on the proton?

450. A weak magnetic field exerts a 1.9 × 10−22 Nforce on an electron moving 3.9 × 106 m/s per-pendicular to the field. What is the magneticfield strength?

451. A 5.0 × 10−5 T magnetic field exerts a 6.1 ×10−17 N force on a 1.60 × 10−19 C charge, whichmoves at a right angle to the field. What is thecharge’s speed?

452. A 14 A current passes through a 2 m wire. A3.6 × 10−4 T magnetic field is at right angles tothe wire. What is the magnetic force on the wire?

453. A 1.0 m printer cable is perpendicular to a 1.3 × 10−4 T magnetic field. What current mustthe cable carry to experience a 9.1 × 10−5 Nmagnetic force?

454. A wire perpendicular to a 4.6 × 10−4 T magneticfield experiences a 2.9 × 10−3 N magnetic force.How long is the wire if it carries a 10.0 A current?

455. A 12 m wire carries a 12 A current. What mag-netic field causes a 7.3 × 10−2 N magnetic forceto act on the wire when it is perpendicular tothe field?

456. A magnetic force of 3.7 × 10−13 N is exerted onan electron moving at 7.8 × 106 m/s perpen-dicular to a sunspot. How large is the sunspot’smagnetic field?

8.0 Ω3.0 Ω

4.0 Ω

2.0 Ω

24.0 V

3.0 Ω

8.0 Ω

4.0 Ω5.0 Ω 2.0 Ω

5.0 Ω

5.0 Ω

5.0 Ω

3.0 Ω

5.0 Ω

5.0 Ω3.0 Ω

5.0 Ω

15.0 V

5.0 Ω

2.0 Ω

5.0 Ω

6.0 Ω4.0 Ω

1.5 Ω3.0 Ω

12.0 V

457. 3.9 × 10−15 N

458. 2 × 105 m/s

459. 1.5 × 105 A

460. 3.7 × 102 N

461. 1.7 × 10−2 T

462. 0.33 V

463. 0.90 s

464. 7.6 × 102 turns

465. 450 V

466. 3 A

467. 1.8 A

468. 6.9 × 104 V = 69 kV

469. 3.4 × 104 V = 34 kV

470. 29 turns

471. 48 turns

472. 6.8 × 10−2 m2

473. 2.5 × 10−3 A = 2.5 mA

474. 149 V

475. 0.85 A

476. 120 V

477. 48:1

478. 3.77 × 10−19 J

479. 3.32 × 10−10 m

480. 5.00 × 1015 Hz

481. 1.00 × 10−13 m

482. 2.9 eV

483. 3.0 × 10−7 m

484. 1.50 × 107 m/s

485. 26 kg

486. 4.62 × 10−19 J


457. An electron moves with a speed of 2.2 ×106 m/s at right angles through a 1.1 × 10−2 Tmagnetic field. How large is the magnetic forceon the electron?

458. A pulsar’s magnetic field is 1 × 10−8 T. How fastdoes an electron move perpendicular to this fieldso that a 3.2 × 10−22 N magnetic force acts onthe charge?

459. A levitation device designed to suspend 75 kguses 10.0 m of wire and a 4.8 × 10−4 T mag-netic field, perpendicular to the wire. Whatcurrent is needed?

460. A power line carries 1.5 × 103 A for 15 km.Earth’s magnetic field is 2.3 × 10−5 T at a 45°angle to the power line. What is the magneticforce on the line?

Chapter 20 Electromagnetic Induction461. A coil with 540 turns and a 0.016 m2 area is

rotated exactly from 0° to 90.0° in 0.050 s. Howstrong must a magnetic field be to induce anemf of 3.0 V?

462. A 550-turn coil with an area of 5.0 × 10−5 m2 is ina magnetic field that decreases by 2.5 × 10−4 T in2.1 × 10−5 s. What is the induced emf in the coil?

463. A 246-turn coil has a 0.40 m2 area in a mag-netic field that increases from 0.237 T to 0.320 T. What time interval is needed to inducean emf of −9.1 V?

464. A 9.5 V emf is induced in a coil that rotatesfrom 0.0° to 90.0° in a 1.25 × 10−2 T magneticfield for 25 ms. The coil’s area is 250 cm2. Howmany turns of wire are in the coil?

465. A generator provides a rms emf of 320 V across100 Ω. What is the maximum emf?

466. Find the rms current in the circuit in prob-lem 465.

467. Some wind turbines can provide an rms cur-rent of 1.3 A. What is the maximum ac current?

468. A transformer has 1400 turns on the primary and140 turns on the secondary. What is the voltageacross the primary if secondary voltage is 6.9 kV?

469. A transformer has 140 turns on the primaryand 840 turns on the secondary. What is thevoltage across the secondary if the primaryvoltage is 5.6 kV?

470. A step-down transformer converts a 3.6 kV volt-age to 1.8 kV. If the primary (input) coil has 58turns, how many turns does the secondary have?

471. A step-up transformer converts a 4.9 kV voltageto 49 kV. If the secondary (output) coil has 480turns, how many turns does the primary have?

472. A 320-turn coil rotates from 0° to 90.0° in a 0.046T magnetic field in 0.25 s, which induces an aver-age emf of 4.0 V. What is the area of the coil?

473. A 180-turn coil with a 5.0 × 10−5 m2 area is in amagnetic field that decreases by 5.2 × 10−4 T in1.9 × 10−5 s. What is the induced current if thecoil’s resistance is 1.0 × 102 W?

474. A generator provides a maximum ac current of1.2 A and a maximum output emf of 211 V.Calculate the rms potential difference.

475. Calculate the rms current for problem 474.

476. A generator can provide a maximum output emfof 170 V. Calculate the rms potential difference.

477. A step-down transformer converts 240 V acrossthe primary to 5.0 V across the secondary.What is the step-down ratio (N1:N2)?

Chapter 21 Atomic Physics478. Determine the energy of a photon of green

light with a wavelength of 527 nm.

479. Calculate the de Broglie wavelength of an electron with a velocity of 2.19 × 106 m/s.

480. Calculate the frequency of ultraviolet (UV)light having a photon energy of 20.7 eV.

481. X-ray radiation can have an energy of 12.4MeV. To what wavelength does this correspond?

482. Light of wavelength 240 nm shines on a potas-sium surface. Potassium has a work function of2.3 eV. What is the maximum kinetic energy ofthe photoelectrons?

483. Manganese has a work function of 4.1 eV. Whatis the wavelength of the photon that will justhave the threshold energy for manganese?

484. What is the speed of a proton with a de Brogliewavelength of 2.64 × 10−14 m?

485. A cheetah can run as fast as 28 m/s. If the cheetahhas a de Broglie wavelength of 8.97 × 10−37 m,what is the cheetah’s mass?

486. What is the energy of a photon of blue lightwith a wavelength of 430.8 nm?

896

487. 4.30 × 1014 Hz

488. 0.40 m

489. 6.0 × 1014 Hz

490. 0.24 eV

491. 4.0 × 10−21 kg

492. 5.64 eV

493. 2.5 × 10−43 m

494. 2.5 × 10−7 m

495. 7.72 × 1014 Hz

496. 7.1 × 106 m/s

497. 333.73 MeV

498. 363.89 MeV

499. 0.543 705 u

500. 20882Pb

501. 168O

502. 42He

503. 15.0 s

504. 1.6 × 10−9 Ci

505. 31.92 h

506. 1020.6 MeV

507. 35.46 MeV

508. 0.600216 u

509. 13153I

510. 42He

511. 11154Xe

512. 1.25 days

513. 924 days

514. 8.1 × 10−9 s−1


487. Calculate the frequency of infrared (IR) lightwith a photon energy of 1.78 eV.

488. Calculate the wavelength of a radio wave thathas a photon energy of 3.1 × 10−6 eV.

489. Light of frequency 6.5 × 1014 Hz illuminates a lithium surface. The ejected photoelectronsare found to have a maximum kinetic energy of 0.20 eV. Find the threshold frequency ofthis metal.

490. Light of wavelength 519 nm shines on a rubid-ium surface. Rubidium has a work function of2.16 eV. What is the maximum kinetic energyof the photoelectrons?

491. The smallest known virus moves across a Petridish at 5.6 × 10−6 m/s. If the de Broglie wave-length of the virus is 2.96 × 10−8 m, what is thevirus’s mass?

492. The threshold frequency of platinum is 1.36 ×1015 Hz. What is the work function of platinum?

493. The ship Queen Elizabeth II has a mass of 7.6 ×107 kg. Calculate the de Broglie wavelength ifthis ship sails at 35 m/s.

494. Cobalt has a work function of 5.0 eV. What isthe wavelength of the photon that will just havethe threshold energy for cobalt?

495. Light of frequency 9.89 × 1014 Hz illuminates a calcium surface. The ejected photoelectronsare found to have a maximum kinetic energy of 0.90 eV. Find the threshold frequency ofthis metal.

496. What is the speed of a neutron with a de Brogliewavelength of 5.6 × 10−14 m?

Chapter 22 Subatomic Physics497. Calculate the binding energy of 39

19K.

498. Determine the difference in the binding energyof 107

47Ag and 6329Cu.

499. Find the mass defect of 5828Ni.

500. Complete this radioactive-decay formula:21284Po ⎯→ ? + 4

2He.

501. Complete this radioactive-decay formula:16

7N ⎯→ ? + −10e + v��.


62Sm ⎯→ 14360Nd + ? .

503. A 3.29 × 10−3 g sample of a pure radioactivesubstance is found after 30.0 s to have only8.22 × 10−4 g left undecayed. What is the half-life of the substance?

504. The half-life of 4824Cr is 21.6 h. A chromium-48

sample contains 6.5 × 106 nuclei. Calculate theactivity of the sample in mCi.

505. How long will it take a sample of lead-212(which has a half-life of 10.64 h) to decay toone-eighth its original strength?

506. Compute the binding energy of 12050Sn.

507. Calculate the difference in the binding energyof 12

6C and 168O.

508. What is the mass defect of 6430Zn?

509. Complete this radioactive-decay formula:? ⎯→ 131

54Xe + −10e + v��.


74W ⎯→ 15672Hf + ? .

511. Complete this radioactive-decay formula:? ⎯→ 107

52Te + 42He.

512. A 4.14 × 10−4 g sample of a pure radioactivesubstance is found after 1.25 days to have only2.07 × 10−4 g left undecayed. What is the sub-stance’s half-life?

513. How long will it take a sample of cadmium-109with a half-life of 462 days to decay to one-fourthits original strength?

514. The half-life of 5526Fe is 2.7 years. What is the

decay constant for the isotope?

AP

PE

ND

IX JA

dvanced Topics

897

Advanced Topics

CONTENTS

Angular Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 898

Tangential Speed and Acceleration . . . . . . . . . . . . . . . . . 902

Rotation and Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 904

Rotational Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 906

Properties of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 908

Fluid Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 910

The Doppler Effect and the Big Bang . . . . . . . . . . . . . . . 912

Special Relativity and Time Dilation . . . . . . . . . . . . . . . 914

Special Relativity and Velocities . . . . . . . . . . . . . . . . . . . . 916

The Equivalence of Mass and Energy . . . . . . . . . . . . . . . 918

General Relativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 920

De Broglie Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 922

Electron Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924

Semiconductor Doping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926

Superconductors and BCS Theory . . . . . . . . . . . . . . . . . . 928

Antimatter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 930

897Appendix J: Advanced Topics

Teacher ResourcesVisit go.hrw.com for teacherresources related to online exten-sions in this appendix.

Keyword HF6APJXT

898

The Language of PhysicsThere is a relationship betweenrotational motion and circularmotion. When a solid object(such as a Ferris wheel) under-goes rotational motion about itsfixed axis, a point on the rotatingobject (such as a light bulb on aFerris wheel) undergoes circularmotion.

Visual Strategy

Figure 1Point out that the angle throughwhich the light bulb moves isrelated to the distance around thecircle that the light bulb moves(the arc length).

If the light bulb movesthrough an angle twice as

large as the one shown in (b),how would the new arc lengthcompare with the arc lengthshown?

The new arc length would betwice as large.

Visual Strategy

Figure 2Strengthen students’ understand-ing of radian measurement byhaving them use the values givenin Figure 2 to estimate the radianmeasures not shown. The stu-dents can then verify theiranswers by using the conversionequation shown on the next page.

What is the radian measureequal to 75°?

⎯1

5

2⎯pA

Q

A

Q

898 Appendix J: Advanced Topics

Angular Kinematics

A point on an object that rotates about a fixed axis undergoes

circular motion around that axis. The linear quantities intro-

duced in the chapter “Motion in One Dimension” cannot be

used for circular motion because we are considering the rota-

tional motion of an extended object rather than the linear

motion of a particle. For this reason, circular motion is

described in terms of the change in angular position. All

points on a rigid rotating object, except the points on the

axis, move through the same angle during any time interval.

Measuring angles with radians

Many of the equations that describe circular motion require

that angles be measured in (rad) rather than in

degrees. To see how radians are measured, consider Figure 1,which illustrates a light bulb on a rotating Ferris wheel. At

t = 0, the bulb is on a fixed reference line, as shown in

Figure 1(a). After a time interval Δt, the bulb advances to a

new position, as shown in Figure 1(b). In this time interval,

the line from the center to the bulb (depicted with a red line

in both diagrams) moved through the angle q with respect to

the reference line. Likewise, the bulb moved a distance s, mea-

sured along the circumference of the circle; s is the arc length.

In general, any angle q measured in radians is defined by

the following equation:

q = = ⎯r

s⎯⎯

Note that if the arc length, s, is equal to the length of the

radius, r, the angle q swept by r is equal to 1 rad. Because q is

the ratio of an arc length (a distance) to the length of the

radius (also a distance), the units cancel and the abbreviation

rad is substituted in their place. In other words, the radian is a

pure number, with no dimensions.

When the bulb on the Ferris wheel moves through an angle

of 360° (one revolution of the wheel), the arc length s is equal

to the circumference of the circle, or 2pr. Substituting this

value for s into the equation above gives the corresponding

angle in radians.

q = ⎯⎯r

s⎯⎯ = ⎯

2pr

r⎯ = 2p rad

arc length⎯

radius

radians

Appendix J: Advanced Topics

Referenceline

rO

Lightbulb

(a)

Referenceline

r

s

Lightbulb

(b)

θO

Figure 1A light bulb on a rotating Ferriswheel (a) begins at a point along areference line and (b) movesthrough an arc length s and there-fore through the angle q .

x

y

360°

330°

315°300°

90°60°

45°

30°

0°

270°240°

225°210°

180°

150°135°

120°23 2

1

31

41

61

345

6ππ π

ππ

π

ππ

πππ

ππ

π

ππ

76

54 4

3 32

53

74

116

2

Figure 2Angular motion is measured in unitsof radians. Because there are 2pradians in a full circle, radians areoften expressed as a multiple of p.

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Thus, 360° equals 2p rad, or one complete revolution. In other words, one

revolution corresponds to an angle of approximately 2(3.14) = 6.28 rad.

Figure 2 on the previous page depicts a circle marked with both radians

and degrees.

It follows that any angle in degrees can be converted to an angle in radians

by multiplying the angle measured in degrees by 2p/360°. In this way, the

degrees cancel out and the measurement is left in radians. The conversion

relationship can be simplified as follows:

q(rad) = ⎯18

p0°⎯q(deg)

Angular displacement

Just as an angle in radians is the ratio of the arc length to the radius, the

traveled by the bulb on the Ferris wheel is the change

in the arc length, Δs, divided by the distance of the bulb from the axis of rota-

tion. This relationship is depicted in Figure 3.

This equation is similar to the equation for linear displacement in that this

equation denotes a change in position. The difference is that this equation

gives a change in angular position rather than a change in linear position.

For the purposes of this textbook, when a rotating object is viewed from

above, the arc length, s, is considered positive when the point rotates counter-

clockwise and negative when it rotates clockwise. In other words, Δq is posi-

tive when the object rotates counterclockwise and negative when the object

rotates clockwise.

ANGULAR DISPLACEMENT

Δq = ⎯Δr

s⎯

angular displacement (in radians) =change in arc length⎯⎯⎯distance from axis

angular displacement

Angular DisplacementEarth has an equatorial radius ofapproximately 6380 km androtates 360° every 24 h.

a. What is the angular displace-ment (in degrees) of a personstanding at the equator for 1.0 h?

b. Convert this angular displace-ment to radians.

c. What is the arc length traveledby this person?

Answersa. 15°b. 0.26 radc. approximately 1700 km

wire, and lay them along the circle youdrew with your compass. Approximatelyhow many pieces of wire do you use to go all the way around the circle? Draw linesfrom the center of the circle to each end of one of the wires. Note that the anglebetween these two lines equals 1 rad. Howmany of these angles are there in this cir-cle? Repeat the experiment with a largercircle, and compare the results of each trial.

Radians and Arc Length

MATERIALS LIST

•drawing compass

•paper

•thin wire

•wire cutters or scissors

SAFETY

Cut ends of wire are sharp. Cut and han-dle wire carefully.

Use the compass to draw a circle on asheet of paper, and mark the center pointof the circle. Measure the radius of the cir-cle, and cut several pieces of wire equal tothe length of this radius. Bend the pieces of

Homework Options

TEACHER’S NOTESStudents should find the sameresults with both circles. In each case, it takes approximately6 pieces of wire (6r) to go around the circle because cir-cumference = 2pr ≈ 6r.

This QuickLab can easilybe performed outside of thephysics lab room.

Referenceline

θ2θ1

O

Figure 3A light bulb on a rotating Ferriswheel rotates through an angulardisplacement of Δq = q2 − q1.

900 900 Appendix J: Advanced Topics

Angular velocity

is defined in a manner similar to that for linear velocity. The

average angular velocity of a rotating rigid object is the ratio of the angular dis-

placement, Δq, to the corresponding time interval, Δt. Thus, angular velocity

describes how quickly the rotation occurs. Angular velocity is abbreviated as

wavg (w is the Greek letter omega).

Angular velocity is given in units of radians per second (rad/s). Sometimes, angu-

lar velocities are given in revolutions per unit time. Recall that 1 rev = 2p rad. The

magnitude of angular velocity is called angular speed.

Angular acceleration

Figure 4 shows a bicycle turned upside down so that a repairperson can work

on the rear wheel. The bicycle pedals are turned so that at time t1 the wheel has

angular velocity w1, as shown in Figure 4(a). At a later time, t2, it has angular

velocity w2, as shown in Figure 4(b). Because the angular velocity is changing,

there is an The average angular acceleration, aavg (a is

the Greek letter alpha), of an object is given by the relationship shown below.

Angular acceleration has the units radians per second per second (rad/s2).

t2

w2

(b)

t1

w1

(a)

ANGULAR ACCELERATION

aavg = ⎯wt2

2

−−

wt1

1⎯ = ⎯ΔΔwt⎯

average angular acceleration =change in angular velocity⎯⎯⎯

time interval

angular acceleration.

ANGULAR VELOCITY

wavg = ⎯⎯ΔΔ

qt

⎯

average angular velocity =angular displacement⎯⎯⎯

time interval

Angular velocity

Figure 4An accelerating bicycle wheelrotates with (a) an angular velocityw1 at time t1 and (b) an angularvelocity w2 at time t2. Thus, thewheel has an angular acceleration.

Angular VelocityAn Indy car can complete 120laps in 1.5 h. Even though thetrack is an oval rather than a cir-cle, you can still find the averageangular speed. Calculate the aver-age angular speed of the Indy car.

Answer0.14 rad/s

Angular Acceleration A top that is spinning at 15 rev/sspins for 55 s before coming to astop. What is the average angularacceleration of the top while it isslowing?

Answer−1.7 rad/s2


The relationships between the signs of angular displacement, angular velocity,

and angular acceleration are similar to those of the related linear quantities. As

discussed earlier, by convention, angular displacement is positive when an object

rotates counterclockwise and negative when an object rotates clockwise. Thus,

by definition, angular velocity is also positive when an object rotates counter-

clockwise and negative when an object rotates clockwise. Angular acceleration

has the same sign as the angular velocity when it increases the magnitude of the

angular velocity, and the opposite sign when it decreases the magnitude.

If a point on the rim of a bicycle wheel had an angular velocity greater

than a point nearer the center, the shape of the wheel would be changing.

Thus, for a rotating object to remain rigid, as does a bicycle wheel or a Ferris

wheel, every portion of the object must have the same angular velocity and

the same angular acceleration. This fact is precisely what makes angular

velocity and angular acceleration so useful for describing rotational motion.

Kinematic equations for constant angular acceleration

All of the equations for rotational motion defined thus far are analogous to

the linear quantities defined in the chapter “Motion in One Dimension.” For

example, consider the following two equations:

wavg = ⎯qt

f

f

−−

qti

i⎯ = ⎯ΔΔ

qt

⎯ vavg = ⎯x

t

f

f

−−

x

ti

i⎯ = ⎯ΔΔ

x

t⎯

The equations are similar, with q replacing x and w replacing v. The correla-

tions between angular and linear variables are shown in Table 1.In light of the similarities between variables in linear motion and those in

rotational motion, it should be no surprise that the kinematic equations of

rotational motion are similar to the linear kinematic equations. The equations

of rotational kinematics under constant angular acceleration are summarized

in Table 2, along with the corresponding equations for linear motion under

constant acceleration. The rotational motion equations apply only for objects

rotating about a fixed axis with constant angular acceleration.

The quantity w in these equations represents the instantaneous angular velocity

of the rotating object rather than the average angular velocity.

Table 2 Rotational and Linear Kinematic Equations

Rotational motion with con- Linear motion with constantstant angular acceleration acceleration

wf = w i + aΔt vf = vi + aΔt

Δq = w iΔt + ⎯21⎯a(Δt)2 Δx = viΔt + ⎯

21⎯a(Δt)2

wf2 = w i

2 + 2aΔq vf2 = vi

2 + 2aΔx

Δq = ⎯21⎯(w i + wf)Δt Δx = ⎯

21⎯(vi + vf)Δt

Table 1Angular Substitutesfor Linear Quantities

Angular Linear

q x

w v

a a

Practice ProblemsVisit go.hrw.com to find sample andpractice problems for angular dis-placement, angular velocity, angularacceleration, and angular kinematics.

Keyword HF6APJX

Angular KinematicsA barrel is given a downhillrolling start of 1.5 rad/s at the topof a hill. Assume a constantangular acceleration of 2.9 rad/s2.

a. If the barrel takes 11.5 s to getto the bottom of the hill, whatis the final angular speed ofthe barrel?

b. What angular displacementdoes the barrel experienceduring the 11.5 s ride?

Answersa. 35 rad/sb. 2.1 × 102 rad

Demonstration

Equal Angular SpeedPurpose Illustrate that angularspeed is constant at any radiusfor a rigid extended object.Materials record player/turntableor bicycle; tape; colored markersProcedure Use the tape andmarkers to make two brightly col-ored flags, and attach them to theturntable so that one flag is nearthe rim and the other is near, butnot at, the center. (Alternatively, ifa turntable is not available, youcan use a bicycle wheel.) Start theturntable at a moderately slow speed (33 ⎯1

3⎯ rpm) so that the flags

are easily observed. Have studentsnote the rotational speed of bothflags. Point out that each flagmakes a complete rotation in thesame amount of time. Changespeeds on the turntable, andrepeat the observations.

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Visual Strategy

Figure 1

Which horse would travel far-ther before hitting the ground

if the horses were released fromthe carousel?

The outer horse would travelfarther because it has a

greater tangential speed.A

Q


Tangential Speed and Acceleration

The chapter on circular motion introduced the concepts of tangential speed and

acceleration. This feature explores these concepts in greater detail. Before read-

ing further, be sure you have read the appendix feature “Angular Kinematics.”

Tangential speed

Imagine an amusement-park carousel rotating about its center. Because a

carousel is a rigid object, any two horses attached to the carousel have the

same angular speed and angular acceleration. However, if the two horses are

different distances from the axis of rotation, they have different

The tangential speed of a horse on the carousel is its speed along a

line drawn tangent to its circular path.

The tangential speeds of two horses at different distances from the center

of a carousel are represented in Figure 1. Note that the two horses travel the

same angular displacement during the same time interval. To achieve this, the

horse on the outside must travel a greater distance (Δs) than the horse on the

inside. Thus, the outside horse at point B has a greater tangential speed than

the inside horse at point A. In general, an object that is farther from the axis

of a rigid rotating body must travel at a higher tangential speed to cover the

same angular displacement as would an object closer to the axis.

If the carousel rotates through an angle q, a horse rotates through an arc

length Δs in the interval Δt. To find the tangential speed, start with the equa-

tion for angular displacement:

Δq = ⎯Δr

s⎯

Next, divide both sides of the equation by the time it takes to travel Δs :

⎯ΔΔ

qt

⎯ = ⎯r

ΔΔs

t⎯

As discussed in the appendix feature “Angular Kinematics,” the left side of the

equation equals wavg. Also, Δs is a linear distance, so Δs divided by Δt is a linear

speed along an arc length. If Δt is very short, then Δs is so small that it is nearly

tangent to the circle; therefore, Δs/Δt is the tangential speed, vt.

In the tangential speed equation, w is the instantaneous angular speed,

rather than the average angular speed, because the time interval is so short.

TANGENTIAL SPEED

vt = rw

tangential speed = distance from axis × angular speed

speeds.tangential

ω

vt,inside

A

Bvt,outside

Figure 1Horses on a carousel move at thesame angular speed but differenttangential speeds.

Demonstration

Tangential Speed VersusAngular SpeedPurpose Show that tangentialspeed depends on radius.Materials two tennis ballsattached to different lengths ofstring (approx. 1.0 m and 1.5 m)Procedure Outside on an athleticfield, hold the ends of bothstrings and whirl the tennis ballsat constant angular speed overyour head. Point out the equalangular speeds of the tennis balls.Ask students to predict the flightsof the tennis balls when thestrings are released.

Aiming away from studentsand any breakable items, releasethe strings and have studentsobserve the flights. Discuss thelonger horizontal displacementof the outer ball as a function ofits greater tangential speed.


This equation is valid only when w is measured in radians per unit of time.

Other measures of angular speed must not be used in this equation.

Tangential acceleration

If a carousel speeds up, the horses on it experience an angular acceleration.

The linear acceleration related to this angular acceleration is tangent to the

circular path and is called the If an object rotating

about a fixed axis changes its angular speed by Δw in the interval Δt, the tan-

gential speed of a point on the object has changed by the amount Δvt . Using

the equation for tangential speed and dividing by Δt results in the following:

Δvt = rΔw

⎯ΔΔv

tt⎯ = r ⎯

ΔΔwt⎯

If the time interval Δt is very small, then the left side of this relationship gives

the tangential acceleration of the point. The angular speed divided by the time

interval on the right side is the angular acceleration. Thus, the tangential acceler-

ation (at) of a point on a rotating object is given by the following relationship:

The angular acceleration in this equation refers to the instantaneous angu-

lar acceleration. This equation must use the unit radians to be valid. In SI,

angular acceleration is expressed as radians per second per second.

Finding total acceleration

Recall that any object moving in a circle has a centripetal acceleration, as dis-

cussed in the chapter on circular motion. When both components of acceler-

ation exist simultaneously, the tangential acceleration is tangent to the

circular path and the centripetal acceleration points toward the center of the

circular path. Because these components of acceleration are perpendicular to

each other, the magnitude of the total acceleration can be found using the

Pythagorean theorem, as follows:

atotal = �at�2�+� a�c2�

The direction of the total acceleration, as shown in Figure 2, depends on

the magnitude of each component of acceleration and can be found using the

inverse of the tangent function. Note that when there is a tangential accelera-

tion, the tangential speed is changing, and thus this situation is not an exam-

ple of uniform circular motion.

TANGENTIAL ACCELERATION

at = ra

tangential acceleration = distance from axis × angular acceleration

tangential acceleration.

at

ac

atotalθ

Figure 2The direction of the total accelera-tion of a rotating object can befound using the inverse tangentfunction.

Tangential SpeedA golfer has a maximum angularspeed of 6.3 rad/s for her swing.She can choose between two dri-vers, one placing the club head1.9 m from her axis of rotationand the other placing it 1.7 mfrom the axis.

a. Find the tangential speed ofthe club head for each driver.

b. All other factors being equal,which driver is likely to hit theball farther?

Answersa. 12 m/s, 11 m/sb. The longer driver will hit

the ball farther because itsclub head has a higher tan-gential speed.

Tangential AccelerationA yo-yo has a tangential accelera-tion of 0.98 m/s2 when it isreleased. The string is woundaround a central shaft of radius0.35 cm. What is the angularacceleration of the yo-yo?

Answer2.8 × 102 rad/s2

Teaching TipTo simulate the large accelerationsinvolved in spaceflight, Mercuryastronauts rode in the U. S.Navy’s centrifuge in Johnsville,Pennsylvania. The astronauts sat in a gondola at the end of a15.2 m arm that spun around acentral axis. During the spin, theastronauts experienced a combi-nation of centripetal and tangen-tial accelerations of the gondolathat ranged from 8 to 10 times theacceleration due to gravity.

Practice ProblemsVisit go.hrw.com to find sample andpractice problems for tangentialspeed and tangential acceleration.

Keyword HF6APJX

In Section 4 of the chapter “Circular Motion and Gravitation,” you learned

that torque measures the ability of a force to rotate an object around some

axis, such as a cat-flap door rotating on a hinge. Locating the axis of rotation

for a cat-flap door is simple: it rotates on its hinges because the house applies

a force that keeps the hinges in place. Now imagine you are playing fetch with

your dog, and you throw a stick up into the air for the dog to retrieve. How

can you determine the point around which the stick will rotate as it travels

through the air?

Center of mass

Unlike the cat-flap door, the stick is not attached to anything.

There is a special point around which the stick rotates if gravity

is the only force acting on the stick. This point is called the stick’s

The center of mass is also the point at which all the mass of the

body can be considered to be concentrated (for translational motion).

This means that the complete motion of the stick is a combination of

both translational and rotational motion. The stick rotates in the air

around its center of mass. The center of mass, in turn, moves as if the

stick were a point mass, with all of its mass concentrated at that point

for purposes of analyzing its transla-

tional motion. For example, the

hammer in Figure 1 rotates about its

center of mass as it moves through

the air. As the rest of the hammer

spins, the center of mass moves along

the parabolic path of a projectile.

For regularly shaped objects, such

as a sphere or a cube, the center of

mass is at the geometric center of the

object. For more complicated objects,

finding the center of mass is more

difficult. Although the center of mass

is the position at which an extended

object’s mass can be treated as a point

mass, the center of gravity is the posi-

tion at which the gravitational force

acts on the extended object as if it

were a point mass. For many situa-

tions, the center of mass and the cen-

ter of gravity are equivalent.

center of mass.

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MisconceptionAlert

Point out that objects rotatearound their center of mass in theabsence of other forces. For exam-ple, a ruler thrown through theair will rotate around its own cen-ter of mass because air resistanceis evenly distributed and produceszero net torque. However, a rulerwith an index card taped to oneend will not rotate around itsown center of mass when it isthrown through the air becausethe card provides air resistance,which in turn produces a torque.

Teaching TipLet students examine objects thathave a center of mass outside theobject itself, such as a doughnut,a coat hanger, or a boomerang.

STOP


Rotation and Inertia

Figure 1The point around which this ham-mer rotates is the hammer’s centerof mass. The center of mass tracesout the parabola that is characteris-tic of projectile motion.

www.scilinks.orgTopic: Center of MassCode: HF60245


Moment of inertia

You may have noticed that it is easier to rotate a baseball bat around some

axes than others. The resistance of an object to changes in rotational motion

is measured by a quantity called the

The moment of inertia, which is abbreviated as I, is similar to mass because

they are both forms of inertia. However, there is an important difference

between them. Mass is an intrinsic property of an object, and the moment of

inertia is not. The moment of inertia depends on the object’s mass and the

distribution of that mass around the axis of rotation. The farther the mass of

an object is, on average, from the axis of rotation, the greater is the object’s

moment of inertia and the more difficult it is to rotate the object.

According to Newton’s second law, when a net force acts on an object, the

resulting acceleration of the object depends on the object’s mass. Similarly,

when a net torque acts on an object, the resulting change in the rotational

motion of the object depends on the object’s moment of inertia. (This law is

covered in more detail in the appendix feature “Rotational Dynamics.”)

Some simple formulas for calculating the moment of inertia of common

shapes are shown in Table 1. The units for moment of inertia are kg•m2. To

get an idea of the size of this unit, note that bowling balls typically have

moments of inertia about an axis through their centers ranging from about

0.7 kg•m2 to 1.8 kg•m2, depending on the mass and size of the ball.

moment of inertia.

A baseball bat can be modeled as arotating thin rod.When a bat is heldat its end, its length is greatest withrespect to the rotation axis, so itsmoment of inertia is greatest.Thus,the bat is easier to swing if you holdthe bat closer to the center. Baseballplayers sometimes do this eitherbecause a bat is too heavy (large M)or too long (large �).

Did you know?

Shape Moment of inertia Shape Moment of inertia

disk or cylinder aboutsymmetry axis ⎯

21⎯MR2R

point mass about axis MR2R

thin hoop aboutdiameter ⎯

21⎯MR2R

thin hoop aboutsymmetry axis MR2R

thin spherical shellabout diameter ⎯2

3⎯MR2R

solid sphere ⎯25

⎯MR2

about diameterR

thin rod about perpendicular axis ⎯

31⎯Ml 2

through end

l

thin rod about perpendicular axis ⎯

112⎯ Ml 2

through center

l

Table 1 The Moment of Inertia for Various Rigid Objects of Mass M

Demonstration

Moment of Inertia of a RodPurpose Give visual examples ofthe two cases of a thin rod andthe case of a cylinder described in Table 1.Materials broomstick or dowelProcedure If possible, let studentvolunteers assist by trying thevarious demonstrations.Thin rod about center: Hold therod in the center with one hand.Rotate the rod back and forththrough half rotations at regulartime intervals. Note the forcerequired to change the motion.Thin rod about end: Hold rod atone end. Rotate the rod throughhalf circles in the same timeinterval as previously used. Notethe increased force required (cor-responding to the larger momentof inertia).Cylinder: Hold the rod verticallybetween your palms with yourfingers extended. Rotate thecylinder by moving your palmsback and forth in the same regu-lar time interval as used previ-ously. Note the much smallerforce required in this case.

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MisconceptionAlert

Students will likely confuse zeronet torque with zero rotation.State explicitly that an objectrotating at constant speed isexperiencing zero net torque.This situation is analogous to lin-ear motion: an object moving ata constant velocity has zero netforce acting on it.

The Language of PhysicsAnother way to state the secondequilibrium condition is to saythat the sum of the clockwisetorques must equal the sum ofthe counterclockwise torques.

Teaching TipStudents may be confused aboutwhich axis of rotation to choosewhen applying the second condi-tion for equilibrium to an object.Tell students that any axis can beused; the resultant torque actingon an object in rotational equi-librium is independent of wherethe axis is placed. This fact is use-ful in solving rotational equilibri-um problems because anunknown force that acts along aline passing through this axis ofrotation will not produce anytorque. Beginning a diagram byarbitrarily setting an axis where aforce acts can eliminate anunknown in the problem.

STOP


Rotational Dynamics

The appendix feature “Angular Kinematics” developed the kinematic equations

for rotational motion. Similarly, the appendix feature “Rotation and Inertia”

applied the concept of inertia to rotational motion. In this feature, you will see

how torque relates to rotational equilibrium and angular acceleration. You will

also learn how momentum and kinetic energy are described in rotational motion.

Rotational equilibrium

If you and a friend push on opposite sides of a table, as

shown in Figure 1, the two forces acting on the table are

equal in magnitude and opposite in direction. You

might think that the table won’t move because the two

forces balance each other. But it does; it rotates in place.

The piece of furniture can move even though the net

force acting on it is zero because the net torque acting

on it is not zero. If the net force on an object is zero, the

object is in translational equilibrium. If the net torque

on an object is zero, the object is in rotational equilibri-

um. For an object to be completely in equilibrium, both

rotational and translational, there must be both zero net

force and zero net torque. The dependence of equilibri-

um on the absence of net torque is called the second

condition for equilibrium.

Newton’s second law for rotation

Just as net force is related to translational acceleration according to Newton’s

second law, there is a relationship between the net torque on an object and the

angular acceleration given to the object. Specifically, the net torque on an object

is equal to the moment of inertia times the angular acceleration. This relation-

ship is parallel to Newton’s second law of motion and is known as Newton's sec-

ond law for rotating objects. This law is expressed mathematically as follows:

This equation shows that a net positive torque corresponds to a positive angular

acceleration, and a net negative torque corresponds to a negative angular accel-

eration. Thus, it is important to keep track of the signs of the torques acting on

the object when using this equation to calculate an object’s angular acceleration.

NEWTON’S SECOND LAW FOR ROTATING OBJECTS

tnet = Ia

net torque � moment of inertia � angular acceleration

Figure 1The two forces exerted onthis table are equal andopposite, yet the tablemoves. How is this possible?


Angular momentum

Because a rotating object has inertia, it also possesses momentum associated

with its rotation. This momentum is called Angular

momentum is defined by the following equation:

The unit of angular momentum is kg•m2/s. When the net external torque

acting on an object or objects is zero, the angular momentum of the object(s)

does not change. This is the law of conservation of angular momentum. For

example, assuming the friction between the skates and the ice is negligible,

there is no torque acting on the skater in Figure 2, so his angular momentum

is conserved. When he brings his hands and feet closer to his body, more of

his mass, on average, is nearer to his axis of rotation. As a result, the moment

of inertia of his body decreases. Because his angular momentum is constant,

his angular speed increases to compensate for his smaller moment of inertia.

Angular kinetic energy

Rotating objects possess kinetic energy associated with their angular speed. This

is called and is expressed by the following equation:

As shown in Table 1, rotational kinetic energy is analogous to the transla-

tional kinetic energy of a particle, given by the expression ⎯12

⎯mv2. The unit of

rotational kinetic energy is the joule, the SI unit for energy.

ROTATIONAL KINETIC ENERGY

KErot = ⎯12

⎯Iw2

rotational kinetic energy = ⎯12

⎯ × moment of inertia × (angular speed)2

rotational kinetic energy

ANGULAR MOMENTUM

L = Iw

angular momentum = moment of inertia × angular speed

angular momentum.

Figure 2When this skater brings his handsand feet closer to his body, hismoment of inertia decreases, andhis angular speed increases to keeptotal angular momentum constant.

Table 1 Comparing Translational and Rotational Motion

Translational motion Rotational motion

Equilibrium ∑F = 0 ∑t = 0

Newton’s second law ∑F = ma ∑t = Ia

Momentum p = mv L = Iw

Kinetic Energy KE = ⎯12

⎯mv2 KE = ⎯12

⎯Iw2

Practice ProblemsVisit go.hrw.com to find sample andpractice problems for rotationalequilibrium, Newton’s second lawfor rotation, conservation of angularmomentum, and angular kinetic energy.

Keyword HF6APJX

Demonstration

Colliding Magnetic MarblesPurpose This demonstrationillustrates conservation of angu-lar momentum and shows thatobjects moving linearly can alsohave angular momentum.Materials 2 magnetic marbles,overhead projector Procedure Place one magneticmarble in the center of the over-head projector. Send the secondmagnetic marble toward the firstto cause a glancing collision. Themarbles will stick together andspin about their center of mass.Point out to students that the lin-ear momentum is not convertedto angular momentum. Repeatthe demonstration and have stu-dents observe the slight transla-tion of the two marbles after thecollision. This is the linearmomentum, which is lost due tofriction. The spinning is due toconservation of the angularmomentum of the moving marblerelative to the stationary marble.

When the density of a gas is sufficiently low, the pressure, volume, and

temperature of the gas tend to be related to one another in a fairly

simple way. This relationship is a good approximation for the behav-

ior of many real gases over a wide range of temperatures and pres-

sures, provided their particles are not charged, as in a plasma. These

observations have led scientists to develop the concept of an ideal gas.

Volume, pressure, and temperature are the three variables that

completely describe the macroscopic state of an ideal gas. One of the

most important equations in fluid mechanics relates these three

quantities to each other.

The ideal gas law

The ideal gas law is an expression that relates the volume, pressure, and tem-

perature of a gas. This relationship can be written as follows:

The symbol kB represents Boltzmann’s constant. Its value has been experi-

mentally determined to be approximately 1.38 × 10−23 J/K. Note that when

applying the ideal gas law, you must express the temperature in the Kelvin

scale. (See the chapter “Heat” to learn about the Kelvin scale.) Also, the ideal

gas law makes no mention of the composition of the gas. The gas particles

could be oxygen, carbon dioxide, or any other gas. In this sense, the ideal gas

law is universally applicable to all gases.

If a gas undergoes a change in volume, pressure, or temperature (or any

combination of these), the ideal gas law can be expressed in a particularly

useful form. If the number of particles in the gas is constant, the initial and

final states of the gas are related as follows:

N1 = N2

⎯P

T1V

1

1⎯ = ⎯P

T2V

2

2⎯

This relation is illustrated in the experiment shown in Figure 1. In this

experiment, a flask filled with air (V1 equals the volume of the flask) at room

temperature (T1) and atmospheric pressure (P1 = P0) is placed over a heat

source, with a balloon placed over the opening of the flask. As the flask sits over

the burner, the temperature of the air inside it increases from T1 to T2.

IDEAL GAS LAW

PV = NkBT

pressure × volume =number of gas particles × Boltzmann’s constant × temperature

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Properties of Gases

Figure 1The balloon is inflated because thevolume and pressure of the airinside are both increasing.

Demonstration

Temperature, Pressure,and VolumePurpose Demonstrate that the same amount of gas occupiesa larger volume at a higher temperature.Materials heat-resistant flask,balloon, Bunsen burner, gogglesCAUTION Be sure to wear safetygoggles at all times when operat-ing the Bunsen burner, and usetongs when handling the heatedglassware.Procedure Cap the flask with theballoon. Emphasize that no gascan flow in or out of the flask orthe balloon. Ask students what theair volume is in this closed system(about equal to that of the flask).Heat the flask until the balloonexpands. Have students estimateand record the balloon’s approxi-mate size. Ask them how much airis in the flask now (same volume,but fewer molecules) and in theballoon and flask together (sameamount as before, but in a largervolume). Place the balloon undercold water (not the flask, to avoidbreakage) and watch it return toits original size.

Teaching TipFor an alternative demonstration,place a helium-filled balloon in a refrigerator. After 15 minutes,remove the balloon and have stu-dents observe how it behaves as itadjusts to room temperature.

www.scilinks.orgTopic: Gas LawsCode: HF60637


A third way of writing the ideal gaslaw may be familiar to you fromyour study of chemistry:

PV = nRT

In this equation, n is the number ofmoles of gas (one mole is equal to6.02 × 1023 particles). The quantityR is a number called the molar (uni-versal) gas constant and has a valueof 8.3 1 J/(mol•K).

Did you know?

Practice ProblemsVisit go.hrw.com to find sample andpractice problems for the ideal gaslaw.

Keyword HF6APJX

According to the ideal gas law, when the temperature increases, either the pres-

sure or the volume—or both—must also increase. Thus, the air inside the flask

exerts a pressure (P2) on the balloon that serves to inflate the balloon. Because

the balloon is expandable, the air expands to a larger volume (V2) to fill the

balloon. When the flask is taken off the burner, the pressure, volume, and tem-

perature of the air inside will slowly return to their initial states.

Another alternative form of the ideal gas law indicates the law’s dependence

on mass density. Assuming each particle in the gas has a mass m, the total

mass of the gas is N × m = M. The ideal gas law can then be written as follows:

PV = NkBT = ⎯M

m

kBT⎯

P = ⎯M

m

k

VBT⎯ = �⎯

M

V⎯� ⎯

k

mBT⎯ = ⎯

rk

mBT⎯

A real gas

An ideal gas is defined as a gas whose behavior is accurately described by the

ideal gas law. Although no real gas obeys the ideal gas law exactly for all tem-

peratures and pressures, the ideal gas law holds for a broad range of physical

conditions for all gases. The behavior of real gases departs from the behavior

of an ideal gas at high pressures or low temperatures, conditions under which

the gas nearly liquefies. However, when a real gas has a relatively high temper-

ature and a relatively low pressure, such as at room temperature and atmo-

spheric pressure, its behavior approximates that of an ideal gas.

For problems involving the motion of fluids, we have assumed that all gases

and liquids are ideal fluids. An ideal fluid is a liquid or gas that is assumed to be

incompressible. This is usually a good assumption because it is difficult to com-

press a fluid—even a gas—when it is not confined to a container. A fluid will

tend to flow under the action of a force, changing its shape while maintaining a

constant volume, rather than compress.

This feature, however, considers confined gases whose pressure, volume,

and temperature may change. For example, when a force is applied to a piston,

the gas inside the cylinder below the piston is compressed. Even though an

ideal gas behaves like an ideal fluid in many situations, it cannot be treated as

incompressible when confined to a container.

Make sure the bottle is empty, andremove the cap. Place the bottle in thefreezer for at least 10 min. Wet the quarterwith water, and place the quarter over thebottle’s opening as you take the bottle outof the freezer. Set the bottle on a nearbytabletop; then observe the bottle and quar-ter while the air in the bottle warms up. As

the air inside the bottle begins to return toroom temperature, the quarter begins tojiggle around on top of the bottle. Whatdoes this movement tell you about thepressure inside the bottle? What causesthis change in pressure? Hypothesize as towhy you need to wet the quarter beforeplacing it on top of the bottle.

Ideal Gas Law

MATERIALS LIST

•1 plastic 1 L bottle

•1 quarter

Key Models andAnalogiesGraphs offer a convenient way torepresent the relationshipbetween temperature, volume,and pressure for the followingspecial cases of the ideal gas law:

Teaching TipIf students are not familiar withthe Kelvin temperature scale,have them read Section 1 of thechapter “Heat” before attemptingthe online practice problems thatcover the ideal gas law.

P

Constant P

Constant T

Constant V

P

V

T

T

V

TEACHER’S NOTESThis activity is meant to demon-strate that pressure increases withtemperature when the volume ofa gas is constant. Eventually, thepressure becomes great enoughto overcome the weight of thequarter, and it jumps up slightly,allowing air to escape.

This QuickLab can easilybe performed outside of thephysics lab room.

Homework Options

909

Mercury

Empty

You learned about fluid pressure and Bernoulli’s principle in the chapter “Fluid

Mechanics.” This feature discusses some additional topics related to fluid pres-

sure, including atmospheric pressure and the kinetic theory of gases. It also cov-

ers Bernoulli’s equation, which is a more general form of Bernoulli’s principle.

Atmospheric pressure

The weight of the air in the upper portion of Earth’s atmosphere

exerts pressure on the layers of air below. This pressure is called

atmospheric pressure. The force that atmospheric pressure exerts on

our bodies is extremely large. (Assuming a body area of 2 m2, this

force is on the order of 200 000 N, or 40 000 lb.) How can we exist

under such tremendous forces without our bodies collapsing? The

answer is that our body cavities and tissues are permeated with flu-

ids and gases that are pushing outward with a pressure equal to that

of the atmosphere. Consequently, our bodies are in equilibrium—

the force of the atmosphere pushing in equals the internal force

pushing out.

An instrument that is commonly used to measure atmospheric

pressure is the mercury barometer. Figure 1 shows a very simple

mercury barometer. A long tube that is open at one end and closed

at the other is filled with mercury and then inverted into a dish of

mercury. Once the tube is inverted, the mercury does not empty into

the bowl. Instead, the atmosphere exerts a pressure on the mercury

in the bowl. This atmospheric pressure pushes the mercury in the

tube to some height above the bowl. In this way, the force exerted on

the bowl of mercury by the atmosphere is equal to the weight of the

column of mercury in the tube. Any change in the height of the col-

umn of mercury means that the atmosphere’s pressure has changed.

Kinetic theory of gases

Many models of a gas have been developed over the years. Almost all of these

models attempt to explain the macroscopic properties of a gas, such as pres-

sure, in terms of events occurring in the gas on a microscopic scale. The most

successful model by far is the kinetic theory of gases.

In kinetic theory, gas particles are likened to a collection of billiard balls that

constantly collide with one another. This simple model is successful in explain-

ing many of the macroscopic properties of a gas. For instance, as these particles

strike a wall of a container, they transfer some of their momentum during the

collision. The rate of transfer of momentum to the container wall is equal to

the force exerted by the gas on the container wall, in accordance with the

impulse-momentum theorem. This force per unit area is the gas pressure.

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Fluid Pressure

Figure 1The height of the mercury in thetube of a barometer indicates theatmospheric pressure. (This illus-tration is not drawn to scale.)

Demonstration

Hydrostatic PressurePurpose Demonstrate that pres-sure increases with depth.Materials 32 oz plastic soda bot-tle, tape, water, bucket; optional:plastic straw or small pieces ofglass or metal tubing and model-ing clay or silicone puttyProcedure Drill three holes atdifferent heights along the side of the bottle, each about 10 cmabove the other with the lowesthole near the bottom of the bottle.Stagger the holes horizontallyabout 1 cm apart so that streamsof water flowing through them donot collide. Insert short segmentsof tubing tightly into the holes toimprove the flow. Cover the holeswith tape, fill the bottle withwater, and place it at the edge of atable. Place it high enough abovethe bucket so that the effects ofdepth on the range of each streamwill be easily observed. Ask stu-dents to predict how the waterstreams will compare. Quicklypull the tape away from all threeholes, and have students observethe shape of each stream. Watershooting out from near the bot-tom of the bottle exits the bottle ata higher speed than water shoot-ing out from near the top does.The reason is that the pressure inthe bottle increases with depth.

www.scilinks.orgTopic: Atmospheric PressureCode: HF60114


Bernoulli’s EquationA camper creates a shower byattaching a tube to the bottom of ahanging bucket that is open to theatmosphere on top. If the waterlevel in the bucket is 3.15 m abovethe end of the tube (the showerhead), then what is the speed ofthe water exiting the tube?

Answer7.86 m/s

A horizontal pipe narrows from a cross section of 2.0 m2 to0.30 m2. If the speed of the waterflowing through the wider area ofthe pipe is 8.0 m/s, what is thespeed of the water flowingthrough the narrow part?

Answer53 m/s

Assuming incompressible flow,what is the change in pressure asthe pipe narrows?

Answer1.4 × 106 Pa

Bernoulli’s equation

Imagine a fluid moving through a pipe of varying cross-sectional

area and elevation, as shown in Figure 2. When the cross-sectional

area changes, the pressure and speed of the fluid can change. This

change in kinetic energy may be compensated for by a change in

gravitational potential energy or by a change in pressure (so ener-

gy is still conserved). The expression for the conservation of ener-

gy in fluids is called Bernoulli’s equation. Bernoulli’s equation is

expressed mathematically as follows:

Bernoulli’s equation differs slightly from the law of conservation of energy.

For example, two of the terms on the left side of the equation look like the

terms for kinetic energy and gravitational potential energy, but they contain

density, r, instead of mass, m. The reason is that the conserved quantity in

Bernoulli’s equation is energy per unit volume, not just energy. This statement

of the conservation of energy in fluids also includes an additional term: pres-

sure, P. If you wish to compare the energy in a given volume of fluid at two

different points, Bernoulli’s equation takes the following equivalent form:

P1 + ⎯12

⎯ rv12 + rgh1 = P2 + ⎯1

2⎯ rv2

2 + rgh2

Comparing Bernoulli’s principle and Bernoulli’s equation

Two special cases of Bernoulli’s equation are worth mentioning here. First, if

the fluid is at rest, then both speeds are zero. This case is a static situation, such

as a column of water in a cylinder. If the height at the top of the column, h1, is

defined as zero and h2 is the depth, then Bernoulli’s equation reduces to the

equation for pressure as a function of depth, introduced in the chapter on fluids:

P1 = P2 + rgh2 (static f luid)

Second, imagine again a fluid flowing through a horizontal pipe with a con-

striction. Because the height of the fluid is constant, the gravitational potential

energy does not change. Bernoulli’s equation then reduces to the following:

P1 + ⎯12

⎯ rv12 = P2 + ⎯1

2⎯ rv2

2 (horizontal pipe)

This equation suggests that if v1 is greater than v2 at two different points in

the flow, then P1 must be less than P2. In other words, the pressure decreases as

speed increases—Bernoulli’s principle. Thus, Bernoulli’s principle is a special

case of Bernoulli’s equation and is strictly true only when elevation is constant.

BERNOULLI’S EQUATION

P + ⎯12

⎯ rv2 + rgh = constant

pressure + kinetic energy per unit volume +gravitational potential energy per unit volume =

constant along a given streamline

P2A2

P1A1

y1

y2

v1

v2

Δx1

Δx2

Figure 2As a fluid flows through this pipe, itmay change velocity, pressure, andelevation.

Practice ProblemsVisit go.hrw.com to find a sampleand practice problems forBernoulli’s equation.

Keyword HF6APJX

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Teaching TipThis feature discusses theDoppler effect with light waves.It explores how the idea of anexpanding universe originatedfrom observed red shifts and howthis in turn suggests that at somepoint in the past the universe wasinfinitely small and dense. Thecurrent scientific model of theevolution of matter from themoment of the big bang to thepresent day is discussed at theend of the chapter “SubatomicPhysics.”


The Doppler Effect and the Big Bang

In the chapter “Sound,” you learned that relative motion between the source

of sound waves and an observer creates a frequency shift known as the

Doppler effect. For visible light, the Doppler effect is observed as a change in

color because the frequency of light waves determines color.

Frequency shifts

Of the colors of the visible spectrum, red light has the lowest frequency and

violet light has the highest. When a source of light waves is moving toward

an observer, the frequency detected is higher than the source frequency. This

corresponds to a shift toward the blue end of the spectrum, which is called a

blue shift. When a source of light waves is moving away from an observer, the

observer detects a lower frequency, which corresponds to a shift toward the

red end of the spectrum, called a red shift. Visible light is one form of electro-

magnetic radiation. Blue shift and red shift can occur with any type of elec-

tromagnetic radiation, not just visible light. Table 1 illustrates blue shift and

red shift.

In astronomy, the light from distant stars or galaxies is analyzed by a

process called spectroscopy. In this process, starlight is passed through a prism

or diffraction grating to produce a spectrum. Dark lines appear in the spec-

trum at specific frequencies determined by the elements present in the

atmospheres of stars. When these lines are shifted toward the blue end of the

spectrum, astronomers know the star is moving toward Earth; when the lines

are shifted toward the red end, the star is moving away from Earth.

The expansion of the universe

As scientists began to study other galaxies with spectroscopy, the results were

astonishing: nearly all of the galaxies that were observed exhibited a red shift,

Table 1 The Doppler Effect for Light

no shift

blue shift

red shift

stationarysource

v = 0

approachingsource

recedingsource

v

v

Key Models andAnalogiesAsk students to visualize a bal-loon with dots placed randomlyon its surface. From the perspec-tive of any single dot, as the bal-loon is inflated, all other dotsappear to be moving away. This isanalogous to the expansion of theuniverse; the fact that we observeall galaxies as moving away fromEarth does not mean that Earth isthe center of the expansion. Thesame phenomenon would beobserved from any other point inthe universe.

EXTENSION• Have students research vari-

ous theories regarding the ageof the universe. Their reportsshould include a discussion ofthe following: How is theHubble constant related to theage of the universe? Whatmethods are used to measurethe distances to stars? How docurrent theories on the age ofthe universe differ?

• Have students research AlanGuth’s inflation theory, a mod-ification of the standard bigbang theory. What difficultieswith the big bang theory doesthe inflation theory resolve?

• Spectroscopy is widely used inastronomy for a variety ofpurposes. Have students studyits uses, along with other prac-tical applications of theDoppler effect.

• Have students explore variouspossibilities for the future ofthe universe. Their researchshould include an investiga-tion into the question of howmuch dark matter the universecontains and how the answerto this question will affect thefate of the universe.


which suggested that they were moving away from Earth. If all galaxies are

moving away from Earth, the universe must be expanding. This does not

imply that Earth is at the center of the expansion; the same phenomenon

would be observed from any other point in the universe.

The expansion of the universe suggests that at some point in the past

the universe must have had infinite density. The eruption of the universe is

often referred to as the big bang, which is generally considered to have

occurred between about 13 billion and 15 billion years ago. Current

models indicate that the big bang involved such great amounts of energy in

such a small space that matter could not form clumps or even individual

atoms. It took about 380,000 years for the universe to cool from around

1032 K to around 3000 K, a temperature cool enough for atoms to begin

forming.

Experimental verification

In the 1960s, a group of scientists at Princeton predicted that the explosion

of the big bang was so momentous that a small amount of radiation—the

leftover glow from the big bang—should still be found in the universe.

Around this time, Arno Penzias and Robert Wilson of Bell Labs

noticed a faint background hiss interfering with satellite-

communications experiments they were conducting. This

signal, which was detected in equal amounts in all direc-

tions, remained despite all attempts to remove it. Penzias

and Wilson learned of the Princeton group’s work and

realized that the interference they were experiencing

matched the characteristics of the radiation expected from

the big bang. Subsequent experiments have confirmed the

existence of this radiation, known as cosmic microwave back-

ground radiation. This background radiation is considered to

be the most conclusive evidence for the big bang theory.

The big bang theory is generally accepted by scientists today.

Research now focuses on more detailed issues. However, there are certain

phenomena that the standard big bang model cannot account for, such as the

uniform distribution of matter on a large scale and the large-scale clustering of

galaxies. As a result, some scientists are currently working on modifications and

refinements to the standard big bang theory.

In March of 2004, astronomers released a new image from the Hubble Space

Telescope. This image, called the Hubble Ultra Deep Field (HUDF), looks further

back in time than any previously-recorded images. The image contains an esti-

mated 10,000 galaxies. Scientists will study the HUDF to search for galaxies that

existed from 400 million to 800 million years after the big bang. Because galaxies

evolved quickly, many important changes happened within a billion years of the

big bang. Scientists hope that studies of the HUDF image will resolve some of

the current questions regarding the origin and evolution of the universe.

Figure 1Penzias and Wilson detectedmicrowave background radiation,presumably left over from the bigbang, with the horn antenna (in thebackground) at Bell TelephoneLaboratories in New Jersey.

Figure 2This image, called the Hubble UltraDeep Field, is a compilation ofimages taken by two cameras onthe Hubble Space Telescope betweenSeptember 2003 and January 2004.It shows the youngest galaxies everto be seen.These galaxies may haveformed as early as 400 million yearsafter the big bang.

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Teaching TipThis feature discusses howEinstein’s special theory of rela-tivity modifies the notion oftime. Although the concept oftime dilation is challenging, thebasic aspects can be understoodconceptually at this level, and theequation for time dilation can beused to see how the theoriesmerge at small speeds but divergeat speeds that are large relative tothe speed of light.

Before Einstein published thespecial theory of relativity in1905, physicists were strugglingwith an apparent contradiction inphysics. The principle of relati-vity, which states that the laws ofphysics are the same in any iner-tial reference frame, did not allowfor the fact that the speed of lightis independent of the referenceframe from which it is measured.(This contradiction is not imme-diately obvious to students; a classdiscussion of the subject will helpto bring out the contradiction.)

While most physicistsassumed that one of these postu-lates must be abandoned,Einstein had the insight that bothpostulates could be true if ourideas of space and time are modi-fied. Time dilation is one of thenecessary modifications.


Special Relativity and Time Dilation

In the kinematics chapters, you worked with equations that describe motion

in terms of a time interval (Δt). Before Einstein developed the special theory

of relativity, everyone assumed that Δt must be the same for any observer,

whether that observer is at rest or in motion with respect to the event being

measured. This idea is often expressed by the statement that time is absolute.

The relativity of time

In 1905, Einstein challenged the assumption that time is absolute in a paper

titled “The Electrodynamics of Moving Bodies,” which contained his special

theory of relativity. The special theory of relativity applies to observers and

events that are moving with constant velocity (in uniform motion) with

respect to one another. One of the consequences of this theory is that Δt does

depend on the observer’s motion. Consider a passenger in a train that is

moving uniformly with respect to an observer standing beside the track, as

shown in Figure 1. The passenger on the train shines a pulse of light toward

a mirror directly above him and measures the amount of time it takes for

the pulse to return. Because the passenger is moving along with the train, he

sees the pulse of light travel directly up and then directly back down, as in

Figure 1(a). The observer beside the track, however, sees the pulse hit the

mirror at an angle, as in Figure 1(b), because the train is moving with

respect to the track. Thus, the distance the light travels according to the

observer is greater than the distance the light travels from the perspective

of the passenger.

One of the postulates of Einstein’s theory of relativity, which follows from

James Clerk Maxwell’s equations about light waves, is that the speed of light

is the same for any observer, even when there is motion between the source

of light and the observer. Light is different

from all other phenomena in this respect.

Although this postulate seems counterintu-

itive, it was strongly supported by an experi-

ment performed in 1851 by Armand Fizeau.

But if the speed of light is the same for both

the passenger on the train and the observer

beside the track while the distances traveled

are different, the time intervals observed by

each person must also be different. Thus, the

observer beside the track measures a longer

time interval than the passenger does. This

effect is known as time dilation.

(b) Observer’s perspective

Mirror

(a)

Passenger’s perspective

Figure 1(a) A passenger on a train sends a pulse of light toward a mirror directly above. (b) Relative to a stationary observer beside the track, the distance the light travels is greater than that measured by the passenger.

EXTENSION• Have students research length

contraction, another conse-quence of the theory of specialrelativity. If L is the lengthbetween two points as meas-ured by an observer at rest withrespect to the points and v is thespeed between two referenceframes, then the contractedlength L� (measured in the sec-ond reference frame) is given by the following equation:

L� = L�1�−� ⎯v

c�2

2

⎯�Ask students to try differentvalues of speed in the equa-tions for time dilation andlength contraction to deter-mine at what speeds the effectsof special relativity becomesignificant.

• Have students research experi-ments that support the theoryof special relativity, includingthe Michelson-Morley experi-ment and Armand Fizeau’smeasurement of the speed oflight in water.

Teaching TipThe theory of special relativity isvalid only when the motionbetween the two reference framesis uniform. When one referenceframe is accelerating in relationto the other, the general theory ofrelativity must be used.

MisconceptionAlert

Because students have no directexperience with time dilation,many students initially think thattime dilation is a theoretical idearather than a physical effect thatcan be observed. Use the experi-ments discussed at the end of thisfeature to stress the fact that timedilation is a real physical effect.

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915 915

Calculating time dilation

Time dilation is given by the following equation, where Δt� represents the

time interval measured by the person beside the track, and Δt represents the

time interval measured by the person on the train:

Δt� =

In this equation, v represents the speed of the train relative to the person

beside the track, and c is the speed of light in a vacuum, 3.00 × 108 m/s. At

speeds with which we are familiar, where v is much smaller than c, the term

⎯v

c2

2

⎯ is such a small fraction that Δt� is essentially equal to Δt. For this reason,

we do not observe the effects of time dilation in our typical experiences. But

when speeds are closer to the speed of light, time dilation becomes more

noticeable. As seen by this equation, time dilation becomes infinite as v

approaches the speed of light.

According to Einstein, the motion between the train and the track is

relative; that is, either system can be considered to be in motion with respect

to the other. For the passenger, the train is stationary and the observer beside

the track is in motion. If the light experiment is repeated by the observer

beside the track, then the passenger would see the light travel a greater dis-

tance than the observer would. So, according to the passenger, it is the

observer beside the track whose clock runs more slowly. Observers see their

clocks running as if they were not moving. Any clocks in motion relative to

the observers will seem to the observers to run slowly. Similarly, by compar-

ing the differences between the time intervals of their own clocks and clocks

moving relative to theirs, observers can determine how fast the other clocks

are moving with respect to their own.


The effects we have been considering hold true for all physical processes,

including chemical and biological reactions. Scientists have demonstrated

time dilation by comparing the lifetime of muons (a type of unstable ele-

mentary particle) traveling at 0.9994c with the lifetime of stationary muons.

In another experiment, atomic clocks on jet planes flying around the world

were compared with identical clocks at the U.S. Naval

Observatory. In both cases, time dilations were

observed that matched the predictions of Einstein’s

theory of special relativity within the limits of

experimental error.

Δt⎯

�1�−� ⎯v

c�2

2

⎯�

www.scilinks.orgTopic: Relativity of TimeCode: HF61289

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Teaching TipThis feature builds on the previ-ous appendix feature, “SpecialRelativity and Time Dilation.” Inthat feature, the assumption thatthe speed of light is the same forall observers was used to explainwhy time measurements dependon an observer’s frame of refer-ence. In this feature, this assump-tion and its consequences arediscussed in greater detail.

The feature begins by compar-ing the behavior of light with atypical case to show how the twodiffer. Next, the need to revise theclassical addition of velocities isdiscussed. Finally, Einstein’s rela-tivistic addition of velocities isintroduced, along with a discus-sion of how all cases are coveredwith this equation.

Today there is overwhelmingphysical evidence that the speedof light is absolute. Experimentsin particle accelerators, in whichparticles reach speeds very closeto c, support the relativistic ratherthan the classical addition ofvelocities.

Teaching TipThe constancy of the speed oflight is difficult for some studentsto grasp because we have nodirect experience of this phenom-enon. Use examples to familiarizeyour students with Einstein’s the-ory of the constancy of the speedof light. For example, ask studentsto compare the speed of soundand light waves as viewed by twodifferent observers, one at restand one moving toward thesource of the waves.


Special Relativity and Velocities

In Section 4 of the chapter “Two-Dimensional Motion and Vectors,” you

learned that velocity measurements are not absolute; every velocity measure-

ment depends on the frame of reference of the observer with respect to the

moving object. For example, imagine that someone riding a bike toward you

at 25 m/s (v) throws a softball toward you. If the bicyclist measures the soft-

ball’s speed (u�) to be 15 m/s, you would perceive the ball to be moving

toward you at 40 m/s (u) because you have a different frame of reference than

the bicyclist does. This is expressed mathematically by the equation u = v + u�,

which is also known as the classical addition of velocities.

The speed of light

As stated in the appendix feature “Special Relativity and Time Dilation,”

according to Einstein’s special theory of relativity, the speed of light is

absolute, or independent of all frames of reference. If, instead of a softball,

the bicyclist were to shine a beam of light toward you, both you and the bicy-

clist would measure the light’s speed as 3.0 × 108 m/s. This would remain true

even if the bicyclist were moving toward you at 99 percent of the speed of

light. Thus, Einstein’s theory requires a different approach to the addition of

velocities. Einstein’s modification of the classical formula, which he derived

in his 1905 paper on special relativity, covers both the case of the softball and

the case of the light beam.

u = ⎯1 +

v

(

+

vu

u

�/

�

c2)⎯

In the equation, u is the velocity of an object in a reference frame, u� is the

velocity of the same object in another reference frame, v is the velocity of one

reference frame relative to another, and c is the speed of light.

The universality of Einstein’s equation

How does Einstein’s equation cover both cases? First we shall

consider the bicyclist throwing a softball. Because c2 is such a

large number, the vu�/c2 term in the denominator is very small

for velocities typical of our everyday experience. As a result, the

denominator of the equation is essentially equal to 1. Hence, for

speeds that are small compared with c, the two theories give

nearly the same result, u = v + u�, and the classical addition of

velocities can be used.

However, when speeds approach the speed of light, vu�/c2

increases, and the denominator becomes greater than 1 but

never more than 2. When this occurs, the difference between the

two theories becomes significant. For example, if a bicyclist

Figure 1According to Einstein’s relativistic equationfor the addition of velocities, material parti-cles can never reach the speed of light.

EXTENSION• Have students research particle-

accelerator experiments involv-ing particles traveling at speedsclose to c. Their reports shouldinclude a discussion of howobservations support the rela-tivistic addition of velocities.

• The ether was originally con-ceived as the medium throughwhich light waves traveled.Have students investigate theconcept of the ether and theMichelson-Morley experiment,which was intended to detectthe ether. Then have a class dis-cussion about why the conceptof the ether was originallybelieved to be necessary andhow the Michelson-Morleyexperiment and the special theory of relativity affected the theory of the ether.


moving toward you at 80 percent of the speed of light were to throw a ball to

you at 70 percent of the speed of light, you would observe the ball moving

toward you at about 96 percent of the speed of light rather than the 150 per-

cent of the speed of light predicted by classical theory. In this case, the differ-

ence between the velocities predicted by each theory cannot be ignored, and

the relativistic addition of velocities must be used.

In this last example, it is significant that classical addition predicts a speed

greater than the speed of light (1.5c), while the relativistic addition predicts a

speed less than the speed of light (0.96c). In fact, no matter how close the

speeds involved are to the speed of light, the relativistic equation yields a

result less than the speed of light, as seen in Table 1.How does Einstein’s equation cover the second case, in which the bicyclist

shines a beam of light toward you? Einstein’s equation predicts that any

object traveling at the speed of light (u� = c) will appear to travel at the

speed of light (u = c) for an observer in any reference frame:

u = ⎯1 +

v

(

+

vu

u

�/

�

c2)⎯ = ⎯

1 +

v

(

+vc

c

/c2)⎯ = ⎯

1 +v +

(v

c

/c)⎯ = ⎯

(c

v

++v

c

)/c⎯ = c

This corresponds with our earlier statement that the bicyclist measures

the beam of light traveling at the same speed that you do, 3.0 × 108 m/s, even

though you have a different reference frame than the bicyclist does. This

occurs regardless of how fast the bicycle is moving because v (the bicycle’s

speed) cancels from the equation. Thus, Einstein’s relativistic equation suc-

cessfully covers both cases. So, Einstein’s equation is a more general case of

the classical equation, which is simply the limiting case.

Table 1 Classical and Relativistic Addition of Velocities

c � 299 792 458 m/s Classical Relativisticaddition addition

Speed between Speed measured Speed measured Speed measuredframes (v) in A (u�) in B (u) in B (u)

000 025 m/s 000 0 15 m/s 000 000 040 m/s 000 000 040 m/s

100 000 m/s 100 000 m/s 000 200 000 m/s 000 200 000 m/s

50% of c 50% of c 299 792 458 m/s 239 833 966 m/s

90% of c 90% of c 539 626 424 m/s 298 136 146 m/s

99.99% of c 99.99% of c 599 524 958 m/s 299 792 457 m/s

www.scilinks.orgTopic: Speed of LightCode: HF61439

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Teaching TipThe equivalence between massand energy is a consequence ofEinstein’s special theory of rela-tivity. Einstein introduced thesubject in 1905 in a paper titled“Does the inertia of a bodydepend on its energy-content?”

Although the derivation ofrelativistic kinetic energy isbeyond the scope of this book,the mass-energy equivalence canbe understood conceptually atthis level and will enhance stu-dents’ understanding of energy.

The equivalence between massand energy in terms of bindingenergy is explored in the chapter“Subatomic Physics.” Fission andfusion are also discussed ingreater detail in that chapter,along with applications such asnuclear power and the nuclearbomb.

EXTENSION• Have students research fission

and fusion as energy sources.In their reports, they shouldinclude a discussion of the fol-lowing questions: Why arefusion reactors considered tobe safer than fission reactors?What are the difficulties asso-ciated with developing afusion reactor? What are theadvantages and disadvantagesof using nuclear reactors assources of energy?

• Einstein was deeply concernedabout the possibility ofnuclear weapons suggested byhis theory. Have studentsresearch Einstein’s opinionson the subject of nuclearweapons to prepare for a class-room debate on the socialresponsibility of scientists.


The Equivalence ofMass and Energy

Einstein’s ER = mc2 is one of the most famous equations of the twentieth cen-

tury. Einstein discovered this equation through his work with relative velocity

and kinetic energy.

Relativistic kinetic energy

In the appendix feature “Special Relativity and Velocities,” you learned how

Einstein’s special theory of relativity modifies the classical addition of veloci-

ties. The classical equation for kinetic energy (KE = ⎯12

⎯ mv2) must also be mod-

ified for relativity. In 1905, Einstein derived a new equation for kinetic energy

based on the principles of special relativity:

KE = − mc2

In this equation, m is the mass of the object, v is the velocity of the object,

and c is the speed of light. Although it isn’t immediately obvious, this equa-

tion reduces to the classical equation KE = ⎯12

⎯ mv2 for speeds that are small

relative to the speed of light, as shown in Figure 1. The graph also illustrates

that velocity can never be greater than 1.0c in the theory of special relativity.

Einstein’s relativistic expression for kinetic energy has been confirmed by

experiments in which electrons are accelerated to extremely high speeds in par-

ticle accelerators. In all cases, the experimental data correspond to Einstein’s

equation rather than to the classical equation. Nonetheless, the dif-

ference between the two theories at low speeds (relative to c) is so

minimal that the classical equation can be used in all such cases

when the speed is much less than c.

Rest energy

The second term of Einstein’s equation for kinetic energy, –mc2, is

required so that KE = 0 when v = 0. Note that this term is indepen-

dent of velocity. This suggests that the total energy of an object

equals its kinetic energy plus some additional form of energy equal

to mc2. The mathematical expression of this additional energy is

the familiar Einstein equation:

ER = mc2

This equation shows that an object has a certain amount of

energy (ER), known as rest energy, simply by virtue of its mass. The

mc2⎯⎯

�1�−� ��⎯v

c2�2

⎯��

Classical case

Relativistic case

0.5 1.5

2.0c

1.5c

1.0c

0.5c

1.0 2.0

Kinetic energy (KE/mc2)

Vel

ocity

Figure 1This graph of velocity versus kinetic energy forboth the classical and relativistic equations showsthat the two theories are in agreement when v ismuch less than c. Note that v is always less than cin the relativistic case.

rest energy of a body is equal to its mass, m, multiplied by the speed of light

squared, c2. Thus, the mass of a body is a measure of its rest energy. This

equation is significant because rest energy is an aspect of special relativity

that was not predicted by classical physics.


The magnitude of the conversion factor between mass and rest energy

(c2 = 9 × 1016 m2/s2) is so great that even a very small mass has a huge

amount of rest energy. Nuclear reactions utilize this relationship by convert-

ing mass (rest energy) into other forms of energy. In nuclear fission, which is

the energy source of nuclear power plants, the nucleus of an atom is split into

two or more nuclei. Taken together, the mass of these nuclei is slightly less

than the mass of the original nucleus, and a very large amount of energy is

released. In typical nuclear reactions, about one-thousandth of the initial

mass is converted from rest energy into other forms of energy. This change

in mass, although very small, can be detected experimentally.

Another type of nuclear reaction that converts mass into energy is fusion,

which is the source of energy for our sun and other stars. About 4.5 million

tons of the sun’s mass is converted into other forms of energy every second, by

fusing hydrogen into helium. Fortunately, the sun has enough mass to continue

to fuse hydrogen into helium for approximately 5 billion more years.

Most of the energy changes encountered in your typical experiences are

much smaller than the energy changes that occur in nuclear reactions. Such

changes are far too small to be detected experimentally. Thus, for typical

cases, the classical equation still holds, and mass and energy can be thought

of as separate.

Before Einstein’s theory of relativity, conservation of energy and con-

servation of mass were regarded as two separate laws. The equivalence

between mass and energy reveals that in fact these two laws are one.

In the words of Einstein, “Prerelativity physics contains two con-

servation laws of fundamental importance. . . . Through relativity

theory, they melt together into one principle.”

Figure 3Our sun uses a nuclear reactioncalled fusion to convert mass toenergy. About 90 percent of thestars, including our sun, fuse hydro-gen, and some older stars fuse helium.

Teaching TipFigure 2 shows the PEP-IICollider at SLAC. Electrons andpositrons (represented by blueand pink lights) travel along twoseparate rings in opposite direc-tions; when they collide, bothdisappear and new particles arecreated in their place.

The Language of PhysicsThe notion of mass in relativityhas been undergoing a transfor-mation in recent years. In earliertreatments of relativity, which youwill still see in some current text-books, the notion of relativisticmass is used. In these treatments,the symbol m represents the rela-tivistic mass, which increases asthe speed of the object increases.As a result, Einstein’s equation isthe commonly seen E = mc2, andthe energy is the relativistic totalenergy. In these treatments, thenotion of rest mass is used to rep-resent the mass of an object whenits speed is zero.

In modern treatments of rela-tivity, the symbol m representssimply the mass of an object,which remains constant. There isno notion of a change of masswith speed. Einstein’s equation isER = mc2, in which the energy isthe rest energy of the object. Thetotal energy of the object is thenexpressed as follows:

E =

There is no use of the term restmass.

mc2⎯

�1�−� ⎯v

c�2

2

⎯�

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Figure 2Electrons in the Stanford LinearAccelerator in California (SLAC)reach 99.999999967 percent of thespeed of light. At such greatspeeds, the difference betweenclassical and relativistic theoriesbecomes significant.

www.scilinks.orgTopic: Nuclear ReactionsCode: HF61053

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Teaching TipEncourage students who areinterested in this feature to readportions of Einstein’s popularaccount of the theories of specialand general relativity, titledRelativity: The Special and theGeneral Theory. In the first fewchapters of Part II of this book,Einstein discusses the equiva-lence between gravitational andinertial mass and introduces thethought experiment described inthis feature.


General Relativity

In the appendix features on Einstein’s theory of special relativity, you studied

situations involving observers in different reference frames, such as one

observer on a moving train and another on the ground. These examples all

assumed that the two reference frames were moving uniformly with respect

to each other. In other words, neither reference frame was accelerating rela-

tive to the other. Special relativity applies only to nonaccelerating reference

frames. Einstein expanded his special theory of relativity into the general the-

ory to cover all cases, including accelerating reference frames.

Gravitational attraction and accelerating reference frames

Einstein began with a simple question: “If we pick up a stone and then let it go,

why does it fall to the ground?”You might answer that it falls because it is attract-

ed by gravitational force. As usual, Einstein was not satisfied with this typical

answer. He was also intrigued by the fact that in a vacuum, all objects in free fall

have the same acceleration, regardless of their mass. As you learned in the chapter

on gravity, the reason is that gravitational mass is equal to inertial mass. Because

the two masses are equivalent, the extra gravitational force from a larger gravita-

tional mass is exactly canceled out by its larger inertial mass, thus producing the

same acceleration. Einstein considered this equivalence to be a great puzzle.

To explore these questions, Einstein used a thought experiment similar to the

one shown in Figure 1. In Figure 1(a), a person in an elevator at rest on Earth’s

surface drops a ball. The ball is in free fall and accelerates downward, as you would

expect. In Figure 1(b), a similar elevator in space is moving upward with a con-

stant acceleration. If an astronaut in this elevator releases a ball, the floor acceler-

Figure 1Einstein discovered that there is noway to distinguish between (a) agravitational field and (b) an accel-erating reference frame.

a

g

(a) (b)


ates up toward the ball. To the astronaut, the ball appears to be accelerating down-

ward, and this situation is identical to the situation described in (a). In other

words, the astronaut may think that his spaceship is on Earth and that the ball falls

because of gravitational attraction. Because gravitational mass equals inertial

mass, the astronaut cannot conduct any experiments to distinguish between the

two cases. Einstein described this realization as “the happiest thought of my life.”

Gravity and light

Now, imagine a ray of light crossing the accelerating elevator. Suppose that the

ray of light enters the elevator from the left side. As the light ray travels across the

elevator from left to right, the floor of the elevator accelerates upward. Thus, to

an astronaut in the elevator, the light ray would appear to follow a parabolic path.

If Einstein’s theory of the equivalence between gravitational fields and acceler-

ating reference frames is correct, then light must also bend this way in a gravita-

tional field. Einstein proposed using the sun’s gravitational field to test this idea.

The effect is small and difficult to measure, but Einstein predicted that it could be

done during a solar eclipse. Einstein published the theory of general relativity in

1916. Just three years later, in 1919, the British astronomer Arthur S. Eddington

conducted observations of the light from stars during an eclipse. This experiment

provided support for Einstein’s theory of general relativity.

Curved spacetime

Although light traveling near a massive object such as the sun

appears to bend, is it possible that the light is actually following the

straightest path? Einstein theorized that the answer to this question

is yes. In general relativity, the three dimensions of space and the

one dimension of time are considered together as four-dimensional

space-time. When no masses are present, an object moves through

“flat” space-time. Einstein proposed that masses change the shape

of space-time, as shown in Figure 2. A light ray that bends near the

sun is following the new shape of the distorted space-time.

For example, imagine rolling a tennis ball across a water bed. If the

water bed is flat, the tennis ball will roll straight across. If you place a

heavy bowling ball in the center, the bowling ball changes the shape of

the water bed. As a result, the tennis ball will then follow a curved path,

which, in Newton’s theory, is due to the gravitational force between the

two. In general relativity, the tennis ball is simply following the curved

path of space-time, which is distorted by the bowling ball. Unlike

Newton’s mathematical theory of gravitation, Einstein’s theory of

curved space-time offers a physical explanation for gravitational force.

Today, Einstein’s theory of general relativity is well accepted.

However, scientists have not yet been able to incorporate it with another well-

accepted theory that describes things at the microscopic level: quantum mechan-

ics. Many scientists are now working toward a unification of these two theories.

Figure 2In the theory of general relativity,masses distort four-dimensionalspace-time, as illustrated here.Thisdistortion creates the effect wedescribe as gravitational attraction.

www.scilinks.orgTopic: Albert EinsteinCode: HF60097

922

Teaching TipLouis de Broglie, a French physi-cist, was awarded the Nobel Prizein 1929 for his prediction of thewave nature of material particles.De Broglie originally proposedthis theory in his doctoral thesisin 1924.

The de Broglie family was anaristocratic French family, andLouis de Broglie held the title ofPrince. De Broglie came late tothe study of theoretical physicsbecause he first studied history.Only after serving as a radiooperator in World War I did hebegin his study of physics.

De Broglie later found a con-nection between the wave natureof electrons and the Bohr modelof hydrogen, in which only cer-tain orbits of the electron are stable. De Broglie postulated thatthe stable orbits are those thatcontain an integral number ofwavelengths; this is analogous tothe standing waves that can existon a vibrating string. This con-nection showed that the stableorbits allowed in Bohr’s model area result of the interference pat-terns of the electrons. De BroglieWaves are also covered in thechapter “Atomic Physics.”

De Broglie Waves

Figure 1This image of cat hairs,produced by an electronmicroscope, is magnified500 times.

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In the chapter on waves, we treated waves and particles as if there were a clear

distinction between the two. For most of the history of science, this was

believed to be the case. However, in the early twentieth century, scientists

were confronted with experimental evidence suggesting that the properties

of matter are not always as clear-cut as everyone had assumed.

The dual nature of light

This scientific revolution began in 1900, when Max Planck introduced the

possibility that energy could come in discrete units. In 1905, Einstein extend-

ed Planck’s theory, suggesting that all electromagnetic waves (such as light)

sometimes behave like particles. According to this theory, light can behave

both like a wave and like a particle; some experiments reveal its wave nature,

and other experiments display its particle nature. Although this idea was ini-

tially greeted with skepticism, it explained certain phenomena that the wave

theory of light could not account for and was soon confirmed empirically in

a variety of experiments.

Matter waves

The idea that light has a dual nature led Louis de Broglie to hypothesize that

perhaps all matter has wavelike characteristics. De Broglie believed that there

should not be two separate branches of physics, one for electromagnetic

waves and another for matter. In his doctoral thesis, submitted in 1924, he

proposed a theory of matter waves to reconcile this discrepancy. At that time,

there was no experimental evidence to support his theory.

De Broglie’s calculations suggested that matter waves had a wavelength, l,

often called the de Broglie wavelength, given by the following equation:

l = ⎯⎯h

p⎯⎯ = ⎯

m

h

v⎯⎯

The variable h in this equation is called Planck’s constant, which is approxi-

mately equal to 6.63 × 10−34 J•s. The variable p is the object’s momentum,

which is equivalent to its mass, m, times its velocity, v. Note that the dual

nature of matter suggested by de Broglie is evident in this equation, which

includes both a wave concept (l) and a particle concept (mv).

De Broglie’s equation shows that the smaller the momentum of an object,

the larger its de Broglie wavelength. But even when the momentum of

an object is very small from our perspective, h is so small that the

wavelength is still much too small for us to detect. In order

to detect a wavelength, one must use an opening

comparable in size to the wavelength because

waves passing through such an opening

Appendix J

Figure 3These color-enhanced images from a scanning electron microscope show, from left to right, a bean weevil emerging from a bean seed, two strips of velcro fastened together, and pollen grains.

Figure 2In this photograph, electron waves arediffracted by a crystal. Experimentssuch as this show the wave nature ofelectrons and thereby provide empiri-cal evidence for de Broglie’s theory ofthe dual nature of matter.

will display patterns of constructive and destructive interference. When the

opening is much larger than the wavelength, waves travel through it without

being affected.

The de Broglie wavelength of a 0.15 kg baseball moving at 30 m/s is about

1.5 × 10−34 m. This is almost a trillion trillion times smaller than the diameter

of a typical air molecule—much smaller than any possible opening through

which we could observe interference effects. This explains why the de Broglie

wavelength of objects cannot be observed in our everyday experience.

However, in the microscopic world, the wave effects of matter can be

observed. Electrons (m = 9.109 × 10−31 kg) accelerated to a speed of 1.4 × 107

m/s have a de Broglie wavelength of about 10−10 m, which is approximately

equal to the distance between atoms in a crystal. Thus, the atoms in a crystal

can act as a three-dimensional grating that should diffract electron waves.

Such an experiment was performed three years after de Broglie’s thesis by

Clinton J. Davisson and Lester H. Germer, and the electrons did create pat-

terns of constructive and destructive interference, such as the pattern in

Figure 2. This experiment gave confirmation of de Broglie’s theory of the

dual nature of matter.

The electron microscope

A practical device that relies on the wave characteristics of matter is the elec-

tron microscope. In principle, the electron microscope is similar to an ordi-

nary compound microscope. But while ordinary microscopes use lenses to

bend rays of light that are reflected from a small object, electron microscopes

use electric and magnetic fields to accelerate and focus a beam of electrons.

Rather than examining the image through an eyepiece, as in an ordinary

microscope, a magnetic lens forms an image on a fluorescent screen. Without

the fluorescent screen, the image would not be visible.

Electron microscopes are able to distinguish details about 100 times small-

er than optical microscopes can. Because of their great resolving power, elec-

tron microscopes are widely used in many areas of scientific research.

www.scilinks.orgTopic: Electron MicroscopesCode: HF60487

EXTENSIONHave students research the elec-tron microscope and its uses. Intheir research, ask students tocompare the electron microscopewith a typical optical microscope.Then have students research thevarious fields in which electronmicroscopes are used and whatkinds of images they produce.Have students share their resultswith the class.

Teaching TipThe equation for the de Brogliewavelength of an electron isderived from the equation for themomentum of a particle of light,also known as a photon (p = h/l).De Broglie postulated that thisequation might apply to the elec-tron as well (l = h/p = h/mv).

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Teaching TipThis feature discusses some basicideas from quantum mechanicsand builds on these ideas toexplore a phenomenon calledelectron tunneling. The featureconcludes with a discussion ofthe scanning tunneling micro-scope, which utilizes electrontunneling to generate highlydetailed images that depict theatomic structure on the surfaceof a material.

MisconceptionAlert

Some students may think that thepotential well shown in Figure 1represents a physical well or area.Be sure they understand that theheight of the well corresponds toan amount of energy rather thanto a physical height. Classically,the electron is confined to thewell if its energy is less than U.

STOP


Electron Tunneling

In the chapter “Electrical Energy and Current,” we discussed current as the

motion of charge carriers, which we treated as particles. But, as discussed in

the “De Broglie Waves” appendix feature, the electron has both particle and

wave characteristics. The wave nature of the electron leads to some strange

consequences that cannot be explained in terms of classical physics. One

example is tunneling, a phenomenon whereby electrons can pass into regions

that, according to classical physics, they do not have the energy to reach.

Probability waves

To see how tunneling is possible, we must explore matter waves in greater detail.

De Broglie’s revolutionary idea that particles have a wave nature raised the

question of how matter waves behave. In 1926, Erwin Schrödinger proposed a

wave equation that described the manner in which de Broglie matter waves

change in space and time. Two years later, in an attempt to relate the wave and

particle natures of matter, Max Born suggested that the square of the amplitude

of a matter wave is proportional to the probability of finding the corresponding

particle at that location. This theory is called quantum mechanics.

Tunneling

Born’s interpretation makes it possible for a particle to be found in a location

that is not allowed by classical physics. Consider an electron with a potential

energy of zero in the region between 0 and L (region II) of Figure 1. We call

this region the potential well. The electron has a potential energy of some

finite value U outside this area (regions I and III). If the energy of the elec-

tron is less than U, then according to classical physics, the electron cannot

escape the well without first acquiring additional energy.

The probability wave for this electron (in its lowest energy

state) is shown in Figure 2 on the next page. Between any two

points of this curve, the area under the corresponding part of

the curve is proportional to the probability of finding the elec-

tron in that region. The highest point of the curve corresponds

to the most probable location of the electron, while the lower

points correspond to less probable locations. Note that the

curve never actually meets the x-axis. This means that the elec-

tron has some finite probability of being anywhere in space.

Hence, there is a probability that the electron will actually be

found outside the potential well. In other words, according to

quantum mechanics, the electron is no longer confined to strict

boundaries because of its energy. When the electron is found

outside the boundaries established by classical physics, it is said

to have tunneled to its new location.

U

Potential well

L0

I II III

Figure 1An electron has a potential energy of zero inside thewell (region II) and a potential energy of U outside thewell. According to classical physics, if the electron’senergy is less than U, it cannot escape the well withoutabsorbing energy.

EXTENSIONHave students form small groupsto brainstorm about possible usesof the STM. Then have themresearch uses of the STM that arecurrently being developed. Haveeach group share its ideas andresearch with the class.

Teaching TipBe sure students understand thatregions I, II, and III of the poten-tial well (Figure 1) correspond toregions I, II, and III of the proba-bility wave (Figure 2). Becausethere is a finite probability thatthe electron will be found inregions I and III of Figure 2,there is a probability that theelectron will be found outside thepotential well.


The scanning tunneling microscope

In 1981, Gerd Binnig and Heinrich Rohrer, at IBM Zurich, discovered a prac-

tical application of tunneling current: a powerful microscope called the scan-

ning tunneling microscope, or STM. The STM can produce highly detailed

images with resolution comparable to the size of a single atom. The image of

the surface of graphite shown in Figure 3 demonstrates the power of the

STM. Note that individual carbon atoms are recognizable. The smallest detail

that can be discerned is about 0.2 nm, or approximately the size of an atom’s

radius. A typical optical microscope has a resolution no better than 200 nm,

or about half the wavelength of visible light, so it could never show the detail

seen in Figure 3.In the STM, a conducting probe with a very sharp tip (about the width of

an atom) is brought near the surface to be studied. According to classical

physics, electrons cannot move between the surface and the tip because they

lack the energy to escape either material. But according to quantum theory,

electrons can tunnel across the barrier, provided the distance is small enough

(about 1 nm). Scientists can apply a potential difference between the surface

and the tip to make electrons tunnel preferentially from surface to tip. In this

way, the tip samples the distribution of electrons just above the surface.

The STM works because the probability of tunneling decreases exponen-

tially with distance. By monitoring changes in the tunneling current as the tip

is scanned over the surface, scientists obtain a sensitive measure of the topog-

raphy of the electron distribution on the surface. The result is used to make

images such as the one in Figure 3. The STM can measure the height of sur-

face features to within 0.001 nm, approximately 1/100 of an atomic diameter.

Although the STM was originally designed for imaging atoms, other

practical applications are being developed. Engineers have greatly reduced

the size of the STM and hope to someday develop a computer in which every

piece of data is held by a single atom or by small groups of atoms and then

read by an STM.

Figure 3A scanning tunneling microscope (STM) was usedto produce this image of the surface of graphite, aform of carbon. The contours represent thearrangement of individual carbon atoms on thesurface. An STM enables scientists to see smalldetails on surfaces with a lateral resolution of0.2 nm and a vertical resolution of 0.00 1 nm.

0 LProbability wave

I II III

Figure 2The probability curve for an elec-tron in its lowest energy stateshows that there is a certain proba-bility of finding the electron outsidethe potential well.

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Key Models andAnalogiesBe sure students understand thedifference between the classifica-tion of materials into categoriesbased on their ability to conductcharge and the physical charac-teristics that lead to the differentclassifications. Remind them thatelectrical attraction and repul-sion can be explained at theatomic level by considering thetransfer of electrons betweenobjects. Then tell them that theability of some materials to con-duct charge better than others isexplained at the atomic level byband theory.

Visual Strategy

Figure 1Be sure students understand thatthe third case, (c), is simply anextension of the second case, (b),in which the energy levels are soclose together that they are repre-sented as continuous bands.

What causes the splitting ofenergy levels in each of these

cases?

When more than one atom ispresent, the electric field pro-

duced by each atom affects theenergy levels of other atoms,causing the splitting shown inthe figure.

Could splitting be observedfor a single atom?

noA

Q

A

Q


Semiconductor Doping

Figure 1Energy levels split when twoatoms are close together (a).Adding a few more nearbyatoms causes further splitting(b). When many atoms interact,the energy levels are so closelyspaced that they can be repre-sented as energy bands (c).

Ener

gy

Atomic separation

Two atoms

(a)

Ener

gy

Atomic separation

Many atoms

(c)

Allowed energy band

Forbidden energy gap

Allowed energy band

Ener

gy

Atomic separation

Four atoms

(b)

Materials can be classified according to their ability to conduct electricity. A

good conductor has a large number of free charge carriers that can move easi-

ly through the material, whereas an insulator has a small number of free

charge carriers that are relatively immobile. Semiconductors exhibit electronic

properties between those of insulators and those of conductors. The develop-

ment of band theory uses basic physical principles to explain some of the

properties of these three categories of materials.

Electron energy levels

As seen in the chapter “Atomic Physics,” the electrons in an atom can possess

only certain amounts of energy. For this reason, the electrons are often said

to occupy specific energy levels. Electrons in a shell sometimes form a set of

closely spaced energy levels. Normally, electrons are in the lowest energy level

available to them. The specific arrangement of electrons in which all are in

the lowest possible energy levels of an atom is called the atom’s

If an atom absorbs sufficient energy from the environment, some of the

atom’s electrons can move to higher energy levels. The atom is then said to be

in an If an electron absorbs so much energy that it is no longer

bound to the atom, it is then called a free electron.

Band theory

Band theory uses the concept of energy levels to explain the mechanisms of

conduction in many solids. When identical atoms are far apart, they have

identical energy-level diagrams. No two electrons in the same system can

occupy the same state. As a result, when two atoms are brought closer togeth-

er, the energy levels of each atom are altered by the influence of the electric

field of the other atom. Figure 1 shows how two energy levels split when there

are two atoms (a), four atoms (b), and many atoms (c) at different separation

distances. In the case of two atoms, each energy level splits into two different

energy levels, as shown in Figure 1(a). Notice that the energy difference

between two new energy levels depends on the distance between the atoms.

excited state.

ground state.


When more atoms are brought close together, each energy level splits into

more levels. If there are many atoms, the energy level splits so many times and

the new energy levels are so closely spaced that they may be regarded as a con-

tinuous band of energies, as in Figure 1(c). The highest band containing occu-

pied energy levels is called the valence band, as shown in Figure 2. The band

immediately above the valence band is called the conduction band.

Electron-hole pairs and intrinsic semiconductors

Imagine that a few electrons are excited from the valence band to the conduc-

tion band by an electric field, as in Figure 3. The electrons in the conduction

band are free to move through the material. Normally, electrons in the

valence band are unable to move because all nearby energy levels are occu-

pied. But when an electron moves from the valence band into the conduction

band, it leaves a vacancy, or in an otherwise filled valence band. The

hole is positively charged because it results from the removal of an electron

from a neutral atom. Whenever another valence electron from this or a near-

by atom moves into the hole, a new hole is created at its former location. So,

the net effect can be viewed as a positive hole migrating through the material

in a direction opposite the motion of the electrons in the conduction band.

In a material containing only one element or compound, there are an

equal number of conduction electrons and holes. Such combinations of

charges are called electron-hole pairs, and a semiconductor that contains such

pairs is called an intrinsic semiconductor. In the presence of an electric field,

the holes move in the direction of the field and the conduction electrons

move opposite the field.

Adding impurities to enhance conduction

One way to change the concentration of charge carriers is to add impurities,

atoms that are different from those of an intrinsic semiconductor. This process

is called Even a few added impurity atoms (about one part in a mil-

lion) can have a large effect on a semiconductor’s resistance. The semiconduc-

tor’s conductivity increases as the doping level increases. When impurities

dominate conduction, the material is called an extrinsic semiconductor.

There are two methods for doping a semiconductor: either add impurities

that have extra valence electrons or add impurities that have fewer valence

electrons compared with the atoms in the intrinsic semiconductor.

Semiconductors used in commercial devices are usually doped silicon or

germanium. These elements have four valence electrons. Semiconductors are

doped by replacing an atom of silicon or germanium with one containing

either three valence electrons or five valence electrons. Note that a doped

semiconductor is electrically neutral because it is made of neutral atoms. The

balance of positive and negative charges has not changed, but the number of

charges that are free and able to move has. These charges are therefore able to

participate in electrical conduction.

doping.

hole,

In-Depth Physics ContentYour students can visit go.hrw.comfor an online chapter that integratesmore in-depth development of theconcepts covered here.

Keyword HF6APJX

Energy

Conduction band

Forbidden gap

Valence band

Figure 2Energy levels of atoms becomeenergy bands in solids. The valenceband is the highest occupied band.

EnergyConduction electrons

Applied E field

− − −

+ + +

Electrons− Holes+

Figure 3An electric field can excite valenceelectrons into the conduction band,where they are free to movethrough the material. Holes in thevalence band can then move in theopposite direction.

Demonstration

Hole FlowPurpose Illustrate hole flow.Materials nine rubber stoppersProcedure Have 10 studentsstand facing the class with theirright palms out. Place a stopper inthe hand of every person exceptthe student on the far right.

Beginning at the far right,have each student look to theirright and place their stopper intheir neighbor’s palm if that per-son does not already have a stop-per. All of the stoppers will shiftone palm to the right, and theperson on the far left will bewithout a stopper. Ask the stu-dents to consider the movementof the empty space as an electron“hole.” Point out that the holemoves to the left as the stoppers(the electrons) move to the right.

www.scilinks.orgTopic: SemiconductorsCode: HF61376

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Superconductors and BCS Theory

The resistance of many solids (other than semiconductors) increases with

increasing temperature. The reason is that at a nonzero temperature, the

atoms in a solid are always vibrating, and the higher the temperature, the

larger the amplitude of the vibrations. It is more difficult for electrons to

move through the solid when the atoms are moving with large amplitudes.

This situation is somewhat similar to walking through a crowded room. It

is much harder to do so when the people are in motion than when they are

standing still.

If the resistance depended only on atomic vibrations, we would expect the

resistance of the material that is cooled to absolute zero to go gradually to

zero. Experiments have shown, however, that this does not happen. In fact,

the resistances of very cold solids behave in two very different ways—either

the substance suddenly begins superconducting at temperatures above

absolute zero or it never superconducts, no matter how cold it gets.

Resistance from lattice imperfections

The graph in Figure 1 shows the temperature dependence of the resistance of

two similar objects, one made of silver and the other made of tin. The tem-

perature dependence of the resistance of the silver object is similar to that of

a typical metal. At higher temperatures, the resistance decreases as the metal

is cooled. This decrease in resistance suggests that the amplitude of the lattice

vibrations is decreasing, as expected. But at a temperature of about 10 K, the

curve levels off and the resistance becomes constant. Cooling the metal fur-

ther does not appreciably lower the resistance, even though the vibrations of

the metal’s atoms have been lessened.

Part of the cause of this nonzero resistance, even at absolute zero, is lattice

imperfection. The regular, geometric pattern of the crystal, or lattice, in a solid

is often flawed. A lattice imperfection occurs when some of the atoms do not

line up perfectly.

Imagine you are walking through a crowded room in which the people are

standing in perfect rows. It would be easy to walk through the room between

two rows. Now imagine that occasionally one person stands in the middle of

the aisle instead of in the row, making it harder for you to pass. This is similar

to the effect of a lattice imperfection. Even in the absence of thermal vibra-

tions, many materials exhibit a residual resistance due to the imperfect geo-

metric arrangement of their atoms.

Figure 1 shows that the resistance of tin jumps to zero below a certain

temperature that is well above absolute zero. A solid whose resistance is zero

below a certain nonzero temperature is called a The tem-

perature at which the resistance goes to zero is the critical temperature of the

superconductor, as described in the chapter “Electrical Energy and Current.”

superconductor.

Tin Object

Silver Object

20

10

0 Tc 10Temperature (K)

Res

ista

nce

(Ω)

20

Figure 1The resistance of silver exhibits thebehavior of a normal metal. Theresistance of tin goes to zero at tem-perature Tc, the temperature atwhich tin becomes a superconductor.

Demonstration

Lattice ImperfectionsPurpose Visually demonstratethe effects of lattice imperfec-tions on electron movement.Materials board, nails, marbleProcedureBefore class: Prepare a “lattice” byplacing nails in a board in a regu-lar pattern (about every 2 cm)with occasional variations. Thereshould be no clear routes fromone edge of the board to theopposite edge.In class: Show students that thereis room for a marble to passunobstructed between the nails.Point out that the occasional mis-placed nails prevent a free paththrough the lattice. Demonstratethis by holding the board at anangle and allowing the marble toroll down. You may want to makean additional lattice with moreimperfections to show studentsthe comparison between the two.


BCS theory

Before the discovery of superconductivity, it was thought that all materials

should have some nonzero resistance due to lattice vibrations and lattice

imperfections, much like the behavior of the silver in Figure 1.The first complete microscopic theory of superconductivity was not devel-

oped until 1957. This theory is called BCS theory after the three scientists

who first developed it: John Bardeen, Leon Cooper, and Robert Schrieffer.

The crucial breakthrough of BCS theory is a new understanding of the spe-

cial way that electrons traveling in pairs move through the lattice of a super-

conductor. According to BCS theory, electrons do suffer collisions in a

superconductor, just as they do in any other material. However, the collisions

do not alter the total momentum of a pair of electrons. The net effect is as if

the electrons moved unimpeded through the lattice.

Cooper pairs

Imagine an electron moving through a lattice, such as electron 1 in Figure 2.There is an attractive force between the electron and the nearby positively

charged atoms in the lattice. As the electron passes by, the attractive force

causes the lattice atoms to be pulled toward the electron. The result is a con-

centration of positive charge near the electron. If a second electron is nearby,

it can be attracted to this excess positive charge in the lattice before the lattice

has had a chance to return to its equilibrium position.

Through the process of deforming the lattice, the first electron gives up

some of its momentum. The deformed region of the lattice attracts the sec-

ond electron, transferring excess momentum to the second electron. The net

effect of this two-step process is a weak, delayed attractive force between the

two electrons, resulting from the motion of the lattice as it is deformed by the

first electron. The two electrons travel through the lattice acting as if they

were a single particle. This particle is called a Cooper pair. In BCS theory,

Cooper pairs are responsible for superconductivity.

The reason superconductivity has been found at only low temperatures so

far is that Cooper pairs are weakly bound. Random thermal motions in the lat-

tice tend to destroy the bonds between Cooper pairs. Even at very low tempera-

tures, Cooper pairs are constantly being formed, destroyed, and reformed in a

superconducting material, usually with different pairings of electrons.

Calculations of the properties of a Cooper pair have shown that this pecu-

liar bound state of two electrons has zero total momentum in the absence of

an applied electric field. When an external electric field is applied, the Cooper

pairs move through the lattice under the influence of the field. However, the

center of mass for every Cooper pair has exactly the same momentum. This

crucial feature of Cooper pairs explains superconductivity. If one electron

scatters, the other electron in a pair also scatters in a way that keeps the total

momentum constant. The net result is that scattering due to lattice imperfec-

tions and lattice vibrations has no net effect on Cooper pairs.

−−

Electron 2 Electron 1

Lattice ionE

Figure 2The first electron deforms the lat-tice, and the deformation affects thesecond electron. The net result is asif the two electrons were looselybound together. Such a two-electronbound state is called a Cooper pair.

In-Depth Physics ContentYour students can visit go.hrw.comfor an online chapter that integratesmore in-depth development of theconcepts covered here.

Keyword HF6APJX

Demonstration

Cooper PairsPurpose Help students visualizeCooper pairs.Materials two tennis balls joinedby a 20 cm stringCAUTION Throw the tennis ballsaway from students.Procedure Explain that thisdemonstration is limited in itsability to show electron-latticeinteractions but that it does showconservation of momentum(horizontally) for a pair of objectslinked together, such as the elec-trons in a Cooper pair. Hold oneof the tennis balls, whirl the freetennis ball over your head, andthen release the pair into the air.The whirling pair shows horizon-tal conservation of momentum(disregarding gravitationaleffects). The momentum of anindividual ball changes, but thepair acts as one object and travelsin a straight line.

www.scilinks.orgTopic: SuperconductorsCode: HF61478

Antimatter

Startling discoveries made in the twentieth century have confirmed that elec-

trons and other particles of matter have antiparticles. Antiparticles have the

same mass as their corresponding particle but an opposite charge.

The discovery of antiparticles

The discovery of antiparticles began in the 1920s with work by the theoreti-

cal physicist Paul Adrien Maurice Dirac (1902–1984), who developed a ver-

sion of quantum mechanics that incorporated Einstein’s theory of special

relativity. Dirac’s theory was successful in many respects, but it had one

major problem: its relativistic wave equation required solutions corre-

sponding to negative energy states. This negative set of solutions suggested

the existence of something like an electron but with an opposite charge,

just as the negative energy states were opposite to an electron’s typical

energy states. At the time, there was no experimental evidence of such

antiparticles.

In 1932, shortly after Dirac’s theory was introduced, evidence of the anti-

electron was discovered by the American physicist Carl Anderson. The anti-

electron, also known as the positron, has the same mass as the electron but is

positively charged. Anderson found the positron while examining tracks cre-

ated by electronlike particles in a cloud chamber placed in a magnetic field.

As described in the chapter “Magnetism,” such a field will cause moving

particles to follow curved paths. The direction in which a particle moves

depends on whether its charge is positive or negative. Anderson noted that

some of the tracks had deflections typical of an electron’s mass, but in the

opposite direction, corresponding to a positively charged particle.

Pair production and annihilation

Since Anderson’s initial discovery, the positron has been observed in a

number of experiments. In perhaps the most common process, a gamma

ray with sufficiently high energy collides with a nucleus, creating

an electron-positron pair. An example of this process, known

as pair production, is shown in Figure 1 on the next page.

During pair production, the energy of the photon is

completely converted into the rest energy and kinetic

energy of the electron and the positron. Thus, pair pro-

duction is a striking verification of the equivalence of

mass (rest energy) and other forms of energy as pre-

dicted by Einstein’s special theory of relativity. (This

equivalence is discussed in the appendix feature “The

Equivalence of Mass and Energy.”)

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Teaching TipThis feature discusses the discov-ery of antimatter particles, theprocesses of pair production andannihilation, and the productionof anti-hydrogen atoms in a par-ticle accelerator. The charge of ananti-electron, also known as apositron, is equal in magnitudebut opposite in sign to the chargeof an electron. The positron andelectron have equal masses andopposite spins. Antimatter wastheoretically predicted by Dirac inthe late 1920s. The anti-electronwas detected experimentally in1932, and the anti-proton andanti-neutron were detected in1955 and 1956, respectively.Today, it is believed that everyparticle has a correspondingantiparticle.

Teaching TipPoint out that charge must beconserved in pair production andannihilation. For example, in pairproduction, the initial photon haszero charge, and the total chargeof the positron and electron is alsozero. It is impossible for a photonto produce a single electron orpositron because this would vio-late the conservation of charge.

930 Appendix J

Teaching TipPoint out that the tracks shownin Figure 1 are created whencharged particles move through aliquid that is just below its boil-ing point. The chamber contain-ing the liquid is placed in amagnetic field. Charged particlescreate bubbles as they movethrough the liquid, and thesebubbles are photographed tocreate images like the one shown.Because the charges are in a mag-netic field, the direction in whicha particle moves depends on theparticle’s charge, and the curva-ture depends on the particle’smomentum. The particles shownin Figure 1 have equal and oppo-site momenta and charges.

EXTENSIONIn April 1997, researchersannounced the exciting discoveryof evidence of a gigantic streamof positrons at the center ofthe galaxy. This discovery was made by NASA’s Earth-orbiting Compton Gamma RayObservatory satellite. This satellitedetected 511 000 eV gamma raysthat are thought to have beenproduced in pair annihilation.The cause and nature of thispositron fountain are currentlybeing researched. Have interestedstudents research this discoveryand any recent developmentsrelated to the discovery, and havethem present their findings to theclass.

931

Once formed, a positron will most likely soon collide with an oppositely

charged electron in a process known as pair annihilation. This process is the

opposite of pair production—an electron-positron pair produces two pho-

tons. In the simplest example of pair annihilation, an electron and a positron

initially at rest combine with each other and disappear, leaving behind two

photons. Because the initial momentum of the electron-positron pair is zero,

it is impossible to produce a single photon. Momentum can be conserved

only if two photons moving in opposite directions, both with the same ener-

gy and magnitude of momentum, are produced.

Antimatter produced in a particle accelerator

After the positron was discovered, physicists began to search for the anti-

proton and anti-neutron. However, because the proton and neutron are

much more massive than the electron, a much greater amount of energy is

required to produce their antiparticles. By 1955, technological advances in

particle accelerators brought evidence of the anti-proton, and evidence of

the anti-neutron was found a year later.

The discovery of other antiparticles leads to the question of whether these

antiparticles can be combined to form antimatter and, if so, how that anti-

matter would behave. In 1995, physicists at the CERN particle accelerator in

Geneva, Switzerland, succeeded in producing anti-hydrogen atoms, that is,

atoms with a single anti-electron orbiting an anti-proton. Researchers

observed nine anti-hydrogen atoms during a three-week period. Unfortun-

ately, the anti-hydrogen atoms had a short lifetime—less than 37 billionths

of a second—because as soon as an anti-hydrogen atom encountered ordi-

nary matter, the two annihilated one another. Attempts to produce antimat-

ter for greater time periods are currently under way.

Figure 1The red and green spirals shownhere are the paths of a positron andan electron moving through a mag-netic field. Note that these pathshave about the same shape but areopposite in direction.


www.scilinks.orgTopic: AntiparticlesCode: HF60084

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BackgroundThis probeware version of theSkills Practice Lab “Free-FallAcceleration” from the chapter“Motion in One Dimension” usesa Vernier motion detector insteadof a motion timer. The motiondetector is easier to use than amotion timer and requires nocalibration.

Safety CautionRemind students to keep the areaclear during the experiment andto prevent the blocks from strik-ing their feet.

Tips and Tricks

• Students should have the pro-gram DataMate® on theirgraphing calculators. Refer toAppendix B for instructions.The motion detector must beclamped tightly so that it can’tmove during the experiment.Students should practice drop-ping the blocks so that theydon’t rotate. Remind studentsto keep their hands out of theway of the motion detector’ssignal.

• Remind students to keep atleast 0.5 m between the blockand the sensor to allow thesensor to work properly.

• Show the students how to usethe arrow keys to trace thegraph in step 12. Explain howto choose the points and howto find the difference betweenthe y-values of the points instep 12. Review what thegraphs of velocity and acceler-ation in steps 13 and 14 revealabout the motion of the block.Remind students to make surethe calculator is turned onbefore they perform each trial.

932 Appendix K: CBL™ Lab Procedures

Free-Fall Acceleration

MATERIALS LIST

• 3 wooden blocks ofdifferent masses

• balance• C-clamp• LabPro® or CBL2™

interface• masking tape

• meterstick• support stand with

V-clamp• thin foam pad• TI graphing calcula-

tor with link cable• Vernier motion

detector

You can use a motion detector and a LabPro orCBL2 interface instead of a recording timer to deter-mine velocity and acceleration. The motion detectormeasures the position of an object by sending soundwaves toward the object and measuring the time thatthe waves take to echo back to the sensor.

SAFETY• Tie back long hair, secure loose clothing, and

remove loose jewelry to prevent its getting caughtin moving or rotating parts.

• Attach masses securely. Falling or dropped massescan cause serious injury.

PROCEDUREPreparation

Follow Preparation steps 1–2 for the Skills PracticeLab “Free-Fall Acceleration” in the chapter “Motionin One Dimension.”

Apparatus Setup

3. Connect the LabPro or CBL2 interface to the calculator with the unit-to-unit link cable.Connect the motion detector to either of the twoDIG/SONIC ports on the interface.

4. Set up the apparatus as shown in Figure 1. Placethe ring stand near the edge of the lab table. Usethe C-clamp to clamp the base of the ring standsecurely to the table. Position the clamp so that itprotrudes as little as possible.

5. Using the V-clamp, securely clamp the motiondetector to the ring stand so that the detectorfaces down, over the edge of the table. Make surethe motion detector is far enough away from theedge of the table so the signal will not hit thetabletop, clamp, or table leg. Cover the floor

under the motion detector with a foam pad toreduce feedback.

6. Use a meterstick to measure a distance 0.5 mbelow the motion detector, and mark the pointwith tape on the ring stand. This point is thestarting position from which the blocks will bedropped from rest.

7. Start the program DataMate® on your graphingcalculator. Press CLEAR to reset the program.

8. Test to be sure the motion detector is positionedproperly.

a. Read the measurement displayed on the cal-culator screen for the distance between themotion detector and the floor. If the readingseems unusually low, adjust the motiondetector to make sure the signal is not hittingthe table.

b. Hold the wooden block directly beneath themotion detector, and move the block up anddown. Note the distance measurement dis-played on the calculator. Make sure themotion detector is not “seeing” other objects,such as the stand base, the tabletop, or thetable leg.

Speed and Acceleration of a Falling Object

9. Measure the mass of the first wooden block, andlabel the block with tape. Record the mass inyour data table. Hold the block horizontallybetween your hands, with your hands flat.Position the block directly below the motiondetector and level with the 0.5 m mark.

10. When the area is clear of people and objects, onestudent should select START to begin data collec-tion. When the motion detector begins to click,the student holding the block should wait twoseconds and then release the block by pullingboth hands out to the side. Releasing the blockthis way will prevent the block from twisting as itfalls, which may affect your results.

11. When the motion detector has stopped clicking,the graph selection screen will appear on the cal-culator. Press ENTER to plot a graph of the dis-tance in meters against time in seconds. (Note: Ifthe graph has spikes or black lines, repeat thetrial to obtain a smooth graph.)

• If the CBL2 yields values thatseem out of range, check thecalculator batteries.

CheckpointsStep 6: Make sure students real-ize that the motion detector is thereference point for all distancemeasurements. The tape markmust be measured below thedetector, not above the ground.

Step 8: Students should be ableto demonstrate that the detectoris working properly. They shouldbe able to explain why the dis-tance measurement increases asthe block moves away anddecreases as the block movestoward the detector.

Step 10: Make sure students candrop the block without letting itturn or getting their hands in theway.

Step 11: The graph should besmooth; dense, black lines orspikes signify “drop outs,” timeswhen the detector could notdetect the block. Check the setup,and have students practice drop-ping the block before repeatingthe trial.

Step 12: Students should identi-fy the portion of the curve thatrepresents the block’s motion.Guide students to choose fourpairs of consecutive points on thegraph.

Step 13: Students should beable to interpret the velocitygraph on the calculator. Theyshould be able to explain that thevelocity of the block increased asthe block fell.

Step 14: Students should beable to use the acceleration graphto explain whether the motion ofthe block meets their expecta-tions. If there are any problemswith the setup, students shouldbe alerted by the graphs.

933Appendix K: CBL™ Lab Procedures 933

15. Repeat this procedure using wooden blocks ofdifferent masses. Drop each block from the samelevel in each trial. Record all data for each trial.

16. Clean up your work area. Put equipment awaysafely so that it is ready to be used again.

ANALYSIS, CONCLUSIONS,AND EXTENSIONComplete the Analysis and Conclusions items forthe Skills Practice Lab “Free-Fall Acceleration.” Yourteacher may also instruct you to complete theExtension exercise. (Note: For Analysis item 1, use0.02 s for the average period of the timer.)

12. Use the arrow keys to trace along the curve. Onthe far left and the far right, the curve representsthe position of the block before and after itsmotion. The middle section of the curve repre-sents the motion of the falling block. Sketch thisgraph in your lab notebook.

a. Choose a point on the curve at the begin-ning of this middle section. Press the rightarrow key once to select the next point. Thedifference between the x-values for thesetwo readings will be 0.05 s, the time intervalbetween successive readings. Find the differ-ence between the y-values of these twopoints, and record it as A–B for Trial 1 inyour data table.

b. Press the right arrow key twice to move toanother point on the curve. Press the rightarrow key once more to select the next point.Find the difference between the y-values ofthese two points, and record it as C–D forTrial 1 in your data table.

c. Choose two more pairs of points along thecurve. Select points at even intervals alongthe curve so that the first pair is at the begin-ning of the block’s motion and the last pairis at the end. Find the difference between they-values of each pair of points. Record themas E–F and G–H for Trial 1. Press ENTER toreturn to the graph selection screen.

13. Press the down arrow key once, and then pressENTER to display a graph of the velocity in m/sagainst time in seconds. Sketch this graph inyour lab notebook. Press ENTER to return to thegraph selection screen.

14. Press the down arrow key once and then pressENTER to display a graph of the acceleration inm/s2 against time in seconds. Sketch this graphin your lab notebook. Press ENTER to return tothe graph selection screen. Select MAIN SCREEN

to return to the main screen of DataMate®.

Figure 1

Step 8b: While you read the distance measurements dis-played on the calculator, move the wooden block up anddown below the motion detector to check the readings.Step 9: Hold the block flat and parallel to the tape mark.Step 10: Release the block by pulling hands straight tothe sides. It may take some practice to release the blockso that it falls straight down without turning. If the blockturns while falling, the motion detector will measure thedistance to the closest part of the block and introduceerror into your results.

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BackgroundThis probeware version of theSkills Practice Lab “Force andAcceleration” from the chapter“Forces and the Laws of Motion”uses a Vernier motion detectorinstead of a motion timer. Themotion detector is easier to usethan a motion timer and requiresno calibration. This version alsoadds a Vernier dual-range forcesensor, which allows students notonly to calculate force from themass but also to measure forcedirectly.

Safety CautionRemind students to fasten massessecurely, to keep the area clearduring the experiment, and toprevent the cart from falling offthe table.

Tips and Tricks

• Narrow, smooth drapery cordworks better than ordinarystring or twine.

• Students should have the pro-gram DataMate® on their cal-culators. Refer to Appendix Bfor instructions. The motiondetector must be stabilized sothat it can’t move.

• Show the students how to usethe arrow keys to trace thegraph. Explain how to choosethe points in step 14 and howto find the difference betweenthe x- and y-values in step 15.

CheckpointsStep 6: Make sure that allclamps are placed so that theyprotrude as little as possible. Ifthe sensor has multiple settings,set it to 10 N. Warn students notto pull on the force sensor.


Force and Acceleration

SAFETY• Tie back long hair, secure loose clothing, and

remove loose jewelry to prevent its getting caughtin moving or rotating parts.

• Attach masses securely. Falling or dropped massescan cause serious injury.


Follow Preparation steps 1–3 for the Skills PracticeLab “Force and Acceleration” in the chapter “Forcesand the Laws of Motion.”

Apparatus Setup

4. Connect the LabPro or CBL2 interface to the cal-culator with the unit-to-unit link cable. Connectthe Dual-Range Force Sensor to the CH1 port onthe interface, and set the switch on the sensor to10N. Connect the motion detector to theDIG/SONIC 1 port on the interface.

5. Turn on the calculator, and start the DataMate®program. Press CLEAR to reset the program.

6. Set up the apparatus as shown in Figure 1.Securely tape the force sensor to the dynamicscart. Tape the poster board to the opposite end ofthe dynamics cart to make a flat, vertical surface.Clamp the pulley to the table edge using a tableclamp, rod, and parallel clamp so that the pulleyis level with the force sensor hook. Position themotion detector so that the cart will move awayfrom it in a straight line. Securely clamp the

motion detector to the ring stand. Place a pieceof tape 0.5 m in front of the motion detector toserve as a starting line for the cart. (Note: Do notpull on the force sensor.)

Constant Mass with Varying Force

7. Carefully measure the mass of the cart assemblyon the platform balance, making sure that thecart does not roll or fall off the balance. Then,load it with masses equal to 0.60 kg. Lightly tapethe masses to the cart to hold them in place.

8. Attach one end of the cord to a small mass hang-er and the other end of the cord to the force sen-sor. Pass the cord over the pulley, and fasten asmall mass to the end to offset the frictionalforce on the cart. The mass is correct when thecar moves forward with a constant velocity whenyou give it a push. The car will have constantvelocity for only a short period after it is pushed,then it will accelerate as the counterweightdrops. This counterweight should stay on the cordthroughout the entire experiment. Add the massof the counterweight to the mass of the cart andmasses, and record the sum as Total Mass in yourdata table.

9. For the first trial, remove a 0.10 kg mass from thecart, and securely fasten it to the end of the cordalong with the counterweight. Record 0.10 kg asthe Accelerating Mass in the data table.

10. Place the cart so that the poster-board end isclosest to the motion detector and is lined upwith the tapeline, 0.5 m in front of the motiondetector. Keep the force sensor cord clear so thatthe cart will be able to move freely.

11. Make sure that the calculator is turned on. Makesure that the area under the falling mass is clearof obstacles. Select START to begin collectingdata, and release the cart simultaneously. Themotion detector will begin to click as it collectsdata.

12. Carefully stop the cart when the 0.10 kg masshits the floor. Do not let the cart fall off the table.

13. When the motion detector has stopped clicking,the graph selection screen will appear on the cal-culator. Press ENTER to plot a graph of the forcesensor reading against time. Use the arrow keys

MATERIALS LIST• balance

• calibrated massesand holder

• cord, smooth

• dynamics cart

• hooked mass, 1000 g

• LabPro® or CBL2™interface

• mass hanger

• meterstick

• pulley with tableclamp

• rod and parallelclamp

• square of poster board, 25 cm � 25 cm

• support stand withV-jaw clamp

• tape

• TI graphing calcula-tor with link cable

• Vernier dual-rangeforce sensor

• Vernier motiondetector

Step 8: Make sure that massesare securely attached. Studentsshould be able to demonstratethat the counterweight allows thecart to move at a constant veloci-ty when the cart is given a smallpush.

Step 10: The force sensor cordmust be kept out of the path ofthe cart; it must also be prevent-ed from dragging behind the cartand slowing it down.

Step 12: It may help to clamp awood block to the edge of thetable to serve as a bumper to keepthe cart from rolling off the table.

Step 13: The graph will not be aperfectly straight line. Studentsmay be confused by the valuesabove and below the line. Guidestudents to select the most fre-quently appearing value.

Step 14: Students should identi-fy the part of the curve thatshows the motion of the cart. Ifthe curve does not show the cartmoving away from the detector,make sure that the detectorpoints toward the cart with nointerference. If there is interfer-ence from the tabletop, tilt thedetector slightly upward. Coverthe tabletop and any objects onthe table, such as gas jets orfaucets, with soft cloths.

Step 15: Students should beable to interpret the velocity andacceleration graphs on the graph-ing calculator. Students should beable to use the graphs to explainwhether the motion of the cartmeets their expectations. If thereare any problems with the setup,the students should be alerted bythe graphs.

Step 18: Students should recog-nize the difference between thesetwo trials and the previous trials.They must add mass to the carteach time without changing themass on the cord.


17. Leave the 0.20 kg mass on the end of the cord,and attach the 0.10 kg mass from the cart secure-ly to the end of the cord. Repeat the procedurefor Trial 3.

Constant Force with Varying Mass

18. For the two trials in this part of the experiment,keep 0.30 kg and the counterweight on the cord.Be sure to include this mass when recording thetotal mass for these trials.

19. Add a 0.50 kg mass to the cart. Tape the mass tothe cart to keep it in place. Run the experimentand record the total mass, accelerating mass,accelerating force, distance, and time under Trial4 in your data table.

20. Tape a 1.00 kg mass to the cart, and repeat theprocedure. Record the data under Trial 5 in yourdata table.


ANALYSIS, CONCLUSIONS,AND EXTENSIONComplete the Analysis and Conclusions items forthe Skills Practice Lab “Force and Acceleration.”Your teacher may also instruct you to complete theExtension exercise. (Note: For Analysis item 1, donot enter new values for Accelerating Force in yourdata table. Instead, compare your calculated valueswith the values you entered earlier.)

to trace along the curve. The y-value is the forcein newtons; it should be fairly constant. Recordthis value as the Accelerating Force in the datatable. Press ENTER to return to the graph selec-tion screen.

14. Press the down arrow key once, and then pressENTER to display a graph of the distance inmeters against time. Use the arrow keys to tracealong the curve. On the far left and the far right,the flat portion of the curve represents the posi-tions of the cart before and after its motion. Themiddle section of the curve represents themotion of the cart. Choose a point on the curvenear the beginning of this middle section (butnot the beginning point itself), and chooseanother point near the end.

15. Find the difference between the y-values of thesetwo points, and record it as the Distance forTrial 1 in your data table. Find the differencebetween the x-values for these two readings tofind the time elapsed between measurements.Record this as the Time Interval for Trial 1 inyour data table. Press ENTER to return to thegraph selection screen.

a. Press the down arrow key once, and thenpress ENTER to display a graph of the veloci-ty in m/s against time. Press ENTER toreturn to the graph selection screen.

b. Press the down arrow key once, and thenpress ENTER to display a graph of the accel-eration in m/s2 against time. Press ENTER toreturn to the graph selection screen. SelectMAIN SCREEN to return to the main screen.

16. Replace the 0.10 kg mass in the cart. Remove the0.20 kg mass from the cart, and attach it securelyto the end of the cord. Repeat the procedure forTrial 2.

Figure 1Step 6: Make sure that the motion detector has a clearview of the cart. For all trials, start the cart from the sameposition.Step 13: Using the data collection interface with thegraphing calculator will allow you to view graphs of themotion after each trial.

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BackgroundThis probeware version of theSkills Practice Lab “Specific HeatCapacity” from the chapter“Heat” uses temperature probesinstead of standard thermome-ters. Because temperature probesare not prone to shattering, theyare safer than thermometers.

Safety CautionThis lab presents several safetyhazards. Remind students neverto leave a hot plate unattendedwhile it is on. Make sure all stu-dents wear the appropriate safetygear at all times.

• Each group must have at leasttwo students. Groups must per-form more than one step at atime, so groups may be largerthan usual. With larger groups,make sure all students areinvolved and paying attentionto safety.

• Bringing water to a boil takes10–20 min; other tasks can becompleted as the water heats.

• Each group must have a levelwork surface large enough sothat students are not too closeto the hot plates.

• Students should place wetmetal shot in designated con-tainers at the end of the lab tobe dried and used again.

Tips and Tricks

• Students should have the pro-gram DataMate® on theirgraphing calculators. Refer toAppendix B for instructions.Students should begin collect-ing data for the calorimeterbefore adding the metal sam-ple to the calorimeter. Whenadding the metal to thecalorimeter, don’t let it touchthe sensor in the water.


Specific Heat Capacity

MATERIALS LIST

• beakers, 2• balance• hot plate• ice


• metal calorimeterand stirring rod

• metal heating vesselwith metal heatingdipper

• samples of variousmetals in shot orbead form

• small plastic dish• stainless steel tem-

perature probe• TI graphing calcula-

tor with link cable

SAFETY• When using a burner or hot plate, always wear

goggles and an apron to protect your eyes andclothing. Tie back long hair, secure loose clothing,and remove loose jewelry. If your clothing catcheson fire, walk to the emergency lab shower and usethe shower to put out the fire.

• Never leave a hot plate unattended while it isturned on.

• Do not heat glassware that is broken, chipped, orcracked. Use tongs or a mitt to handle heatedmetal, glassware, and other equipment because itdoes not always look hot when it is hot. Allow allequipment to cool before storing it.

• Never put broken glass or ceramics in a regularwaste container. Use a dustpan, brush, and heavygloves to carefully pick up broken pieces and dis-pose of them in a container specifically providedfor this purpose.


Follow Preparation steps 1–3 for the Skills PracticeLab “Specific Heat Capacity” in the chapter “Heat.”

Finding the Specific Heat Capacity of a Metal

4. Choose a location where you can set up theexperiment away from the edge of the table andaway from other groups. Make sure the hot plateis in the “off” position before you plug it in.

5. Fill a metal heating vessel with 200 mL of water,and place it on the hot plate. Turn on the hotplate, and adjust the heating controls to heat thewater.

6. Set up the calculator and interface for data collection:

a. Connect the LabPro or CBL2 interface tothe calculator with the unit-to-unit linkcable.

b. Connect the temperature probe to the CH1port on the interface. Turn on the calculatorand start the DataMate® program.

c. Press CLEAR to reset the program.

d. Hold the temperature probe in the air withnothing touching it. Observe the tempera-ture readings displayed on the calculator.When the readings are stable, record the dis-played temperature in your data table as theroom temperature.

7. Obtain about 100 g of the metal shot. First, meas-ure the mass of the small plastic dish. Place themetal shot in the dish, and determine the mass ofthe shot. Record the number and the mass of theshot in your data table. Place the temperatureprobe in the metal heating dipper, and carefullypour the shot into the metal heating dipper. Makesure the temperature probe is surrounded by themetal shot.

8. Place the dipper containing the shot into the topof the heating vessel, as shown in Figure 1. Makesure the temperature probe cable does not touchthe hot plate or any heated surface.

9. While the sample is heating, find the mass of theempty inner cup of the calorimeter and the stir-ring rod. Record the mass in your data table. Donot leave the hot plate unattended.

10. For the water in the calorimeter, you will needabout 100 g of cold water. The water must becolder than room temperature. (Do not usewater colder than 5°C below room temperature.If ice is used to cool the water, make sure all theice has melted before pouring the water into thecalorimeter.)

• Demonstrate how to carefullypour the shot into the heatingdipper so that it surrounds thetemperature probe. The probemust not touch the dipper’ssides.

CheckpointsStep 4: All hot plates and con-tainers of liquids must be keptaway from the edge of the table.Metal shot must be kept in a con-tainer at all times.

Step 5: The hot plate must beturned up to the highest level.

Step 6: The graphing calculator,interface, and cords must be keptaway from the hot plate.

Step 8: Students must exercisecare when placing the heating dip-per into the heating vessel.

Step 9: For the next few steps,make sure the hot plate is not leftunattended.

Step 10: Only a small amountof ice is required; if too much isused, it should be removedbefore students measure the massof the water.

Step 16: Students should usemitts when handling the hotmetal heating dipper. Remindstudents not to let the hot metalshot touch the temperatureprobe in the calorimeter.

Step 17: Students should beable to use the graph to find thetemperature at specific timesduring the experiment.

Step 18: If students are notdoing more trials, make sure allhot plates are turned off. If theyremain on, make sure that heatingvessels have enough water in themand that hot plates are attendedto. The vessel, water, and hot plateare still hot; student groupsshould move away from the hotplate and exercise caution.


16. Quickly transfer the metal shot to the cold waterin the calorimeter and replace the cover. Use amitt when handling the metal heating dipper.Use the stirring rod to gently agitate the sampleand to stir the water in the calorimeter. If you arenot doing any more trials, make sure the hotplate is turned off. Otherwise, make sure there isplenty of water in the heating vessel, and do notleave the hot plate unattended.

17. When data collection is finished, a graph of tem-perature and time will be displayed. Time in sec-onds is graphed on the x-axis, and thetemperature readings are graphed on the y-axis.Use the arrow keys to trace along the curve.Record the highest temperature reading dis-played on your graph in the data table. PressENTER to return to the main screen.

18. If time permits, perform additional trials withother samples. Record data for all trials in yourdata table.

ANALYSIS, CONCLUSIONS,AND EXTENSIONComplete the Analysis and Conclusions items forthe Skills Practice Lab “Specific Heat Capacity.”Yourteacher may also instruct you to complete theExtension exercise.

11. Place the calorimeter and stirrer on the balance,and carefully add 100 g of the water. Record themass of the water in your data table. Replace thecup in its insulating shell, and cover.

12. Determine the temperature of the metal shot byobserving the temperature readings in degreesCelsius displayed on the calculator. When the read-ings are stable, record the displayed temperature inyour data table as the metal’s initial temperature.

13. Carefully remove the temperature probe fromthe dipper.

14. Use the stirring rod to stir the water in thecalorimeter. Place the temperature probe in thecalorimeter. Determine the initial temperature ofthe water in the calorimeter by observing thetemperature readings displayed on the calcula-tor. When the readings are stable, record the dis-played temperature in your data table as thecalorimeter’s initial temperature. Leave theprobe in the calorimeter.

15. From the DataMate® main screen, select START

to begin collecting the temperature readings forthe water in the calorimeter.

Figure 1

Step 5: Start heating the water beforeyou set up the calculator and temperatureprobe. Never leave a hot plate unattendedwhen it is turned on.Step 7: Be very careful when pouring themetal shot into the dipper around thetemperature probe.Step 15: Begin taking temperature read-ings a few seconds before adding the shotto the calorimeter.Step 17: Record the highest temperaturereached by the water, shot, and calorime-ter combination, not the final temperature.

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BackgroundThis probeware version of theSkills Practice Lab “Speed ofSound” from the chapter“Sound” uses a microphone,which is positioned above anopen cardboard tube, instead of a resonance apparatus.

Tips and Tricks

• Students should have the pro-gram DataMate® on theirgraphing calculators. Refer toAppendix B for instructions.The sound used should not betoo loud, because a loudsound causes the graph tohave too many peaks. Thegraphs for different types ofsounds will look very similar;the tube is dominated by itsown low-resonance frequencyfor most sounds.

• Show students how to identifythe second peak on the graphand how to determine whetherthe sound was too loud.

CheckpointsStep 5: Make sure students weargoggles if there are clamps andsupport rods at eye level.

Step 7: Remind students towork quietly; random noises mayaffect other groups’ results. Anyquiet, sharp sound will workwell. After making the initialsound, students must be silentuntil the CBL2 displays DONE.

Step 8: Students should realizethat the sound traveled twice thelength of the tube during thetime interval.

Step 9: Guide students to findthe time values for each peak.


Speed of Sound

MATERIALS LIST• cardboard tube• LabPro® or CBL2™

interface

• masking tape

• meterstick

• stainless-steel tem-perature probe

• support stand withclamp


• Vernier microphone

In this lab, you will determine the speed of sound.You will place a microphone directly above theopening of a large tube, where the microphone willrecord a short, sharp noise. After the sound travelsdown the tube and reflects back, the microphonewill record the sound again. You can use the timebetween recordings and the distance that the soundtraveled to determine the speed of sound in air.

SAFETY• Put on goggles before you install clamps at eye

level.

PREPARATIONFollow Preparation steps 1–2 for the Skills PracticeLab “Speed of Sound” in the chapter “Sound,”but for the data table, label the first through fourth columns Trial, Temperature (˚C), Distancefrom microphone to bottom of tube (m), and Timeinterval(s).

PROCEDUREFinding the Speed of Sound

3. Connect the LabPro or CBL 2 interface to the cal-culator with the unit-to-unit link cable. Connectthe microphone to the CH1 port on the interface.Connect the temperature probe to the CH2 porton the interface.

4. Turn on the calculator, and start the DataMate®program. Observe the temperature readings dis-played on the calculator. When the readings are

stable, record the displayed temperature in yourdata table. Unplug the temperature probe, andpress CLEAR to reset the program.

5. Set up the microphone, ring stand, and tube asshown in Figure 1. Tape or clamp the tubesecurely in place. Clamp the microphone to theedge of the table or to a ring stand so that themicrophone points down and is directly abovethe open end of the tube.

6. Set up the calculator for data collection.

a. Select SETUP from the main screen.

b. Press the up arrow key once to select MODE

and press ENTER.

c. Select TIME GRAPH from the SELECT MODE

screen.

d. Select ADVANCED from TIME GRAPH

SETTINGS.

e. Select CHANGE TRIGGERING from ADV.TIME GRAPH SETTINGS.

f. Select CH1-MICROPHONE from SELECT

TRIGGERING.

g. Select INCREASING from TRIGGER TYPE.

h. Enter “0.1” for a trigger threshold.

i. Enter “0” for the pre-store.

j. Select OK three times to return to the mainscreen.

7. Select START to prepare for data collection. Make aloud, short noise—such as a snap of the fingers—directly above the tube. This noise will trigger theinterface to collect the sound data.

8. Use the meterstick to measure the length fromthe bottom of the microphone to the bottom ofthe tube. Record this length to the nearest mil-limeter in the data table.

ANSWERS

Analysis1. Sample data:distance = 1.962 m

2. Sample data:Trial 1: v = 327.0 m/sTrial 2: v = 280.3 m/sTrial 3: v = 392.4 m/sTrial 4: v = 327.0 m/s


9. Look at the graph on the graphing calculator,which shows the sound plotted against time inseconds. There should be two peaks on thegraph, one near the beginning and one a littlelater. The first peak is the sound, and the secondpeak is the echo of the sound. Use the arrow keysto trace the graph. (Note: If the graph has spikesor black lines, repeat the trial to obtain a smoothgraph.)

10. Find the difference between the x-values of thetwo peaks to find the time interval between them.Record the time interval in your data table. Sketchthe graph in your lab notebook. Press ENTER onthe calculator to return to the main screen.

11. Repeat the procedure for several trials. Try differ-ent sounds, such as a soft noise, a loud noise, ahigh-pitched sound, and a low-pitched sound.Record all data in your data table.

12. Clean up your work area. Put equipment awaysafely so that it is ready to be used again. Recycleor dispose of used materials as directed by yourteacher.

ANALYSIS1. Organizing Data For each trial, multiply the

measured distance by 2 to find the total distancethat the sound traveled.

2. Organizing Data Use the values for the dis-tance traveled and the time interval from yourdata table to find the speed for each trial.

CONCLUSIONSComplete the Conclusions items for the SkillsPractice Lab “Speed of Sound” in the chapter“Sound.”

Figure 1Step 7: The interface will begin collecting sound data assoon as you make a sound, so work quietly until you areready to begin the experiment. Remain quiet until datacollection has finished. Background noise may affect yourresults.Step 9: On the graph, the first and second peaks may notbe the same height, but they should both be noticeablyhigher than the other points on the graph. If the soundwas too loud, the graph will show many high and lowpoints. Repeat with a softer sound for better results.

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BackgroundThis probeware version of theSkills Practice Lab “MagneticField of a Conducting Wire”from the chapter “Magnetism”uses a voltage probe instead of astandard ammeter or multimeter.This version also adds a magneticfield sensor, which allows stu-dents to measure the strength ofthe magnetic field (B).

Tips and Tricks

• Students should have the pro-gram DataMate® on theirgraphing calculators. Refer toAppendix B for instructions.Zeroing the magnetic fieldsensor in step 9 will set thesensor to read Earth’s magnet-ic field strength as zero, so thereadings taken during the labwill not include Earth’s mag-netic field.

• Show students how to set upthe wire coil on the gal-vanometer apparatus and howto use tape to mark the refer-ence circle for the magneticfield sensor. Make sure stu-dents know the correct posi-tion of the magnetic fieldsensor.

CheckpointsStep 5: Make sure that thepower supply, resistor, and switchare connected properly and are atthe proper settings. Studentsshould be able to demonstratethat all connections are properlymade.

Step 8: Make sure that the mag-netic field sensor is securelyclamped in the right position.


Magnetic Field of a Conducting Wire

MATERIALS LIST• alligator clips• galvanometer


• masking tape

• power supply

• resistor

• support stand andburet clamp

• double-pole, double-throw switch


• Vernier magneticfield sensor

• voltage probe

• insulated connectingwires and bare cop-per wire

In this lab, you will construct a circuit with a current-carrying wire and will use a magnetic field sensor toinvestigate the relationship between the magneticfield and the current in the wire. You will be able todetermine the magnitude and direction of the mag-netic field that surrounds the wire.

SAFETY• Never close a circuit until it has been approved

by your teacher. Never rewire or adjust any ele-ment of a closed circuit. Never work with electrici-ty near water; be sure the floor and all worksurfaces are dry.

• Do not attempt this exercise with any batteries,electrical devices, or magnets other than thoseprovided by your teacher for this purpose.

• Wire coils may heat up rapidly during this experi-ment. If heating occurs, open the switch immedi-ately and handle the equipment with a hot mitt.Allow all equipment to cool before storing it.


1. Read the Safety cautions for the Skills PracticeLab “Magnetic Field of a Conducting Wire” inthe chapter “Magnetism.”

2. Read the entire lab, and plan what steps you willtake. Prepare a table with 4 columns and 7 rows.In the first row, label the columns Trial, CurrentDirection, ΔVR (V), and BMeasured (T). Label thesecond through seventh rows as 1 through 6.

Magnetic Field Strength

3. Set up the apparatus as shown in Figure 1. Use 1 m of copper wire to make a square loop aroundthe coil support pins on the galvanometer appa-ratus. Attach alligator clips to the ends of thewire. Label one clip A, and label the other B.Place the galvanometer apparatus so that you arefacing the plane of the coil.

4. Use masking tape to mark a line on the stand ofthe galvanometer directly under the top of thecoil. Make another tape line perpendicular to thefirst, as shown in Figure 1. The two tape segmentsshould cross in the middle of the apparatus. Onthe second tape, on the side away from you, marka point 2 cm from the center. Using this point asthe center point, draw a circle with a 1 cm radius.

5. Construct a circuit that contains the power sup-ply and a 1 Ω resistor wired in series through themiddle set of posts on the switch. Place theswitch so that it moves from left to right.Connect the front right post of the switch to theend of the coil marked A, and connect the rearright post of the switch to the end of the coilmarked B. Now, connect the front left post of theswitch to the end of the coil marked B, and con-nect the rear left post of the switch to the end ofthe coil marked A. Do not close the switch orturn on the power supply until your teacher hasapproved your circuit.

6. Connect the interface to the calculator with theunit-to-unit link cable. Connect the voltageprobe to the CH1 port and the magnetic fieldsensor to the CH2 port on the interface. Set theswitch on the magnetic field sensor to HIGH.Connect the voltage probe to measure the volt-age across the resistor.

7. Turn on the calculator, and start the DataMate®program. Press CLEAR to reset the program.

8. Set up a support stand with a buret clamp tohold the magnetic field sensor vertically. Positionthe magnetic sensor securely so that the whitedot is facing you and the sensor is directly abovethe 1 cm circle marked on the tape.

Step 11: Some students mayneed help determining theCurrent Direction. Studentsshould be able to demonstratethat they have recorded the cor-rect direction.

ANSWERS

Analysis1. Student answers will vary.For sample data, values rangefrom I = 1.07 A to I = 3.10 A.

2. Student graphs should showthat the magnetic field strengthincreases as the current increases.

Conclusions3. For each position, the mag-netic field strength increases asthe current increases.

4. When the current directionchanges, the direction of themagnetic field also changes.


14. Repeat step 13 with the power supply set to one-third of the maximum setting. Record all data inyour data table as Trials 5 and 6.


ANALYSIS1. Organizing Data For each trial, use the equa-

tion ΔV = IR to find the current.

2. Constructing Graphs Use the data fromTrials 1, 3, and 5 to plot a graph of Bwire in teslasagainst the current in amperes in the circuit.Also, plot graphs for Trials 2, 4, and 6. Use acomputer, graphing calculator, or graph paper.

CONCLUSIONS3. Drawing Conclusions For each position, what

is the relationship between the current in thewire loop and the magnetic field strength?

4. Drawing Conclusions What is the relation-ship between the direction of current in the wireand the direction of the magnetic field? Explain.

9. Set the magnetic field sensor to read zero.

a. Select SETUP from the main screen ofDataMate®.

b. Select ZERO from the setup screen.

c. Select CH2-MAGNET F(MT) from theSELECT CHANNEL menu.

d. Monitor the reading displayed on the calcu-lator screen. When the reading appears to bestable, press ENTER to zero the sensor.

e. Keep the magnetic field sensor in this sameposition for the remainder of the experiment.

10. Make sure the dial on the power supply is turnedcompletely counterclockwise. When your teacherhas approved your circuit, turn the dial on thepower supply about halfway to its full value.

11. Close the switch briefly. Read the potential dif-ference across the resistor and the strength of themagnetic field. Open the switch as soon as youhave made your observations. Record ΔVR (V)and BMeasured (T) for Trial 1 in your data table.Determine and record the Current Direction (A to B or B to A).

12. Reverse the direction of the current by closingthe switch in the opposite direction. Read andrecord the potential difference and the strengthof the magnetic field for Trial 2. Open the switchas soon as you have made your observations.Determine and record the Current Direction (A to B or B to A).

13. Increase the setting on the power supply to abouttwo-thirds of the maximum setting on the dial.Repeat the procedure in steps 11 through 12.Record all data in your data table as Trials 3 and 4.

Figure 1

Step 3: Loop the copper wire around the support pins,and attach alligator clips to the ends. Place the galvanome-ter with one support pin on the left and one on the right.Step 4: Use two pieces of tape to mark perpendicularlines, and mark a circle to use as a reference for placingthe sensor.Step 5: Place the switch in front of you so that it movesfrom left to right. Check all connections carefully.

33. 1.51 h

35. a. 2.00 minb. 1.00 minc. 2.00 min

37. 931 m

39. −26 m/s; 31 m

41. 1.6 s

43. 5 s; 85 s; +60 m/s

45. −1.5 × 103 m/s2

47. a. 3.40 sb. −9.2 m/sc. −31.4 m/s; −33 m/s

49. a. 4.6 s after stock car starts

b. 38 mc. +17 m/s (stock car),

+21 m/s (race car)

51. 4.44 m/s

CHAPTER

Practice A, p. 891. a. 23 km

b. 17 km to the east

3. 15.7 m at 22° to the sideof downfield

Practice B, p. 921. 95 km/h

3. 21 m/s, 5.7 m/s

Practice C, p. 941. 49 m at 7.3° to the right of

downfield

3. 13.0 m at 57° north ofeast

Practice D, p. 991. 0.66 m/s

3. 7.6 m/s

Practice E, p. 1011. yes, Δy = −2.3 m

3. 2.0 s; 4.8 m

3

Practice B, p. 491. 2.2 s

3. 5.4 s

5. a. 1.4 m/sb. 3.1 m/s

Practice C, p. 531. 21 m

3. 9.1 s

Practice D, p. 551. 9.8 m/s; 29 m

3. −7.5 m/s; 19 m

Practice E, p. 581. +2.51 m/s

3. a. 16 m/sb. 7.0 s

5. +2.3 m/s2

Practice F, p. 641. a. −42 m/s

b. 11 s

3. a. 8.0 m/sb. 1.63 s

2 Review, pp. 68–731. 5.0 m; +5.0 m

3. t1: negative; t2: positive;t3: positive; t4: negative;t5: zero

7. 10.1 km to the east

9. a. +70.0 mb. +140.0 mc. +14 m/sd. +28 m/s

11. 0.2 km west of the flagpole

17. 0.0 m/s2; +1.36 m/s2;+0.680 m/s2

19. 110 m

21. a. −15 m/sb. −38 m

23. 17.5 m

25. 0.99 m/s

31. 3.94 s

CHAPTER

Practice A, p. 151. 5 × 10−5 m

3. a. 1 × 10−8 mb. 1 × 10−5 mmc. 1 × 10−2 μm

5. 1.440 × 103 kg

1 Review, pp. 27–3111. a. 2 × 102 mm

b. 7.8 × 103 sc. 1.6 × 107 μgd. 7.5 × 104 cme. 6.75 × 10−4 gf. 4.62 × 10−2 cmg. 9.7 m/s

13. 1.08 × 109 km

19. a. 3b. 4c. 3d. 2

21. 228.8 cm

23. b, c

29. 4 × 108 breaths

31. 5.4 × 108 s

33. 2 × 103 balls

35. 7 × 102 tuners

37. a. 22 cm; 38 cm2

b. 29.2 cm; 67.9 cm2

39. 9.818 × 10−2 m

41. The ark (6 × 104 m3) was about 100 times as large as a typical house(6 × 102 m3).

43. 1.0 × 103 kg

45. a. 0.618 g/cm3

b. 4.57 × 1016 m2

CHAPTER

Practice A, p. 441. 2.0 km to the east

3. 680 m to the north

5. 0.43 h

2

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Advanced Topics in Physics

943Selected Answers 943

29. 51 N

35. 0.70, 0.60

37. 0.816

39. 1.0 m/s2

41. 13 N down the incline

43. 64 N upward

45. a. 0.25 m/s2 forwardb. 18 mc. 3.0 m/s

47. a. 2 sb. The box will never

move. The force exertedis not enough to over-come friction.

49. −1.2 m/s2; 0.12

51. a. 2690 N forwardb. 699 N forward

53. 13 N, 13 N, 0 N, −26 N

CHAPTER

Practice A, p. 1621. 1.50 × 107 J

3. 1.6 × 103

Practice B, p. 1661. 1.7 × 102 m/s

3. the bullet with the greatermass; 2 to 1

5. 1.6 × 103 kg

Practice C, p. 1681. 7.8 m

3. 5.1 m

Practice D, p. 1721. 3.3 J

3. a. 785 Jb. 105 Jc. 0.00 J

Practice E, p. 1771. 20.7 m/s

3. 14.1 m/s

5. 0.18 m

5

55. a. 32.5 mb. 1.78 s

57. a. 57.7 km/h at 60.0° westof the vertical

b. 28.8 km/h straight down

59. 18 m; 7.9 m

61. 6.19 m/s downfield

CHAPTER

Practice B, p. 1281. Fx = 60.6 N; Fy = 35.0 N

3. 557 N at 35.7° west ofnorth

Practice C, p. 1321. 2.2 m/s2 forward

3. 4.50 m/s2 to the east

5. 14 N

Practice D, p. 1391. 0.23

3. a. 8.7 × 102 N, 6.7 × 102 Nb. 1.1 × 102 N, 84 Nc. 1 × 103 N, 5 × 102 Nd. 5 N, 2 N

Practice E, p. 1411. 2.7 m/s2 in the positive

x direction

3. a. 0.061b. 3.61 m/s2 down the

ramp

4 Review, pp. 145–14911. a. F1 (220 N) and F2

(114 N) both pointright; F1 (220 N) pointsleft, and F2 (114 N)points right.

b. first situation: 220 N tothe right, 114 N to theright; second situation:220 N to the left, 114 Nto the right

21. 55 N to the right

4

Practice F, p. 1051. 0 m/s

3. 3.90 m/s at (4.0 × 101)°north of east

3 Review, pp. 108–1137. a. 5.20 m at 60.0° above

the positive x-axisb. 3.00 m at 30.0° below

the positive x-axisc. 3.00 m at 150° counter-

clockwise from the posi-tive x-axis

d. 5.20 m at 60.0° belowthe positive x-axis

9. 15.3 m at 58.4° south ofeast

19. if the vector is oriented at45° from the axes

21. a. 5 blocks at 53° north ofeast

b. 13 blocks

23. 61.8 m at 76.0° S of E (or Sof W), 25.0 m at 53.1° S ofE (or S of W)

25. 2.81 km east, 1.31 kmnorth

31. 45.1 m/s

33. 11 m

35. a. clears the goal by 1 mb. falling

37. 80 m; 210 m

41. a. 70 m/s eastb. 20 m/s

43. a. 10.1 m/s at 8.53° east ofnorth

b. 48.8 m

45. 7.5 min

47. a. 41.7 m/sb. 3.81 sc. vy,f = −13.5 m/s,

vx,f = 34.2 m/s,vf = 36.7 m/s

49. 10.5 m/s

51. a. 2.66 m/sb. 0.64 m

53. 157 km

37. a. 0.0 kg•m/sb. 1.1 kg•m/s upward

39. 23 m/s

41. 4.0 × 102 N

43. 2.36 × 10−2 m

45. 0.413

47. −22 cm/s, 22 cm/s

49. a. 9.9 m/s downwardb. 1.8 × 103 N upward

CHAPTER

Practice A, p. 2361. 2.5 m/s

3. 1.5 m/s2

Practice B, p. 2381. 29.6 kg

3. 40.0 N

Practice C, p. 2421. 0.692 m

3. a. 651 Nb. 246 Nc. 38.5 N

Practice D, p. 2511. Earth: 7.69 × 103 m/s,

5.51 × 103 s; Jupiter:4.20 × 104 m/s,1.08 × 104 s; moon:1.53 × 103 m/s,8.63 × 103 s

Practice E, p. 2581. 0.75 N•m

3. 133 N

7 Review, pp. 263–2679. 2.7 m/s

11. 62 kg

19. 1.0 × 10−10 m (0.10 nm)

7

Practice C, p. 2031. 5.33 s; 53.3 m to the west

3. a. 1.22 × 104 N to the eastb. 53.3 m to the west


3. a. 12.0 m/sb. 9.6 m/s

Practice E, p. 2141. 3.8 m/s to the south

3. 4.25 m/s to the north

5. a. 3.0 kgb. 5.32 m/s

Practice F, p. 2161. a. 0.43 m/s to the west

b. 17 J

3. a. 4.6 m/s to the southb. 3.9 × 103 J

Practice G, p. 2191. a. 22.5 cm/s to the right

b. KEi = 6.2 × 10−4 J = KEf

3. a. 8.0 m/s to the rightb. KEi = 1.3 × 102 J = KEf

6 Review, pp. 223–22711. a. 8.35 × 10−21 kg•m/s

upwardb. 4.88 kg•m/s to the

rightc. 7.50 × 102 kg•m/s to the

southwestd. 1.78 × 1029 kg•m/s

forward

13. 18 N

23. 0.037 m/s to the south

29. 3.00 m/s

31. a. 0.81 m/s to the eastb. 1.4 × 103 J

33. 4.0 m/s

35. 42.0 m/s toward second base

Practice F, p. 1811. 66 kW

3. 2.61 × 108 s (8.27 years)

5. a. 7.50 × 104 Jb. 2.50 × 104 W

5 Review, pp. 184–1897. 53 J, −53 J

9. 47.5 J

19. 7.6 × 104 J

21. 2.0 × 101 m

23. a. 5400 J, 0 J; 5400 Jb. 0 J, –5400 J; 5400 Jc. 2700 J, –2700 J; 5400 J

33. 12.0 m/s

35. 17.2 s

37. a. 0.633 Jb. 0.633 Jc. 2.43 m/sd. 0.422 J, 0.211 J

39. 5.0 m

41. 2.5 m

45. a. 61 Jb. −45 Jc. 0 J

47. a. 28.0 m/sb. 30.0 m above the

ground

49. 0.107

51. a. 66 Jb. 2.3 m/sc. 66 Jd. −16 J

CHAPTER

Practice A, p. 1991. 2.5 × 103 kg•m/s to the

south

3. 46 m/s to the east

Practice B, p. 2011. 3.8 × 102 N to the left

3. 16 kg•m/s to the south

6




Practice C, p. 3551. 0.1504

3. a. 0.247b. 4.9 × 104 J

5. 755 J

10 Review, pp. 360–3633. b, c, d, e

9. 1.08 × 103 J; done by thegas

15. a. none (Q, W, and ΔU > 0)

b. ΔU < 0, Q < 0 for refrig-erator interior (W = 0)

c. ΔU < 0 (Q = 0, W > 0)

17. a. 1.7 × 106 J, to the rodb. 3.3 × 102 J; by the rodc. 1.7 × 106 J; it increases

27. 0.32

29. a. 188 Jb. 1.400 × 103 J

CHAPTER

Practice A, p. 3711. a. 15 N/m

b. less stiff

3. 2.7 × 103 N/m

Practice B, p. 3791. 1.4 × 102 m

3. 3.6 m

Practice C, p. 3811. 2.1 × 102 N/m

3. 39.7 N/m

5. a. 1.7 s, 0.59 Hzb. 0.14 s, 7.1 Hzc. 1.6 s, 0.62 Hz

Practice D, p. 3871. 0.081 m ≤ l ≤ 12 m

3. 4.74 × 1014 Hz

11

CHAPTER

Practice A, p. 3031. −89.22°C, 183.93 K

3. 37.0°C, 39°C

5. −195.81°C, −320.5°F

Practice B, p. 3111. 755 J

3. 0.96 J

Practice C, p. 3161. 47°C

3. 390 J/kg•°C

9 Review, pp. 322–3259. 57.8°C, 331.0 K

25. a. 2.9 Jb. It goes into the air,

the ground, and thehammer.

31. 25.0°C

33. a. TR = TF + 459.7, or TF = TR − 459.7

b. T = ⎯59

⎯ TR, or TR = ⎯95

⎯ T

35. a. TTH = ⎯32

⎯TC + 50, or

TC = ⎯23

⎯(TTH − 50)

b. −360° TH

37. 330 g

39. 5.7 × 103 J/min = 95 J/s

CHAPTER

Practice A, p. 3381. a. 6.4 × 105 J

b. −4.8 × 105 J3. 3.3 × 102 J

Practice B, p. 3461. 33 J

3. 1.00 × 104 J

5. 1.74 × 108 J

10

927. vt = 1630 m/s; T =5.78 × 105 s

29. Jupiter (m = 1.9 × 1027 kg)

33. F2

37. 26 N•m

39. 12 m/s

41. 220 N

43. 1800 N•m

45. 2.0 × 102 N47. 72%

49. a. 2.25 daysb. 1.60 × 104 m/s

51. a. 6300 N•mb. 550 N

53. 6620 N; no (Fc = 7880 N)

CHAPTER

Practice A, p. 2791. a. 3.57 × 103 kg/m3

b. 6.4 × 102 kg/m3

3. 9.4 × 103 N

Practice B, p. 2821. a. 1.48 × 103 N

b. 1.88 × 105 Pa

3. a. 1.2 × 103 Pab. 6.0 × 10−2 N

8 Review, pp. 288–2919. 2.1 × 103 kg/m3

15. 6.28 N

21. 1.01 × 1011 N

23. 6.11 × 10−1 kg

25. 17 N, 31 N

27. a. 1.0 × 103 kg/m3

b. 3.5 × 102 Pac. 2.1 × 103 Pa

29. 1.7 × 10−2 m

31. 0.605 m

33. 6.3 m

35. a. 0.48 m/s2

b. 4.0 s

37. 1.7 × 10−3 m

8

51. p = 11.3 cm

55. R = −25.0 cm

57. concave, R = 48.1 cm;M = 2.00; virtual

CHAPTER

Practice A, p. 4931. 18.5°3. 1.47

Practice B, p. 5011. 20.0 cm, M = −1.00;

real, inverted image

3. −6.67 cm, M = 0.333;virtual, upright image

Practice C, p. 5081. 42.8°3. 49.8°

14 Review, pp. 514–51911. 26°13. 30.0°, 19.5°, 19.5°, 30.0°23. yes, because nice > nair

25. 3.40; upright

37. a. 31.3°b. 44.2°c. 49.8°

39. 1.31

41. 1.62; carbon disulfide

43. 7.50 cm

45. a. 6.00 cmb. A diverging lens cannot

form an image largerthan the object.

47. a. 3.01 cmb. 2.05 cm

49. blue: 47.8°, red: 48.2°51. 48.8°53. 4.54 m

55. ⎯1

9

0⎯ f

14

43. a. 5.0 × 104 Wb. 2.8 × 10−3 W

CHAPTER

Practice A, p. 4491. 1.0 × 10−13 m

3. 85.7 m–10.1 m; The

wavelengths are shorter

than those of the AM

radio band.

5. 5.4 × 1014 Hz

Practice B, p. 4621. p = 10.0 cm: no image

(infinite q); p = 5.00 cm:q = −10.0 cm, M = 2.00;virtual, upright image

3. R = 1.00 × 102 cm;M = 2.00; virtual image

Practice C, p. 4661. p = 46.0 cm; M = 0.500;

virtual, upright image;h = 3.40 cm

3. p = 45 cm; h = 17 cm;M = 0.41; virtual, uprightimage

5. q = −1.31 cm; M = 0.125;virtual, upright image

13 Review, pp. 476–4807. 3.00 × 108 m/s

11. 1 × 10−6 m

13. 9.1 × 10−3 m (9.1 mm)

21. 1.2 m/s; The image movestoward the mirror’s surface.

35. q = 26 cm; real, inverted;M = −2.0

47. inverted; p = 6.1 cm;f = 2.6 cm; real

49. q2 = 6.7 cm; real;M1 = −0.57, M2 = −0.27;inverted

13

11 Review, pp. 396–3999. 580 N/m

11. 4A

19. 9.7 m

21. a. 0.57 sb. 1.8 Hz

27. 1/3 s; 3 Hz

35. 0.0333 m

39. a. 0.0 cmb. 48 cm

43. a, b, and d (l = 0.5L, L, and2L, respectively)

45. 1.7 N

47. 446 m

49. 9.70 m/s2

51. 9:48 A.M.

CHAPTER

Practice A, p. 4151. a. 8.0 × 10−4 W/m2

b. 1.6 × 10−3 W/m2

c. 6.4 × 10−3 W/m2

3. 2.3 × 10−5 W

5. 4.8 m

Practice B, p. 4271. 440 Hz

3. a. 82.1 Hzb. 115 Hzc. 144 Hz

12 Review, pp. 434–43723. 7.96 × 10−2 W/m2

25. a. 4.0 mb. 2.0 mc. 1.3 md. 1.0 m

29. 3 Hz

35. 3.0 × 103 Hz

37. 5 beats per second

39. 0.20 s

41. Lclosed = 1.5 (Lopen)

12




b. 4.2 × 106 m/s

CHAPTER

Practice A, p. 5991. 6.4 × 10−19 C

3. 2.3 × 10−16 J

Practice B, p. 6071. a. 4.80 × 10−5 C

b. 4.50 × 10−6 J

3. a. 9.00 Vb. 5.0 × 10−12 C

Practice C, p. 6091. 4.00 × 102 s

3. 6.00 × 102 s

5. a. 2.6 × 10−3 Ab. 1.6 × 1017 electronsc. 5.1 × 10−3 A

Practice D, p. 6151. 0.43 A

3. a. 2.5 Ab. 6.0 A

5. 46 Ω

Practice E, p. 6211. 14 Ω3. 1.5 V

5. 5.00 × 102 A

17 Review, pp. 626–6319. −4.2 × 105 V

19. 0.22 J

23. vavg >> vdrift

33. a. 3.5 minb. 1.2 × 1022 electrons

41. 3.4 A

49. 3.6 × 106 J

51. the 75 W bulb

53. 2.0 × 1016 J

17

Practice B, p. 5681. 47 N, along the negative

x-axis; 157 N, along the positive x-axis;11.0 × 101 N, along the negative x-axis

Practice C, p. 5701. x = 0.62 m

3. 5.07 m

Practice D, p. 5751. 1.66 × 105 N/C, 81.1°

above the positive x-axis

3. a. 3.2 × 10−15 N, along the negative x-axis

b. 3.2 × 10−15 N, alongthe positive x-axis

16 Review, pp. 581–58515. 3.50 × 103 N

17. 91 N (repulsive)

19. 1.48 × 10−7 N, along the +xdirection

21. 18 cm from the 3.5 nCcharge

33. 5.7 × 103 N/C, 75° abovethe positive x-axis

35. a. 5.7 × 10−27 N, in a direction opposite E

b. 3.6 × 10−8 N/C

37. a. 2.0 × 107 N/C, along the positive x-axis

b. 4.0 × 101 N

41. 7.2 × 10−9 C

43. velectron = 4.4 × 106 m/s;vproton = 2.4 × 103 m/s

45. 5.4 × 10−14 N

47. 2.0 × 10−6 C

49. 32.5 m

51. a. 5.3 × 1017 m/s2

b. 8.5 × 10−4 mc. 2.9 × 1014 m/s2

53. a. positiveb. 5.3 × 10−7 C

55. a. 1.3 × 104 N/C

57. a. 24.7°b. It will pass through the

bottom surface because qi < qc (qc = 41.8°).

59. 1.38

61. 58.0 m

63. a. 4.83 cm

b. The lens must be moved0.12 cm.

65. 1.90 cm

CHAPTER

Practice A, p. 5311. 5.1 × 10−7 m =

5.1 × 102 nm

3. 0.125°

Practice B, p. 5381. 0.02°, 0.04°, 0.11°3. 11

5. 6.62 × 103 lines/cm

15 Review, pp. 548–5515. q would decrease because l

is shorter in water.

9. 630 nm

11. 160 �m

19. 3.22°21. a. 10.09°, 13.71°, 14.77°

b. 20.51°, 28.30°, 30.66°29. 432.0 nm

31. 1.93 × 10–3 mm = 3 l ;a maximum

CHAPTER

Practice A, p. 5661. 230 N (attractive)

3. 0.393 m

16

15

CHAPTER

Practice A, p. 6891. 3.57 × 106 m/s

3. 6.0 × 10−12 N west

Practice B, p. 6921. 1.7 × 10−7 T in +z

direction

3. 1.5 T

19 Review, pp. 695–69931. 2.1 × 10−3 m/s

33. 2.00 T

39. 2.1 × 10−2 T, in the negative y direction

41. 2.0 T, out of the page

43. a. 8.0 m/s

b. 5.4 × 10−26 J

45. 2.82 × 107 m/s

CHAPTER

Practice A, p. 7141. 0.30 V

3. 0.14 V

Practice B, p. 7261. 4.8 A; 6.8 A, 170 V

3. a. 7.42 Ab. 14.8 Ω

5. a. 1.10 × 102 Vb. 2.1 A

Practice C, p. 7291. 55 turns

3. 25 turns

5. 147 V

20 Review, pp. 739–74311. 0.12 A

27. a. 2.4 × 102 Vb. 2.0 A

20

19Rd: 1.0 A, 4.0 V

Re: 1.0 A, 4.0 V

Rf : 2.0 A, 4.0 V

18 Review, pp. 666–67117. a. 24 Ω

b. 1.0 A

19. a. 2.99 Ωb. 4.0 A

21. a. seven combinations

b. R, 2R, 3R, ⎯R

2⎯, ⎯

R

3⎯, ⎯

2

3

R⎯, ⎯

3

2

R⎯

23. 15 Ω25. 3.0 Ω: 1.8 A, 5.4 V

6.0 Ω: 1.1 A, 6.5 V9.0 Ω: 0.72 A, 6.5 V

27. 28 V

29. 3.8 V

31. a. 33.0 Ωb. 132 Vc. 4.00 A, 4.00 A

33. 10.0 Ω35. a. a

b. cc. dd. e

37. 18.0 Ω: 0.750 A, 13.5 V6.0 Ω: 0.750 A, 4.5 V

39. 4.0 Ω41. 13.96 Ω43. a. 62.4 Ω

b. 0.192 Ac. 0.102 Ad. 0.520 We. 0.737 W

47. a. 5.1 Ωb. 4.5 V

49. a. 11 A (heater), 9.2 A (toaster), 12 A (grill)

b. The total current is 32.2A, so the 30.0 A circuitbreaker will open thecircuit if these appli-ances are all on.

55. 93 Ω57. 3.000 m; 2.00 × 10−7 C

59. 4.0 × 103 V/m

61. a. 4.11 × 10−15 Jb. 2.22 × 106 m/s

63. a. 1.13 × 105 V/mb. 1.81 × 10−14 Nc. 4.39 × 10−17 J

65. 0.545 m, −1.20 m

67. a. 7.2 × 10−13 Jb. 2.9 × 107 m/s

69. a. 3.0 × 10−3 Ab. 1.1 × 1018 electrons/min

71. a. 32 Vb. 0.16 V

73. 1.0 × 105 W

75. 3.2 × 105 J

77. 13.5 h

79. 2.2 × 10−5 V

CHAPTER

Practice A, p. 6501. a. 43.6 Ω

b. 0.275 A

3. 1.0 V, 2.0 V, 2.5 V, 3.5 V

5. 0.5 Ω

Practice B, p. 6551. 4.5 A, 2.2 A, 1.8 A, 1.3 A

3. a. 2.2 Ωb. 6.0 A, 3.0 A, 2.00 A

Practice C, p. 6591. a. 27.8 Ω

b. 26.6 Ωc. 23.4 Ω

Practice D, p. 662Ra: 0.50 A, 2.5 V

Rb: 0.50 A, 3.5 V

Rc: 1.5 A, 6.0 V

18




APPENDIX

Additional Problems1. 11.68 m

3. 6.4 × 10−2 m3

5. 6.7 × 10−5 ps

7. 2.80 h = 2 h, 48 min

9. 4.0 × 101 km/h

11. 48 m/h

13. +25.0 m/s = 25.0 m/s,upward

15. 44.8 m/s

17. −21.5 m/s2 = 21.5 m/s2,backward

19. 38.5 m

21. 126 s

23. 1.27 s

25. 11 km/h

27. 2.74 s

29. 10.5 m, forward

31. 5.9 s

33. 8.3 s

35. 7.4 s

37. −490 m/s2 = 490 m/s2,backward

39. 17.3 s

41. 7.0 m

43. 2.6 m/s

45. −11.4 m/s = 11.4 m/s,downward


49. 5.0° south of west

51. 770 m

53. −33 km/h = 33 km/h,downward

55. 18.9 km, 76° north of west

57. 17.0 m

59. 52.0°

61. 79 s

63. 15.8 m, 55° below the horizontal

65. 0.290 m/s, east; 1.16 m/s,north

67. 2.6 km

ICHAPTER

Practice A, p. 7961. 160.65 MeV; 342.05 MeV

3. 7.933 MeV

Practice B, p. 8021. 12

6C

3. 146C

5. 6328Ni → 63

29Cu + 0−1e + v��

Practice C, p. 8051. 4.23 × 103 s−1, 0.23 Ci

3. 9.94 × 10−7 s−1,6.7 × 10−7 Ci

5. a. about 5.0 × 107 atomsb. about 3.5 × 108 atoms

22 Review, pp. 820–8231. 79; 118; 79

7. 92.162 MeV

9. 8.2607 MeV/nucleon;8.6974 MeV/nucleon

21. a. 42He

b. 42He

23. 560 days

27. a. −eb. 0

33. 1.2 × 10−14

35. 3.53 MeV

37. a. 10n + 197

79Au → 19880Hg +

0−1e + v��

b. 7.885 MeV

39. 32He

41. 2.6 × 1021 atoms43. a. 8

4Be

b. 126C

45. 3.8 × 103 s

47. 1.1 × 1016 fission events

2229. a. 8.34 Ab. 119 V

35. 221 V

37. a. a step-down transformer

b. 1.2 × 103 V

43. 790 turns

45. a. a step-up transformerb. 440 V

47. 171:1

49. 300 V

CHAPTER

Practice A, p. 7551. 2.0 Hz

3. 1.2 × 1015 Hz

Practice B, p. 7581. 4.83 × 1014 Hz

3. 2.36 eV

Practice C, p. 7691. 4.56 × 1014 Hz; Line 4

3. 1.61 × 1015 Hz

5. E6 to E2; Line 1


3. 8.84 × 10−27 m/s

5. 1.0 × 10−15 kg

21 Review, pp. 779–78111. 4.8 × 1017 Hz

13. 1.2 × 1015 Hz

23. a. 2.46 × 1015 Hzb. 2.92 × 1015 Hzc. 3.09 × 1015 Hzd. 3.16 × 1015 Hz

33. 1.4 × 107 m/s

35. 2.00 eV

37. 0.80 eV

21

229. 2.0 × 101 m2

231. 4.30 kg

233. 374°F to −292°F

235. 6.6 × 10−2°C

237. 1.29 × 104 J

239. 4.1 × 10−2 kg

241. 1.200 × 103°C

243. 315 K

245. 1.91 × 10−2 kg = 19.1 g

247. 530 J/kg•°C

249. −930°C

251. 1.50 × 103 Pa = 1.50 kPa

253. 873 J

255. 244 J

257. 5.3 × 103 J

259. 2.4 × 103 Pa = 2.4 kPa

261. 5895 J

263. 5.30 × 102 kJ = 5.30 × 105 J

265. 1.0 × 104 J

267. −18 N

269. −0.11 m = −11 cm

271. 4.0 × 10−2 m = 4.0 cm

273. 0.2003 Hz

275. 730 N/m

277. 1.4 × 103 m/s

279. 2.2 × 104 Hz

281. 8.6 × 103 N/m

283. 3.177 s

285. 82 kg

287. 1.2 s

289. 1.5 × 103 m/s

291. 1.1 W/m2

293. 294 Hz

295. 408 m/s

297. 0.155 m

299. 0.211 m = 21.1 cm

301. 2.9971 × 108 m/s

303. 3.2 × 10−7 m = 320 nm

305. −0.96 cm

307. −1.9 cm

309. 3.8 m

311. 0.25

313. 38 cm

149. 1.58 × 103 kg•m/s, north

151. 3.38 × 1031 kg

153. 18 s

155. 637 m, to the right

157. 7.5 g

159. 0.0 m/s

161. −5.0 × 101 percent

163. 16.4 m/s, west

165. 5.33 × 107 kg•m/s

167. 1.0 × 101 m/s

169. 560 N, east

171. −3.3 × 108 N = 3.3 × 108

N, backward

173. 52 m

175. 24 kg

177. 90.6 km/h, east

179. 26 km/h, 37° north of east

181. −157 J

183. 0.125 kg

185. −4.1 × 104 J

187. 9.8 kg

189. 1.0 m/s, 60° south of east

191. 4.04 × 103 m/s2

193. 42 m/s

195. 8.9 kg

197. 1.04 × 104 m/s = 10.4km/s

199. 1.48 × 1023 kg

201. 1.10 × 1012 m

203. 6.6 × 103 m/s = 6.6 km/s

205. 0.87 m

207. 254 N

209. 0.42 m = 42 cm

211. 25 N

213. 165 kg

215. 5.09 × 105 s = 141 h

217. 5.5 × 109 m = 5.5 × 106

km

219. 1.6 N•m

221. 6.62 × 103 N

223. 0.574 m

225. 8.13 × 10−3 m2

227. 2.25 × 104 kg/m3

69. 66 km, 46° south of east

71. 10.7 m

73. 3.0 s

75. 76.9 km/h, 60.1° west ofnorth

77. 7.0 × 102 m, 3.8° abovethe horizontal

79. 47.2 m

81. 6.36 m/s

83. 13.6 km/h, 73° south ofeast

85. 58 N

87. 14.0 N; 2.0 N

89. 9.5 × 104 kg

91. 258 N, up the slope

93. 15.9 N

95. 2.0 m/s2

97. Fx = 8.60 N; Fy = 12.3 N

99. −448 m/s2 = 448 m/s2,backward

101. 15 kg

103. 0.085

105. 1.7 × 108 N

107. 24 N, downhill

109. 1.150 × 103 N

111. 1.2 × 104 N

113. 0.60

115. 38.0 m

117. 2.5 × 104 J

119. 247 m/s

121. −5.46 × 104 J

123. 3.35 × 106 J

125. 1.23 J

127. 12 s

129. 0.600 m

131. 133 J

133. 53.3 m/s

135. 72.2 m

137. 0.13 m = 13 cm

139. 7.7 m/s

141. 8.0 s

143. 230 J

145. 7.96 m

147. 6.0 × 101 m/s




445. 1.6 A

447. 1.45 A

449. 4 × 10−12 N

451. 7.6 × 106 m/s

453. 0.70 A

455. 5.1 × 10−4 T

457. 3.9 × 10−15 N

459. 1.5 × 105 A

461. 1.7 × 10−2 T

463. 0.90 s

465. 450 V

467. 1.8 A

469. 3.4 × 104 V = 34 kV

471. 48 turns

473. 2.5 × 10−3 A = 2.5 mA

475. 0.85 A

477. 48:1

479. 3.32 × 10−10 m

481. 1.00 × 10−13 m

483. 3.0 × 10−7 m

485. 26 kg

487. 4.30 × 1014 Hz

489. 6.0 × 1014 Hz

491. 4.0 × 10−21 kg

493. 2.5 × 10−43 m

495. 7.72 × 1014 Hz

497. 333.73 MeV

499. 0.543 705 u

501. 168O

503. 15.0 s

505. 31.92 h

507. 35.46 MeV

509. 13153I

511. 11154Xe

513. 924 days

379. 260 N from either charge

381. 1.6 × 10−12 C

383. 0.585 m = 58.5 cm

385. 3.97 × 10−6 N, upward

387. 0.073 m = 7.3 cm

389. 7.5 × 10−6 N, along the+y-axis

391. 4.40 × 105 N/C, 89.1°above the −x-axis

393. −7.4 C

395. 1.6 × 10−19 C397. 36 cm

399. 4.4 × 10−4 J

401. 7.1 × 10−4 F

403. 4.0 A

405. 160 Ω407. 1.7 × 106 W = 1.7 MW

409. 6.4 × 102 N/C

411. 12 V

413. 1.2 × 10−5 m

415. 1.4 × 102 C

417. 7.2 s

419. 4.8 V

421. 116 V

423. 5.0 × 105 W = 0.50 MW

425. 7.5 × 106 V

427. 3.00 × 102 Ω429. 6.0 Ω431. 13 Ω433. 6.0 Ω435. 0.056 A = 56 mA

437. 1.6 A (refrigerator);1.3 A (oven)

439. 12.6 Ω441. 2.6 V

443. 9.4 Ω

315. 2.40

317. 0.98 cm

319. 10.5 cm

321. 64.0 cm in front of themirror

323. 8.3 cm

325. 0.40

327. −11 cm

329. 2.9979 × 108 m/s

331. 33.3 cm

333. 0.19

335. 32.2°

337. −10.4 cm

339. 18 cm

341. 1.63

343. 39.38°

345. 58°

347. −21 cm

349. ∞351. 1.486

353. 1.54

355. 4.8 cm

357. 1.73 to 1.83

359. 5.18 × 10−4 m = 0.518 mm

361. 0.137°

363. 9.0 × 10−7 m =9.0 × 102 nm

365. 11.2°

367. 0.227°

369. 1.445 × 104 lines/cm

371. 140 N attractive

373. 2.2 × 10−17 C

375. 0.00 N

377. 4.0 × 10−8 N, 9.3° belowthe negative x-axis

holt physics appendix

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