physics 120-01 class notes thru 012314 lecture

8/12/2019 Physics 120-01 Class Notes Thru 012314 Lecture

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Physics and Technological Progress The laws of the physics of motion were developed by Galileo

(15641642) and Newton (16421727)

These laws form the basis for: Understanding the motion of objects and fluids on earth and in space

Steam and gasoline engines, modern transportation

Aerodynamics and the development of the airplane

Rocketry and space travel

The laws of electricity and magnetism were assimilated by

Maxwell (18311879)

These laws form the basis for:

Electrical power generation and transmission

Electronics and computer industry Electrical machinery

Lasers and Lighting

In PHY 120 you will learn the laws of physics of motion, conservation of energy and

momentum, thermodynamics and heat transfer and how to apply these laws

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Basic Approach to Understanding Physical

Phenomena Physicists establish hypotheses(theories to be tested)

explaining observations

These hypotheses are tested against observations over an extended

period of time (say, decades). They must be stated in such a way that

they can be proven wrong.

If the hypotheses prove to accurately predict observed phenomena overthis extended period of time they are called theories and if the

theories become widely accepted over a century or two they are called

laws

The observations that stand the test of time are called facts

If a theory is found incorrect by more accurate observations it issuperseded by a new set of hypotheses

If the new hypotheses prove accurate over an extended period of time

they are accepted as theories (may become laws) and the supporting

observations are called facts. and the cycle continues2


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Measurements Play a Key Role in Physics

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The way in which theories are tested is by observations made in

experiments designed by scientists to test the theories. Theseobservations must be accurately measured to determine the

degree to which the theories explain the experimental findings.

Among the key measurements that physicists make are:

1.The time interval between two events.2.The dimensions of an object

3.The mass of an object

4.The electric current through an electrical circuit.

5.The temperature of an object.6.The luminous intensity of a light source.


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xper menta servat ons are a e yMeasurements in the International (SI) Units

The Scientific Method and Experiments Hypotheses must be supported by experiments made with precise

measurements

This requires precise instrumentation and diagnostic equipment, as well as

precisely defined standard measures for mass, length, time, electric current,

temperature and luminosity

The International System (SI) Units are the accepted standard

today, the units we will use in this course are:

Unit Standard Basis for Standard

Length Meter Distance between two marks on a Pt/Ir bar kept at 0 degrees C

Mass Kilogram Mass of a standard cylinder of Pt/Ir

Time Second Time for 9,192,631,770 wave cycles of Cs atomic clock to occur

In this course we will be concerned primarily with the units of length (meters),

time (seconds) and mass (kilograms) 4

A i l B li f h M i f Obj


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Aristotles Beliefs on the Motion of Objects

Held Sway for 2000 Years

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Aristotle (~350 B.C.) believed that there were two types of motion.

1. Natural Motion. Light things like smoke rise and heavy things likedropped boulders fall.

2. Violent Motion Motions that resulted from a push or a pull.

Although Aristotles beliefs were ultimately proven wrong by Galileo and

Newton and there were Greeks who lived at the same time as Aristotle thathad different beliefs, Aristotle was so highly respected for his intellect that his

beliefs were accepted without challenge.

As a result of his belief in Natural Motion he thought that a heavy object

would fall to earth faster than a lighter one.

Aristotle and Ptolemy (~ 150 A.D.) also believed that the sun, planets and

stars revolved around the earth (geocentric hypothesis) rather than the sun

(heliocentric hypothesis)


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With the advent of telescopes, experiments designed to test hypotheses and methods

of measuring time durations Galileo was able to disprove Aristotles hypotheses

Aristotles (incorrect) hypotheses were

thought correct for 2000 Years

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Aristotles thoughts held the upper hand for 2000 years until Galileo and Kepler

made more accurate observations in the 16thcentury A.D.1. With the advent of the telescope Galileo was able to make more accurate

observations and show that not all motion was geocentrica. He observed that Jupiter had moons which orbited around Jupiter, not

around Earth

2. Galileo was also able to show that any two objects fell to earth at the same

speed, independent of their weight, opposed to Aristotles hypotheses, and he

performed accurate experiments on motion of objects using pendulums and

water clocks.

3. Keplers observations on the motion of the planets superseded Ptolemys

predictions

Galileo and Keplers observations were the foundation of Newtons theories ofmotion and gravitation. With Newtons theories in hand there was a theoretical

basis for the observations of Galileo and Kepler and the ideas of Aristotle and

Ptolemy on motion were finally discarded.


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Galileos Experiment Proved Aristotles

Hypothesis Wrong

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A key to the scientific method is making accurate measurements.

A scientific assertion (hypothesis) can be disproven by a well-

designed experiment which allows for accurate observations.

Aristotle (384 B.C.322 B.C.) believed that an object falls at a

speed proportional to its weight. He was so highly respected his

assertion was held to be true for 2000 years.

According to historians, in 1589 Galileo dropped two balls from the

leaning tower of Pisa simultaneously and found they both

impacted the ground at the same time, disproving Aristotles

assertion.

Why do you suppose Aristotle was wrong and so many peoplebelieved Aristotle for so long?

Would the same result have occurred if Galileo had

dropped a feather and a heavier rock at the same time in a

second experiment?

What causes the difference in the two experiments?

Galileo was the first true physicist because he developed well-designed

experiments and made precise measurements of his observations


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Units of Measure

We will be primarily using the SI (International)System of Units of measures in this course The key units for studying motion and the familiar English

unit equivalent are: 1 Meter = 39.37 inches = 3.281 feet

1 Kilogram = 2.205 pounds (mass equivalent)

The acceleration of gravity on earth in SI units The mass of the earth creates a gravitational field that

exerts a force directed toward the center of the earth

g = 9.80 meters/second/second

A body free-falling under the force of gravity will gainspeed at the rate of 9.80 meters per second

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Mass and Weight

Mass is a measure of the inertia (resistance to changein motion) of an object It is related to the quantity of matter in an object

Objects which are subjected to a (net) force will accelerate in

inverse proportion to their mass Weight and Mass are different

The weight of an object is the force with which it is pulled tothe center of the earth (or any other astronomical body) bythe force of gravity

Weight = mass x acceleration of gravity

Mass is an inherent quantity, weight varies with thegravitational field (e.g., objects weigh less on the moon thanon earth, whereas mass is same)

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Converting from one system of units to

another

Write down the units explicitly For example 65 (miles/hour) (convert to feet/second)

Find the conversion factor from one system

to the other 1 mile = 5280 feet, one hour = 3600 seconds

Set up as multiplicative factors so that only

the new units appear in the result 65 miles/hour x 5280 feet/mile x 1 hour/3600 secs

= 95.33 feet/sec

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Check Dimensions

For motion studies the fundamental units are Mass, Length and Time

The answer to a motion problem must have the correctdimensions (check numerator and denominator)

For example Distance is always going to be in [L], (meters, feet, miles)

Speed is always going to be in [L]/[T] (meters/sec, feet/sec,miles/hour)

Acceleration is always going to be in [L]/[T]2

(meter/sec2

,feet/sec2, miles/hour2)

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Common Physics Units (Motion)

Physical Parameter Units Example (SI Units)

Displacement (x) [L] meter, (m)

Time (t) [T] second, (s)

Mass (m) [M] kilogram, (kg)

Speed (v) [L]/ [T] meter per second, (m/s)

Acceleration (a) [L]/ [T]2 meter/sec2, m/s2

Force (F) [M] [L]/ [T]2 newton, kg m/s2

Energy (E) [M] [L]2/ [T]2 joule, kg m2/s2

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Check Dimensions: Example

Check for consistency of dimensions the eq. v = at2

The units of v are [L]/[T], the units of a are [L]/[T]2 , and the

units of time are [T]

We ask the question do the dimensions on the right = thedimensions on the left?

The dimension on the rhs of the equation are:

[L]/[T]2x [T]2= [L], (length)

The dimension of velocity are length/time, [L]/[T], so the two

dont match and the equation is invalid

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Terminology of a right triangle

90

h = hypotenuse

h0 = length of sideopposite the angle

ha = length of side

adjacent to the angle

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Trigonometric Relationships

90

h h0= h sin

ha= h cos

h2= h02 + ha

2; sin = h0/h, cos = ha/h, tan = h0/ha

= sin-1 (h0/h) = cos-1(ha/h) = tan

-1(h0/ha)

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Additional Info on Trig/Geom.

hh0

ha

90 + + = 180

+ = 90 (angles are complementary)

= 90 -

h cos = h sin = ha

h sin = h cos = h0

tan = h0/ha

tan = ha/h0

h

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Scalars and Vectors

A scalar quantity is one that can be described with a singlenumber specifying the size or magnitude of a measurement

The mass of a rock is 30 kg

The height of a tree is 20 m

A vector quantity is one where both the magnitude and the

direction are essential characteristics

The acceleration of gravity is 9.80 m/sec2directed toward the center

of the earth

A boat is headed due north at 1 meter/second (velocity)

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Vector Addition (two vectors @ 90)

Start

N

A

B

R

R = A + B, R ={(275m)2 + (125m)2}1/2= 302m is the magnitude of the

vector (Pythagorean Theorem).

We know that tan = 125m/275m =125/275= 0.454, so = tan-1(0.454) =

24.4, so that the vector points 24.4 north of due east.

275 m due E

125 mdue N

90

Travel due east for 275 m (A) and travel due north for 125 m (B). What isthe resultant vector?

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Any Vector can be resolved into

components related to a co-ordinate system

X

Y

V

Vx

VY

V = VX+ VY= 1X VX+ 1Y VYwhere 1Xand 1Yare unit vectors along the X and Yaxes respectively and VXand VYare the magnitudes of the respective vectors.

VX= V cos , VY= V sin , where V is the magnitude of the vector. This gives us a

method of adding together many vectors, by resolving them into the X and Y

components and simply adding the magnitudes of the X and Y components andaligning them along the X-axis and Y-axis respectively

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Add multiple vectors together by adding their

x and y components (Ref. X-Y Origin)

Y

X

V1V2

V1XV2X

V2YV1Y

V2XV1X+

V2Y V1Y+ V1+ V2

Resultant Vector Magnitude = {(V1x+ V2x)2+ (V1Y+ V2Y)

2}1/2,

Resultant Vector Direction = tan-1{(V1Y+ V2Y)/(V1X+ V2X)}

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Add multiple vectors together by adding their x

and y components (Head to Tail)

Y

X

V1

V2

V1XV2X

V2Y

V1Y

V2XV1X+

V2Y V1Y+ V1+ V2

Resultant Vector Magnitude = {(V1x+ V2x)2+ (V1Y+ V2Y)

2}1/2,

Resultant Vector Direction = tan-1{(V1Y+ V2Y)/(V1X+ V2X)}

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Positive Vector and its Negative

V

- V

-VX= -V cos

VX= V cos

VY= V sin

- VY= - V sin

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Problem 1-9

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Determine the number of milliliters in one ounce given:1 gallon = 128 ounce; 1 gallon = 3.785 x 10-3 m3 and

1mL = 10-6m3

From the first equation: 1 ounce = 1 gallon/128;

From the second equation: 1 gallon = 3.785 x 10-3

m3

So, 1 ounce = 3.785 x 10-3 m3 /128 = 2.96 x 10-5m3

In order to get this into mL, from the third equation we have 1 mL = 10-6 m3or

1m3= 106mL

So, 1 ounce = 2.96 x 10-5m3 = 2.96 x 10-5 m3x 106mL/m3=

2.96 x 101 mL = 29.6 mL


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Problem 1-16

1Line of

Sight

2

85.0 meters

1= 35.0 = line of sight to top of building

2 = 38.0 = line of sight to top of

antenna

What is height of

Antenna onBuilding?

ha

hb

hb= height of buildingha= height of antenna

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Problem 1-16

1Line of

Sight

2

85.0 meters

1= 35.0

2 = 38.0

What is height of

Antenna onBuilding?

ha

hb

Tan (1) = hb/85, hb= 85 tan (35.0)

Tan (2) =(ha+ hb)/85, ha+ hb = 85 tan (38.0)

ha= (ha+ hb)hb= 85 {tan (38.0)tan (35.0)} =

85.0(0.7810.700)

ha = 6.89 meters

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Problem 1-23(a)

N

B = 325 N

A = 445 N

325

R

Rmagnitude = (3252+ 4452)1/2 = 551 N

tan = (325/445) = 0.730

= tan-1 (325/445) = tan-1 (0.730) =

36.1 north of due west

Two forces on a crateOne force is 445 N due west

The other is 325 N due north

What is resultant R?

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P bl 1 23 (b)


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Problem 1-23 (b)

N

B = 325 N due north

A = 445 Ndue west

325

1

R1

R1mag = (3252+ 4452)1/2 = 551 N

With B pointing due north

1= tan-1 (325/445) = tan-1 (0.730) = 36.1north of due west

If B is reversed, it points due south

2= 36.1 degrees south of due west

Direction of resultant vector R2 is changed, but

the magnitude (551N) is the same.

-B = 325 N due south2

R2

325

What if B is reversed?

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P bl 1 36


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Problem 1 - 36

Soccer

Net

A

X

Y

AY

AX

Player #1

Player #2

30.0

N

The direction of AYis due south

The direction of AX is due east

The magnitude of A is 8.6 meters, what are the

magnitudes of AXand AY?

so the magnitude of AX = A sin (30) = 8.6 x 0.5

= 4.3 meters. Its direction is due east

and the magnitude of AY= A cos (30) = 8.6 x

0.866 = 7.4 meters. Its direction is due south.

90

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Problem 147 (Prev. Edition)

35.0

5.0 m

15.0 m

18.0 m

R = A + B + C

N

A football player runs 5.0 m due N, then 15.0 m due E, and then

18.0 m 35south of due east, what is the resultant vector(magnitude and direction)?

A

B

C

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Problem 147 (components of vector A)

5.0 m

N

To solve the problem we will find the x and y components

of the vectors A, B and C individually. We will then add the x-components together for each vector to find the x-component

of the resultant vector R and the y-components of each vector

to find the y-component of the resultant vector R. That is,

Rx= Ax+ Bx+ Cx, Ry= Ay+ By+ Cy , tan = Ry/ Rx

= tan-1 (Ry/ Rx) R = (Rx2+ Ry

2)1/2

Starting with the vector A Football player runs 5.0 m in a

direction due N

The components of A are Axand Ay

The value of Ax= 0 (there is no x component because the

vector is pointing North (+y) direction)

The value of Ay= +5.0 m

A

+Y

+X

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Problem 1 - 47 (Components of Vector B)

15.0 m

N

Continuing with the vector B. Football player runs 15.0 m in a

direction due E. The vector B then has a component in the +xdirection, but none in the +y direction.

Bx= + 15.0 m. and By= 0

B

+Y

+X

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Problem 147 (Components of Vector C)

35.0

18.0 m

N

Continuing with the vector C. Player runs 18.0 m in a direction

35south of due eastCx= + 18.0 m cos (35) Cy= - 18.0 m sin (35)

Cx= + 14.7 m, Cy= - 10.3 m

C

Cx= + 18.0 m cos (35)

Cy=-

18.0ms

in(35)

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physics 120-01 class notes thru 012314 lecture

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