chapter 11 mastering physics

7232019 Chapter 11 Mastering Physics

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1082015 Chapter 11 Homework Assignment

httpssessionmasteringphysicscommyctassignmentPrintViewassignmentID=3673484

Chapter 11 Homework Assignment

Due 1100pm on Saturday October 10 2015

You will receive no credit for items you complete after the assignment is due Grading Policy

Video Tutor Balancing a Meter Stick

First launch the video below You will be asked to use your knowledge of physics to predict the outcome of an

experiment Then close the video window and answer the question at right You can watch the video again at any

point

Part A

Suppose we replace the mass in the video with one that is four times heavier How far from the free end must we

place the pivot to keep the meter stick in balance

Hint 1 How to approach the problem

For the meter stick to be in equilibrium the net torque on it must be zero Torques about the fulcrum may

be exerted by the mass hanging from the end of the stick and by the stickrsquos own weight

Use the condition that the net torque must be equal to zero to obtain a relationship involving 1) the distance

between the left end of the stick and the fulcrum and 2) the distance between the center of mass of the

stick and the fulcrum These two distances must add up to a constant You should get two equations that

you can solve for the location of the fulcrum

ANSWER

Correct

90 cm (10 cm from the weight)

75 cm (25 cm from the weight)

10 cm

50 cm (in the middle)

25 cm

Typesetting math 20




httpssessionmasteringphysicscommyctassignmentPrintViewassignmentID=3673484 2

A Bar Suspended by Two Wires

A nonuniform horizontal bar of mass is supported by two massless wires against gravity The left wire makes an

angle with the horizontal and the right wire makes an angle The bar has length

Part A

Find the position of the center of mass of the bar measured from the bars left end

Express the center of mass in terms of and

Hint 1 Nature of the problem

This is a statics problem There is no net force or torque acting on the bar

Hint 2 Torques about left end of bar

The net torque is zero about any point you select Here we ask you to find the net torque of the system

about the left end of the bar Label the tension in the left wire and label the other wires tension The

weight of the bar is Note that the vector sum of and is zero Using the sign

convention shown in the picture express the sum of the torques about the left end of the bar

Answer in terms of andor Note that not all of these quantities will appear

in your answer

ANSWER

Hint 3 Forces x components

Assume that the tensions in the left and right wires are and respectively What is the sum of the x

components of the forces Because this is a statics problem these forces will sum to zero

Use the sign convention indicated in the figure and express your answer in terms of

andor Note that not all of these quantities will appear in your answer

983149

ϕ

1

ϕ

2

L

983160

L ϕ

1

ϕ

2

T

1

T

2

W =

983149 983143 T

1

T

2

W

L 983160 W T

2

T

1

ϕ

2

ϕ

1

=sum = 0 τ

l e f t

L s i n ( ) minus W 983160 T

2

ϕ

2

T

1

T

2

Σ F

983160

L 983160 W T

2

T

1

ϕ

2

ϕ

1





ANSWER

Hint 4 Forces y components

Assuming that the tensions in the left and right wires are and respectively what is the sum of the y




ANSWER

Hint 5 Eliminate weight from your equations

You should have found three equations by now It is possible to eliminate two variables and solve for interms of the others As an intermediate step solve your torque equation for in terms of texttipT_rm

2T_2 texttipLL etc and then solve your y -component force equation for texttipWW and substitute

back into your expression for texttipxx In other words find an expression for texttipxx

Answer in terms of texttipT_rm 1T_1 texttipT_rm 2T_2 texttipphi _rm 1phi_1

texttipphi _rm 2phi_2 and texttipLL

ANSWER

Hint 6 A useful trig identity

The dimensions for the expression you just found for texttipxx are correct since the units of the

tensions cancel out leaving the units of length in the numerator If you now solve the x -component force

equation for texttipT_rm 1T_1 in terms of texttipT_rm 2T_2 and substitute into your equation for

texttipxx you should find the following trig identity useful

sin(a+b) = sin(a)cos(b)+cos(a)sin(b)

Alternatively you could express your answer in terms of tan(phi_1) and tan(phi_2)

ANSWER

Correct

Precarious Lunch

=sum = 0 F

983160

minus c o s ( ) + c o s ( ) T

1

ϕ

1

T

2

ϕ

2

T

1

T

2

Σ F

983161

L 983160 W T

2

T

1

ϕ

2

ϕ

1

=sum = 0 F

983161

s i n ( ) + s i n ( ) minus W T

1

ϕ

1

T

2

ϕ

2

983160

983160 W

texttipxx

=

largefracT_2 sinleft(phi_2right) LT_1 sinleft(phi_1right)+T_2

sinleft(phi_2right)

texttipxx = largeL fractanleft(phi_2right)tanleft(phi_2right)+tanleft(phi_1right)





A uniform steel beam of length texttipLL and mass texttipm_rm 1m_1 is attached via a hinge to the side of a

building The beam is supported by a steel cable attached to the end of the beam at an angle texttiptheta theta a

shown Through the hinge the wall exerts an unknown force texttipFF on the beam A workman of mass

texttipm_rm 2m_2 sits eating lunch a distance texttipdd from the building

Part A

Find texttipTT the tension in the cable Remember to account for all the forces in the problem

Express your answer in terms of texttipm_rm 1m_1 texttipm_rm 2m_2 texttipLL texttipd

d texttiptheta theta and texttipgg the magnitude of the acceleration due to gravity

Hint 1 Pick the best origin

This is a statics problem so the sum of torques about any axis a will be zero In order to solve for texttipT

T you want to pick the axis such that texttipTT will give a torque but as few as possible other

unknown forces will enter the equations So where should you place the origin for the purpose of calculating

torques

ANSWER

Hint 2 Calculate the sum torques

Now find the sum of the torques about center of the hinge Remember that a positive torque will tend to

rotate objects counterclockwise around the origin

Answer in terms of texttipTT texttipLL texttipdd texttipm_rm 1m_1

texttipm_rm 2m_2 texttiptheta theta and texttipgg

At the center of the bar

At the hinge

At the connection of the cable and the bar

Where the man is eating lunch





ANSWER

ANSWER

Correct

Part B

Find texttipF_mit xF_x the texttipxx-component of the force exerted by the wall on the beam ( texttipF

F) using the axis shown Remember to pay attention to the direction that the wall exerts the force

Express your answer in terms of texttipTT and other given quantities

Hint 1 Find the sign of the force

The beam is not accelerating in the texttipxx-direction so the sum of the forces in the texttipxx-

direction is zero Using the given coordinate system is texttipF_mit xF_x going to have to be positive

or negative

ANSWER

Correct

Part C

Find texttipF_mit yF_y the y-component of force that the wall exerts on the beam ( texttipFF) using the

axis shown Remember to pay attention to the direction that the wall exerts the force

Express your answer in terms of texttipTT texttiptheta theta texttipm_rm 1m_1

texttipm_rm 2m_2 and texttipgg

ANSWER

Sigmatau_a = 0 = largeleft(fracm_1 L2+m_2 dright) g-T L sinleft(thetaright)

texttipTT = largefracgleft(m_1 fracL2+m_2 dright)L sinleft(thetaright)

texttipF_mit xF_x = -T cosleft(thetaright)

texttipF_mit yF_y = gleft(m_1+m_2right)-T sinleft(thetaright)





Correct

If you use your result from part (A) in your expression for part (C) youll notice that the result simplifies

somewhat The simplified result should show that the further the luncher moves out on the beam the lower

the magnitude of the upward force the wall exerts on the beam Does this agree with your intuition

Three-Legged Table

The top view of a table with weight texttipW_rm tW_t is shown in the figure The table has lost the leg at

(texttipL_mit xL_x texttipL_mit yL_y) in the upper

right corner of the diagram and is in danger of tipping over

Company is about to arrive so the host tries to stabilize the

table by placing a heavy vase (represented by the green

circle) of weight textt ipW_rm vW_v at ( texttipXX

texttipYY) Denote the magnitudes of the upward forces

on the table due to the legs at (0 0) (texttipL_mit xL_x

0) and (0 texttipL_mit yL_y) as texttipF_rm 0F_0

texttipF_mit xF_x and texttipF_mit yF_y

respectively

Part A

Find texttipF_mit xF_x the magnitude of the upward force on the table due to the leg at (texttipL_mit x

L_x 0)

Express the force in terms of texttipW_rm vW_v texttipW_rm tW_t texttipXX texttipYY

texttipL_mit xL_x andor texttipL_mit yL_y Note that not all of these quantities may appear in

the answer

Hint 1 Find the useful vector relations

This is a statics problem so about any point texttippp we have sum vectau_p = 0 and at texttipp

p sum vecF = 0 Each is a vector equation yielding algebraic expressions for each of the three

components x y and z Using the coordinate system shown in the figure indicate for each component of these equations whether it gives a useful relationship among the various forces

Check all that apply

Hint 1 Direction of applied forces

All of the forces are in the z direction Figure out what nonzero components of the torque these could

possibly generate

ANSWER





Hint 2 Find the y component of the torque

What is sum tau_y the y component of the torque equation

Express the y component of the torque in terms of texttipW_rm vW_v texttipW_rm tW_t

texttipF_mit xF_x texttipF_mit yF_y texttipL_mit xL_x texttipL_mit yL_y

texttipXX andor texttipYY Note that not all terms may appear in the answer

Hint 1 Choice of origin for torques

Take torques around the origin of the coordinate system (ie this becomes point texttippp in the

torque equations) as this will result in an equation with one and only one unknown force

Hint 2 Sign convention

If you do not get the sign from the cross-product expression tau_y = z F_x - x F_z remember that

the y torque is positive for clockwise rotation about the y axis when looking in the positive y

direction This is a subtle point--normally you look down on the x -y axes from above that is towardnegative z and so positive torque (about the z axis) is counterclockwise

ANSWER

ANSWER

x component of the forces

y component of the forces

z component of the forces

x component of the torques

y component of the torques

z component of the torques

sum tau_y=0 = large-W_v X+left(F_x-fracW_t2right) L_x





Correct

Part B

Find texttipF_mit yF_y the magnitude of the upward force on the table due to the leg at (0 texttipL_mit y

L_y)

Express the force in terms of texttipW_rm vW_v texttipW_rm tW_t texttipYY texttipXX

texttipL_mit yL_y andor texttipL_mit xL_x Note that not all of these quantities may appear in

the final answer

Hint 1 Find the x component of torque

What is sum tau_x the x component of the torque equation

Answer in terms of texttipW_rm vW_v texttipW_rm tW_t texttipF_mit xF_x

texttipF_mit yF_y texttipL_mit xL_x texttipL_mit yL_y texttipXX andor

texttipYY Note that not all terms may appear in the answer Remember to pay attention to the

signs of the torque

ANSWER

ANSWER

Correct

Part C

Find texttipF_rm 0F_0 the magnitude of the upward force on the table due to the leg at (0 0)

Express the force in terms of texttipW_rm vW_v texttipW_rm tW_t texttipL_mit xL_x texttipL_mit yL_y texttipXX andor texttipYY Note that not all terms may appear in the

answer

Hint 1 Summing the total forces

What is sum F_z the total vertical force on the table


texttipF_mit yF_y andor texttipF_rm 0F_0 Remember to pay attention to signs (with a

positive force being upward on the table)

texttipF_mit xF_x = largefracW_v XL_x+fracW_t2

sum tau_x=0 = large-W_v Y-left(fracW_t2-F_yright) L_y

texttipF_mit yF_y = largefracW_v YL_y+fracW_t2





ANSWER

ANSWER

Correct

While the host is greeting the guests the cat (of weight texttipW_rm cW_c) gets on the table and walks until her

position is (xy)=(L_xL_y)

Part D

Find the maximum weight texttipW_rm maxW_max of the cat such that the table does not tip over and brea

the vase

Express the cats weight in terms of texttipW_rm vW_v texttipXX texttipYY texttipL_mit

xL_x and texttipL_mit yL_y

Hint 1 Force when table begins to tip

When the table is just beginning to tip over what is the force texttipF_rm critF_crit supplied by the leg

at the origin

sum F_z= 0 = F_0+F_x+F_y-W_v-W_t

texttipF_rm 0F_0 = largeW_v left(1-fracXL_x-fracYL_yright)





ANSWER

Hint 2 Putting it all together

Go back to your previous equations put in the weight andor torque from the weight of the cat

(texttipW_rm cW_c) to obtain a new expression for texttipF_rm 0F_0 (checking your signs

carefully) equate this to the value you just obtained for texttipF_rm critF_crit and solve for

textt ipW_rm cW_c (now texttipW_rm maxW_max)

ANSWER

Correct

Exercise 1122

You are doing exercises on a Nautilus machine in a gym to strengthen your deltoid (shoulder) muscles Your arms are

raised vertically and can pivot around the shoulder joint and you grasp the cable of the machine in your hand 640 rm

rm cm from your shoulder joint The deltoid muscle is attached to the humerus 150 rm rm cm from the shoulde

oint and makes a 120 rm ^circ angle with that bone (See the figure below )

texttipF_rm critF_crit = 0

texttipW_rm maxW_max = largeW_v left(1-fracXL_x-fracYL_yright)





Part A

If you have set the tension in the cable of the machine to 360 rm rm N on each arm what is the tension ineach deltoid muscle if you simply hold your outstretched arms in place

ANSWER

Correct

Exercise 1130

A vertical solid steel post of diameter texttipdd = 22 rm rm cm and length texttipLL = 240 rm rm m is

required to support a load of mass texttipmm = 7700 rm rm kg You can ignore the weight of the post Take

free fall acceleration to be g=98 rm ms^2

Part A

What is the stress in the post

Express your answer using two significant figures

ANSWER

Correct

Part B

T = 605 rm N

P = 20times106 rm Pa





What is the strain in the post


ANSWER

Correct

Part C

What is the change in the posts length when the load is applied


ANSWER

Correct

Spinning Mass on a Spring

An object of mass texttipMM is attached to a spring with spring constant texttipkk whose unstretched length is

texttipLL and whose far end is fixed to a shaft that is rotating with angular speed texttipomega omega Neglec

gravity and assume that the mass rotates with angular speed texttipomega omega as shown When solving this

problem use an inertial coordinate system as drawn here

Part A

largefrac Delta L L = 99times10minus6

Delta L = 24times10minus5 rm m





Given the angular speed texttipomega omega find the radius texttipRleft(omega right)R(omega) at whic

the mass rotates without moving toward or away from the origin

Express the radius in terms of texttipkk texttipLL texttipMM and texttipomega omega

Hint 1 Acceleration at a certain point

Find texttipaleft(omega right)a(omega) the x component of the acceleration of mass texttipMM at

the instant pictured in the figure

Answer in terms of texttipRR and texttipomega omega

Hint 1 Expression for centripetal acceleration

Recall that the magnitude of centripetal acceleration for constant circular motion is a_rm

cent=v^2R This expression can be converted into texttipomega omega and texttipRR with

the standard equation for velocity of distancetime

ANSWER

Hint 2 Graphical description of forces

Which figure describes correctly the inertial force(s) acting on mass texttipMM

ANSWER

Hint 3 Spring force at a particular instant

texttipaleft(omega right)a(omega) = R omega^2

Figure 1

Figure 2

Figure 3

Figure 4





What is texttipF_rm springleft(Rright)F_spring(R) the x component of the force exerted by the spring

at the instant pictured in the figure

Answer in terms of texttipkk texttipRR and texttipLL

Hint 1 What determines the spring force

Recall that the amount of force a spring exerts is proportional to the distance it is stretched or

compressed with respect to its equilibrium length ( texttipLL in this case)

ANSWER

Hint 4 Newtons 2nd law

The mass is moving on a circular path and so there is a constant change in the direction of its velocity

vector and hence an inward radial acceleration The vector sum of all forces acting on the object

textt ipvecF_rm totalF_total_vec must act to keep the mass on the circular trajectory at constantspeed Hence the forces are radial and

F_rm total = ma_rm radial

Keep in mind that we are (as always) using an inertial reference system

ANSWER

Correct

Part B

Assume that at a certain angular s peed texttipomega _rm 2omega_2 the radius texttipRR becomes

twice texttipLL Find texttipomega _rm 2omega_2

Answer in terms of texttipkk and texttipMM


In Part A you have obtained the formula largeR(omega)=frackLk-Momega^2 Now use the additional

piece of information R(omega_2)=2L to solve for texttipomega _rm 2omega_2

ANSWER

texttipF_rm springleft(Rright)F_spring(R) = k left(R-Lright)

texttipRleft(omega right)R(omega) = largefrackLk-Momega^2

texttipomega _rm 2omega_2 = largesqrtfrack2M





Correct

Part C

You probably have noticed that as you increase texttipomega omega there will be a value texttipomega

_rm critomega_crit for which texttipRleft(omega right)R(omega) goes to infinity Find texttipomega

_rm critomega_crit


Hint 1 What equation to use

Note that the denominator of your expression for texttipRleft(omega right)R(omega) is a function of

texttipomega omega The critical point(s) of texttipRleft(omega right)R(omega) will coincide with

the root(s) of the denominator Before answering analyze the physical validity of your answer

ANSWER

Correct

Part D

What is happening to the spring as the angular velocity approaches texttipomega _rm critomega_crit

Choose the best option

Hint 1 Graphical help

Any spring behaves as shown in the two graphs

textt ipomega _rm critomega_crit = largesqrtfrackM





ANSWER

Correct

plusmn Youngs Modulus

Learning Goal

To understand the meaning of Youngs modulus to perform some real-life calculations related to stretching steel a

common construction material and to introduce the concept of breaking stress

Hookes law states that for springs and other elastic objects

F=kDelta x

where texttipFF is the magnitude of the stretching force Delta x is the corresponding elongation of the spring from

equilibrium and texttipkk is a constant that depends on the geometry and the material of the spring If the

deformations are small enough most materials in fact behave like springs Their deformation is directly proportional

the external force Therefore it may be useful to operate with an expression that is similar to Hookes law but describe

the properties of various materials as opposed to objects such as springs Such an expression does exist Consider

for instance a bar of initial length texttipLL and cross-sectional area texttipAA stressed by a force of magnitud

texttipFF As a result the bar stretches by Delta L

Let us define two new terms

Tensile stress is the ratio of the stretching force to the

cross-sectional area

largerm stress=fracFA

Tensile strain is the ratio of the elongation of the rod to

the initial length of the bar

largerm strain=fracDelta LL

It turns out that the ratio of the tensile stress to the tensile

strain is a constant as long as the tensile stress is not too

large That constant which is an inherent property of a

material is called Youngs modulus and is given by

largeY=f racFADelta LL

Part A

What is the SI unit of Youngs modulus

Hint 1 Look at the dimensions

If you look at the dimensions of Youngs modulus you will see that they are equivalent to the dimension of

The spring streches linearly then breaks at omega = omega_rm crit

The value of texttipomega _rm critomega_crit is so large that the spring will behave linearly for any

practically attainable texttipomega omega

As texttipomega omega approaches texttipomega _rm critomega_crit the spring stops

behaving linearly and begins to act more like an unstretchable rod until it eventually breaks





pressure Use the SI unit of pressure

ANSWER

Correct

Part B

Consider a metal bar of initial length texttipLL and cross-sectional area texttipAA The Youngs modulus of

the material of the bar is texttipYY Find the spring constant texttipkk of such a bar for low values of

tensile strain

Express your answer in terms of texttipYY texttipLL and texttipAA

Hint 1 Use the definition of Youngs modulusConsider the equation defining texttipYY Then isolate texttipFF and compare the result with Hookes

law F = kDelta x

ANSWER

Correct

Part C

Ten identical steel wires have equal lengths texttipLL and equal spring constants texttipkk The wires are

connected end to end so that the resultant wire has length 10L What is the spring constant of the resulting wire

Hint 1 The spring constant

Use the expression for the spring constant determined in Part B From the expression derived in the Part B

you can determine what happens to the spring constant when the length of the spring increases

ANSWER

Pa

texttipkk = largeY fracAL

01k

texttipkk

10k

100k





Correct

Part D


slightly twisted together so that the resultant wire has length texttipLL and its cross-sectional area is ten time

that of the individual wire What is the spring constant of the resulting wire


Use the expression for the spring constant determined in Part B From the expression derived in Part B you

can determine what happens to the spring constant when the area of the spring increases

ANSWER

Correct

Part E

Ten identical steel wires have equal lengths texttipLL and equal spring constants texttipkk The Youngsmodulus of each wire is texttipYY The wires are connected end to end so that the resultant wire has length

10L What is the Youngs modulus of the resulting wire

ANSWER

Correct

Part F

Ten identical steel wires have equal lengths texttipLL and equal spring constants texttipkk The Youngs

modulus of each wire is texttipYY The wires are slightly twisted together so that the resultant wire has length

texttipLL and is ten times as thick as the individual wire What is the Youngs modulus of the resulting wire

ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y





CorrectBy rearranging the wires we create a new object with new mechanical properties However Youngs modulus

depends on the material which remains unchanged To change the Youngs modulus one would have to

change the properties of the material itself for instance by heating or cooling it

Part G

Consider a steel guitar string of initial length L=100 meter and cross-sectional area A=0500 square millimeters

The Youngs modulus of the steel is Y=20 times 10^11 pascals How far ( texttipDelta LDelta L) would suc

a string stretch under a tension of 1500 newtons

Use two significant figures in your answer Express your answer in millimeters

ANSWER

Correct

Steel is a very strong material For these numeric values you may assume that Hookes law holds However

for greater values of tensile strain the material no longer behaves elastically If the strain and stress are large

enough the material deteriorates The final part of this problem illustrates this point and gives you a sense of the stretching limit of steel

Part H

Although human beings have been able to fly hundreds of thousands of miles into outer space getting inside the

earth has proven much more difficult The deepest mines ever drilled are only about 10 miles deep To illustrate t

difficulties associated with such drilling consider the following The density of steel is about 7900 kilograms per

cubic meter and its breaking stress defined as the maximum stress the material can bear without deteriorating

about 20 times 10^9 pascals What is the maximum length of a steel cable that can be lowered into a mine

Assume that the magnitude of the acceleration due to gravity remains constant at 98 meters per second per

second

Use two significant figures in your answer expressed in kilometers

Hint 1 Why does the cable break

The cable breaks because of the stress exerted on it by its own weight At the moment that the breaking

stress is reached the stress at the top of the cable reaches its maximum and the material begins to

deteriorate

Introduce an arbitrary cross-sectional area of the cable (which will cancel out of the final answer) The mass

of the cable below the top point can be found as the product of its volume and its density Use this to find

01Y

texttipYY

10Y

100Y

texttipDelta LDelta L = 15 rm mm





the force at the top that will lead to the breaking stress

Hint 2 Find the stress in the cable

Assume that the cable has cross-sectional area texttipAA and length texttipLL The density is

texttiprho rho The maximum stress in the cable is at the very top where it has to support its own

weight What is this maximum stress

Express your answer in terms of texttiprho rho texttipLL and texttipgg the magnitude

of the acceleration due to gravity Recall that the stress is the force per unit area so the area will notappear in your expression

ANSWER

ANSWER

Correct

This is only about 16 miles and we have assumed that no extra load is attached By the way this length is

small enough to justify the assumption of virtually constant acceleration due to gravity When making such

assumptions one should always check their validity after obtaining a result

Problem 1176

You are trying to raise a bicycle wheel of mass texttipmm and radius texttipRR up over a curb of height

texttiphh To do this you apply a horizontal force texttipvecFF_vec

Part A

maximum stress = rho L g

26 rm km





What is the least magnitude of the force texttipvecFF_vec that will succeed in raising the wheel onto the cu

when the force is applied at the center of the wheel

ANSWER

Correct

Part B


when the force is applied at the top of the wheel

ANSWER

Correct

Part C

In which case is less force required

ANSWER

Correct

Problem 1194

Compressive strength of our bones is important in everyday life Youngs modulus for bone is about 14 times 10^10

rm Pa Bone can take only about a 10 change in its length before fracturing

Part A

What is the maximum force that can be applied to a bone whose minimum cross-sectional area is 30rm

cm^2 (This is approximately the cross-sectional area of a tibia or shin bone at its narrowest point)


ANSWER

F = largefracmg sqrt2Rh-h^2R-h

F = largemg sqrtfrach2R-h

case A

case B





Correct

Part B

Estimate the maximum height from which a 70rm -kg man could jump and not fracture the tibia Take the time

between when he first touches the floor and when he has stopped to be 0006 rm s and assume that the stress

is distributed equally between his legs


ANSWER

Correct

Problem 1183

A garage door is mounted on an overhead rail (the f igure )

The wheels at A and B have rusted so that they do not roll

but rather slide along the track The coefficient of kinetic

friction is 049 The distance between the wheels is 200 rm

m and each is 050 rm m from the vertical sides of the

door The door is uniform and weighs 945 rm rm N It is

pushed to the left at constant speed by a horizontal force

vec F

Part A

If the distance h is 148 rm rm m what is the vertical component of the force exerted on the wheel A by the

track

ANSWER

F_rm max = 42times104 rm N

h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B

If the distance h is 148 rm rm m what is the vertical component of the force exerted on the wheel B by the

track

ANSWER

Correct

Part C

Find the maximum value h can have without causing one wheel to leave the track

ANSWER

Correct

Score Summary

Your score on this assignment is 101

You received 1106 out of a possible total of 11 points

F_V = 815 rm N

h_rm max = 204 rm m





A Bar Suspended by Two Wires

A nonuniform horizontal bar of mass is supported by two massless wires against gravity The left wire makes an

angle with the horizontal and the right wire makes an angle The bar has length

Part A

Find the position of the center of mass of the bar measured from the bars left end

Express the center of mass in terms of and

Hint 1 Nature of the problem

This is a statics problem There is no net force or torque acting on the bar

Hint 2 Torques about left end of bar

The net torque is zero about any point you select Here we ask you to find the net torque of the system

about the left end of the bar Label the tension in the left wire and label the other wires tension The

weight of the bar is Note that the vector sum of and is zero Using the sign

convention shown in the picture express the sum of the torques about the left end of the bar

Answer in terms of andor Note that not all of these quantities will appear

in your answer

ANSWER

Hint 3 Forces x components

Assume that the tensions in the left and right wires are and respectively What is the sum of the x




983149

ϕ

1

ϕ

2

L

983160

L ϕ

1

ϕ

2

T

1

T

2

W =

983149 983143 T

1

T

2

W

L 983160 W T

2

T

1

ϕ

2

ϕ

1

=sum = 0 τ

l e f t

L s i n ( ) minus W 983160 T

2

ϕ

2

T

1

T

2

Σ F

983160

L 983160 W T

2

T

1

ϕ

2

ϕ

1





ANSWER






ANSWER







ANSWER








ANSWER

Correct

Precarious Lunch

=sum = 0 F

983160


1

ϕ

1

T

2

ϕ

2

T

1

T

2

Σ F

983161

L 983160 W T

2

T

1

ϕ

2

ϕ

1

=sum = 0 F

983161


1

ϕ

1

T

2

ϕ

2

983160

983160 W

texttipxx

=


sinleft(phi_2right)










Part A








torques

ANSWER







At the hinge







ANSWER

ANSWER

Correct

Part B







or negative

ANSWER

Correct

Part C





ANSWER









Correct




Three-Legged Table











respectively

Part A


L_x 0)



the answer








possibly generate

ANSWER

















ANSWER

ANSWER












Correct

Part B


L_y)



the final answer






signs of the torque

ANSWER

ANSWER

Correct

Part C



answer













ANSWER

ANSWER

Correct



Part D


the vase





at the origin







ANSWER






ANSWER

Correct

Exercise 1122











Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





ANSWER






ANSWER







ANSWER








ANSWER

Correct

Precarious Lunch

=sum = 0 F

983160


1

ϕ

1

T

2

ϕ

2

T

1

T

2

Σ F

983161

L 983160 W T

2

T

1

ϕ

2

ϕ

1

=sum = 0 F

983161


1

ϕ

1

T

2

ϕ

2

983160

983160 W

texttipxx

=


sinleft(phi_2right)










Part A








torques

ANSWER







At the hinge







ANSWER

ANSWER

Correct

Part B







or negative

ANSWER

Correct

Part C





ANSWER









Correct




Three-Legged Table











respectively

Part A


L_x 0)



the answer








possibly generate

ANSWER

















ANSWER

ANSWER












Correct

Part B


L_y)



the final answer






signs of the torque

ANSWER

ANSWER

Correct

Part C



answer













ANSWER

ANSWER

Correct



Part D


the vase





at the origin







ANSWER






ANSWER

Correct

Exercise 1122











Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m









Part A








torques

ANSWER







At the hinge







ANSWER

ANSWER

Correct

Part B







or negative

ANSWER

Correct

Part C





ANSWER









Correct




Three-Legged Table











respectively

Part A


L_x 0)



the answer








possibly generate

ANSWER

















ANSWER

ANSWER












Correct

Part B


L_y)



the final answer






signs of the torque

ANSWER

ANSWER

Correct

Part C



answer













ANSWER

ANSWER

Correct



Part D


the vase





at the origin







ANSWER






ANSWER

Correct

Exercise 1122











Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





ANSWER

ANSWER

Correct

Part B







or negative

ANSWER

Correct

Part C





ANSWER









Correct




Three-Legged Table











respectively

Part A


L_x 0)



the answer








possibly generate

ANSWER

















ANSWER

ANSWER












Correct

Part B


L_y)



the final answer






signs of the torque

ANSWER

ANSWER

Correct

Part C



answer













ANSWER

ANSWER

Correct



Part D


the vase





at the origin







ANSWER






ANSWER

Correct

Exercise 1122











Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





Correct




Three-Legged Table











respectively

Part A


L_x 0)



the answer








possibly generate

ANSWER

















ANSWER

ANSWER












Correct

Part B


L_y)



the final answer






signs of the torque

ANSWER

ANSWER

Correct

Part C



answer













ANSWER

ANSWER

Correct



Part D


the vase





at the origin







ANSWER






ANSWER

Correct

Exercise 1122











Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m

















ANSWER

ANSWER












Correct

Part B


L_y)



the final answer






signs of the torque

ANSWER

ANSWER

Correct

Part C



answer













ANSWER

ANSWER

Correct



Part D


the vase





at the origin







ANSWER






ANSWER

Correct

Exercise 1122











Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





Correct

Part B


L_y)



the final answer






signs of the torque

ANSWER

ANSWER

Correct

Part C



answer













ANSWER

ANSWER

Correct



Part D


the vase





at the origin







ANSWER






ANSWER

Correct

Exercise 1122











Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





ANSWER

ANSWER

Correct



Part D


the vase





at the origin







ANSWER






ANSWER

Correct

Exercise 1122











Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





ANSWER






ANSWER

Correct

Exercise 1122











Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





Part A


ANSWER

Correct

Exercise 1130




Part A



ANSWER

Correct

Part B

T = 605 rm N








ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m







ANSWER

Correct

Part C



ANSWER

Correct






Part A


















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m
















ANSWER



ANSWER



Figure 1

Figure 2

Figure 3

Figure 4











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m











ANSWER







ANSWER

Correct

Part B







ANSWER








Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





Correct

Part C



_rm critomega_crit






ANSWER

Correct

Part D










ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





ANSWER

Correct


Learning Goal




F=kDelta x




















Part A














ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m






ANSWER

Correct

Part B



tensile strain



law F = kDelta x

ANSWER

Correct

Part C






ANSWER

Pa


01k

texttipkk

10k

100k





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





Correct

Part D







ANSWER

Correct

Part E



ANSWER

Correct

Part F




ANSWER

01k

texttipkk

10k

100k

01Y

texttipYY

10Y

100Y








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m








Part G





ANSWER

Correct




Part H







second





deteriorate



01Y

texttipYY

10Y

100Y













ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m












ANSWER

ANSWER

Correct




Problem 1176



Part A


26 rm km







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m







ANSWER

Correct

Part B



ANSWER

Correct

Part C


ANSWER

Correct

Problem 1194



Part A




ANSWER



case A

case B





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m





Correct

Part B





ANSWER

Correct

Problem 1183








vec F

Part A


track

ANSWER


h_rm max = 26 rm m

F_V = 130 rm N




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m




Correct

Part B


track

ANSWER

Correct

Part C


ANSWER

Correct

Score Summary



F_V = 815 rm N

h_rm max = 204 rm m

chapter 11 mastering physics

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