Formal Homework Assignment 3

10/26/14 10:39 PMFormal Homework Assignment 3

Page 1 of 31http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3179126

Formal Homework Assignment 3Due: 11:15am on Tuesday, October 14, 2014

To understand how points are awarded, read the Grading Policy for this assignment.

A message from your instructor...

For "Conceptual Question 2.7" the +x direction is assumed to be directed to the right.

Conceptual Question 2.7

The figure shows the position-versus-time graph for a moving object. At which lettered point orpoints:

Part A

Is the object moving the fastest?

ANSWER:

Correct

Part B

Is the object speeding up?

ANSWER:

A

B

C

D

E

F



Correct

Part C

Is the object moving to the left and turning around?

ANSWER:

Correct

What Velocity vs. Time Graphs Can Tell You

A common graphical representation of motion along a straight line is the v vs. t graph, that is, the graph of (instantaneous) velocity as a function of time. Inthis graph, time is plotted on the horizontal axis and velocity on the vertical axis. Note that by definition, velocity and acceleration are vector quantities. Instraight-line motion, however, these vectors have only a single nonzero component in the direction of motion. Thus, in this problem, we will call the velocityand the acceleration, even though they are really the components of the velocity and acceleration vectors in the direction of motion, respectively.

Here is a plot of velocity versus time for a particle that travels along a straight line with a varying velocity. Refer to this plot to answer the following questions.

A

B

C

D

E

F

t vv

a



Part A

What is the initial velocity of the particle, ?

Express your answer in meters per second.

Hint 1. Initial velocity

The initial velocity is the velocity at .

Hint 2. How to read a v vs. t graph

Recall that in a graph of velocity versus time, time is plotted on the horizontal axis and velocity on the vertical axis. For example, in the plotshown in the figure, at .

ANSWER:

Correct

Part B

What is the total distance traveled by the particle?

Express your answer in meters.

Hint 1. How to approach the problem

Recall that the area of the region that extends over a time interval under the v vs. t curve is always equal to the distance traveled in .Thus, to calculate the total distance, you need to find the area of the entire region under the v vs. t curve. In the case at hand, the entire regionunder the v vs. t curve is not an elementary geometrical figure, but rather a combination of triangles and rectangles.

Hint 2. Find the distance traveled in the first 20.0 seconds

What is the distance traveled in the first 20 seconds of motion, between and ?


Hint 1. Area of the region under the v vs. t curve

The region under the v vs. t curve between and can be divided into a rectangle of dimensions by ,

and a triangle of base and height , as shown in the figure.

v0

t = 0 s

v = 2.00 m/s t = 30.0 s

= 0.5 v0 m/s

x

t t

x1 t = 0.0 s t = 20.0 s

t = 0.0 s t = 20.0 s 20.0 s 0.50 m/s20.0 s 1.50 m/s



ANSWER:

Hint 3. Find the distance traveled in the second 20.0 seconds

What is the distance traveled in the second 20 seconds of motion, from to ?



The region under the v vs. t curve between and is a rectangle of dimensions by , as shown inthe figure.

ANSWER:

Hint 4. Find the distance traveled in the last 10.0 seconds

What is the distance traveled in the last 10 seconds of motion, from to ?



= 25 x1 m

x2 t = 20.0 s t = 40.0 s

t = 20.0 s t = 40.0 s 20.0 s 2.00 m/s

= 40 x2 m

x3 t = 40.0 s t = 50.0 s

t = 40.0 s t = 50.0 s 10.0 s 2.00 m/s



The region under the v vs. t curve between and is a triangle of base and height , as shown inthe figure.

ANSWER:

ANSWER:

Correct

Part C

What is the average acceleration of the particle over the first 20.0 seconds?

Express your answer in meters per second per second.

Hint 1. Definition and graphical interpretation of average acceleration

The average acceleration of a particle that travels along a straight line in a time interval is the ratio of the change in velocity experienced by the particle to the time interval , or

.

In a v vs. t graph, then, the average acceleration equals the slope of the line connecting the two points representing the initial and final velocities.

Hint 2. Slope of a line

The slope of a line from point A, of coordinates , to point B, of coordinates , is equal to the "rise" over the "run," or

.

ANSWER:

t = 40.0 s t = 50.0 s 10.0 s 2.00 m/s

= 10 x3 m

= 75 x m

aav

aav t vt

=aav vt

m ( , )xA yA ( , )xB yB

m= yB yAxB xA

= 0.075 aav m/s2



CorrectThe average acceleration of a particle between two instants of time is the slope of the line connecting the two corresponding points in a v vs. tgraph.

Part D

What is the instantaneous acceleration of the particle at ?

Hint 1. Graphical interpretation of instantaneous acceleration

The acceleration of a particle at any given instant of time or at any point in its path is called the instantaneous acceleration. If the v vs. t graph ofthe particle's motion is known, you can directly determine the instantaneous acceleration at any point on the curve. The instantaneousacceleration at any point is equal to the slope of the line tangent to the curve at that point.

Hint 2. Slope of a line

The slope of a line from point A, of coordinates , to point B, of coordinates , is equal to the "rise" over the "run," or

.

ANSWER:

CorrectThe instantaneous acceleration of a particle at any point on a v vs. t graph is the slope of the line tangent to the curve at that point. Since in the last10 seconds of motion, between and , the curve is a straight line, the tangent line is the curve itself. Physically, this meansthat the instantaneous acceleration of the particle is constant over that time interval. This is true for any motion where velocity increases linearlywith time. In the case at hand, can you think of another time interval in which the acceleration of the particle is constant?

Now that you have reviewed how to plot variables as a function of time, you can use the same technique and draw an acceleration vs. time graph, that is,the graph of (instantaneous) acceleration as a function of time. As usual in these types of graphs, time is plotted on the horizontal axis, while the verticalaxis is used to indicate acceleration .

Part E

Which of the graphs shown below is the correct acceleration vs. time plot for the motion described in the previous parts?

a t = 45.0 s

m ( , )xA yA ( , )xB yB

m= yB yAxB xA

=

1

0.20

-0.20

0.022

-0.022

a

m/s2

m/s2

m/s2

m/s2

m/s2

t = 40.0 s t = 50.0 s

ta



Hint 1. How to approach the problem

Recall that whenever velocity increases linearly with time, acceleration is constant. In the example here, the particle's velocity increases linearlywith time in the first 20.0 of motion. In the second 20.0 , the particle's velocity is constant, and then it decreases linearly with time in the last10 . This means that the particle's acceleration is constant over each time interval, but its value is different in each interval.

Hint 2. Find the acceleration in the first 20

What is , the particle's acceleration in the first 20 of motion, between and ?


Hint 1. Constant acceleration

Since we have already determined that in the first 20 of motion the particle's acceleration is constant, its constant value will be equal tothe average acceleration that you calculated in Part C.

ANSWER:

Hint 3. Find the acceleration in the second 20

What is , the particle's acceleration in the second 20 of motion, between and ?


Hint 1. Constant velocity

In the second 20 of motion, the particle's velocity remains unchanged. This means that in this time interval, the particle does notaccelerate.

ANSWER:

s ss

s

a1 s t = 0.0 s t = 20.0 s

s

= 0.075 a1 m/s2

s

a2 s t = 20.0 s t = 40.0 s

s



Hint 4. Find the acceleration in the last 10

What is , the particle's acceleration in the last 10 of motion, between and ?


Hint 1. Constant acceleration

Since we have already determined that in the last 10 of motion the particle's acceleration is constant, its constant value will be equal tothe instantaneous acceleration that you calculated in Part D.

ANSWER:

ANSWER:

CorrectIn conclusion, graphs of velocity as a function of time are a useful representation of straight-line motion. If read correctly, they can provide you withall the information you need to study the motion.

Relative Velocity vs. Time Graph Ranking Task

Two cars travel on the parallel lanes of a two-lane road. The cars are at the same location at time , and move in such a way as to produce thevelocity (relative to the ground) vs. time graph shown in the figure. On the graph, one verticalblock is equivalent to one velocity unit.

Part A

Rank car #1s speed relative to the ground at the lettered times (A through E).

= 0 a2 m/s2

s

a3 s t = 40.0 s t = 50.0 s

s

= -0.20 a3 m/s2

Graph A

Graph B

Graph C

Graph D

t = 0 s



Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. Speed relative to the ground

The graph displays the velocity of each car relative to the ground. Since you are asked only about car #1, simply ignore car #2 and interpret thegraph. Recall that speed is the magnitude of velocity.

Hint 2. Find the speed of car #1 at time A

What is the speed of car #1 relative to the ground at time A? Remember that speed is the magnitude of velocity.

ANSWER:

ANSWER:

Correct

Part B

Rank car #1s speed relative to car #2 at the lettered times.


Hint 1. Relative speed

The speed of car #1 relative to car #2 is the speed of car #1 that would be measured by someone in car #2, for example a passenger aiming aradar gun at car #1. If you know the velocity of car #1 relative to a fixed point and the velocity of car #2 relative to that same fixed point, then themagnitude of the difference in velocities gives the speed of one car relative to the other.

Hint 2. Determine the directions of travel at time A

At time A, in which direction(s) are car #1 and car #2 headed?

ANSWER:

greater than 1 unit

between 0 and 1 unit

less than 0 units



Hint 3. Determine the directions of travel at time D

At time D, in which direction(s) are car #1 and car #2 headed?

ANSWER:

ANSWER:

Correct

Part C

Rank the distance between the cars at the lettered times.


Hint 1. Determine the directions of travel at time A

The problem states that the two cars begin at the same location, so their initial separation is zero. At time A, are the cars moving in the samedirection or in opposite directions?

ANSWER:

car #1: positive, car #2: positive

car #1: positive, car #2: negative

car #1: negative, car #2: positive

car #1: negative, car #2: negative

car #1: positive, car #2: positive

car #1: positive, car #2: negative

car #1: negative, car #2: positive

car #1: negative, car #2: negative

same

opposite



CorrectSince the two cars are headed in opposite directions, their separation distance must be increasing.

Hint 2. When do the cars stop traveling in opposite directions?

At what time do the two cars stop traveling in opposite directions?

ANSWER:

Hint 3. How is the separation distance changing?

Is the separation between the cars increasing or decreasing between time C and time E?

ANSWER:

ANSWER:

Correct

Adding and Subtracting Vectors Conceptual Question

Six vectors (A to F) have the magnitudes and directions indicated in the figure.

A

B

C

D

E

Increasing

Decreasing



Part A

Which two vectors, when added, will have the largest (positive) x component?

Hint 1. Largest x component

The two vectors with the largest x components will, when combined, give the resultant with the largest x component. Keep in mind that positive xcomponents are larger than negative x components.

ANSWER:

Correct

Part B

Which two vectors, when added, will have the largest (positive) y component?

Hint 1. Largest y component

The two vectors with the largest y components will, when combined, give the resultant with the largest y component. Keep in mind that positive ycomponents are larger than negative y components.

ANSWER:

C and E

E and F

A and F

C and D

B and D

C and D

A and F

E and F

A and B

E and D



Correct

Part C

Which two vectors, when subtracted (i.e., when one vector is subtracted from the other), will have the largest magnitude?

Hint 1. Subtracting vectors

To subtract two vectors, add a vector with the same magnitude but opposite direction of one of the vectors to the other vector.

ANSWER:

Correct


For "Components of Vectors" use the grid to measure the length of each vector's x and y component. The arrows for the vectors are randomly placed on thegrid with no common origin. Therefore, the direction of the vector will be important to note but not necessarily where it starts (i.e. it is irrelevant if the vectorstarts at the graph's origin or not). Similarly, the vectors drawn in "Tactics Box 3.1 Determining the Components of a Vector" and " Resolving VectorComponents with Trigonometry" are drawn to help you identify the components of a vector. For " Resolving Vector Components with Trigonometry" theywant the answers to 3 significant digits.

Components of Vectors

Shown is a 10 by 10 grid, with coordinate axes x and y .The grid runs from -5 to 5 on both axes. Drawn on this grid are four vectors, labeled through

. This problem will ask you various questions about these vectors. All answers should be indecimal notation, unless otherwise specified.

Part A

What is the x component of ?

Express your answer to two significant figures.

A and F

A and E

D and B

C and D

E and F

A

D

A



Hint 1. How to derive the component

A component of a vector is its length (but with appropriate sign) along a particular coordinate axis, the axes being specfied in advance. You areasked for the component of that lies along the x axis, which is horizontal in this problem. Imagine two lines perpendicular to the x axis running

from the head (end with the arrow) and tail of down to the x axis. The length of the x axis between the points where these lines intersect is the

x component of . In this problem, the x component is the x coordinate at which the perpendicular from the head of the vector hits the origin(because the tail of the vector is at the origin).

ANSWER:

Correct

Part B

What is the y component of ?

Express your answer to the nearest integer.

ANSWER:

Correct

Part C

What is the y component of ?


Hint 1. Consider the direction

Don't forget the sign.

ANSWER:

Correct

Part D

What is the component of ?


Hint 1. How to find the start and end points of the vector components

A vector is defined only by its magnitude and direction. The starting point of the vector is of no consequence to its definition. Therefore, you needto somehow eliminate the starting point from your answer. You can run two perpendiculars to the x axis, one from the head (end with the arrow)

A

A

A

= 2.5Ax

A

= 3Ay

B

= -3By

x C



of , and another to the tail, with the x component being the difference between x coordinates of head and tail (negative if the tail is to the right

of the head). Another way is to imagine bringing the tail of to the origin, and then using the same procedure you used before to find the

components of and . This is equivalent to the previous method, but it might be easier to visualize.

ANSWER:

Correct

The following questions will ask you to give both components of vectors using the ordered pairs method. In this method, the x component is written first,followed by a comma, and then the y component. For example, the components of would be written 2.5,3 in ordered pair notation.

The answers below are all integers, so estimate the components to the nearest whole number.

Part E

In ordered pair notation, write down the components of vector .

Express your answers to the nearest integer.

ANSWER:

Correct

Part F

In ordered pair notation, write down the components of vector .

Express your answers to the nearest integer.

ANSWER:

Correct

Part G

What is true about and ? Choose from the pulldown list below.

ANSWER:

Correct

C

C

A B

= -2Cx

A

B

, = 2,-3Bx By

D

, = 2,-3Dx Dy

B D

They have different components and are not the same vectors.

They have the same components but are not the same vectors.

They are the same vectors.



Tactics Box 3.1 Determining the Components of a Vector

Learning Goal:

To practice Tactics Box 3.1 Determining the Components of a Vector.

When a vector is decomposed into component vectors and parallel to the coordinate axes, we can describe each component vector with a single

number (a scalar) called the component. This tactics box describes how to determine the x component and y component of vector , denoted and .

TACTICS BOX 3.1 Determining the components of a vector

1. The absolute value of the x component is the magnitude of the component vector .

2. The sign of is positive if points in the positive x direction; it is negative if points in the negative x direction.3. The y component is determined similarly.

Part A

What is the magnitude of the component vector shown in the figure?

Express your answer in meters to one significant figure.

ANSWER:

Correct

Part B

What is the sign of the y component of vector shown in the figure?

A A x A yA Ax Ay

| |Ax Ax A xAx A

x A

x

Ay

A x

= 5 | |Ax m

Ay A



ANSWER:

Correct

Part C

Now, combine the information given in the tactics box above to find the x and y components, and , of vector shown in the figure.

Express your answers, separated by a comma, in meters to one significant figure.

ANSWER:

Correct

Resolving Vector Components with Trigonometry

Often a vector is specified by a magnitude and a direction; for example, a rope with tension exerts a force of magnitude in a direction 35 north of east. This is a good way to think ofvectors; however, to calculate results with vectors, it is best to select a coordinate system andmanipulate the components of the vectors in that coordinate system.

Part A

positive

negative

Bx By B

, = -2,-5 , Bx By m m

T T



Find the components of the vector with length = 1.00 and angle =20.0 with respect to the x axis as shown.

Enter the x component followed by the y component, separated by a comma.

Hint 1. What is the x component?

Look at the figure shown.

points in the positive x direction, so is positive. Also, the magnitude is

just the length .

ANSWER:

Correct

Part B

Find the components of the vector with length = 1.00 and angle =20.0 with respect to the x axis as shown.


Hint 1. What is the x component?

The x component is still of the same form, that is, .

ANSWER:

Correct

The components of still have the same form, that is, , despite 's placement with respect to the y axis on the drawing.

Part C

Find the components of the vector with length = 1.00 and angle 35.0 as shown.


A a

A x Ax | |AxOL = OMcos()

= 0.940,0.342A

B b

Lcos()

= 0.940,0.342B

B (Lcos(),L sin()) B

C c =



Hint 1. Method 1: Find the angle that makes with the positive x axis

Angle = 0.611 differs from the other two angles because it is the angle between the vector and the y axis, unlike the others, which are with

respect to the x axis. What is the angle that makes with the positive x axis?

Express your answer numerically in degrees.

ANSWER:

Hint 2. Method 2: Use vector addition

Look at the figure shown.

1. .

2. .

3. , the x component of is negative, since points in the negative xdirection.

Use this information to find . Similarly, find .

ANSWER:

Correct

Vector Addition Ranking Task

Six vectors ( through ) have the magnitudes and directions indicated in the figure.

C

C

125

= +C C x C y

| | = length(QR) = c sin()C xCx C

C x

Cx Cy

= -0.574,0.819C

a f



Part A

Rank the vector combinations on the basis of their magnitude.


Hint 1. Adding vectors graphically

To add two vectors together, imagine sliding one vector (without rotating it) until its tail coincides with the tip of the second vector. The sum of thetwo vectors, termed the resultant vector , is the vector that goes from the tail of the first vector to the tip of the second vector. The magnitude of

the resultant, , is determined by the sum of the squares of its x and y components, that is,

ANSWER:

Correct

Part B

Rank the vector combinations on the basis of their angle, measured counterclockwise from the positive x axis. Vectors parallel to the positive x axishave an angle of 0 . All angle measures fall between 0 and 360 .


Hint 1. Angle of a vector

The angle of a vector is to be measured counterclockwise from the x axis, with the x axis as 0 . The following vectors are at the angleslisted and are shown on the graph below.

R

| |R

| | = .R +R2x R2y

V

=V 1 0

= 2



Notice that the magnitude of the vector is irrelevant when determining its angle

ANSWER:

Correct


For " Vector Addition and Subtraction" all of the vectors are written such that they have a common origin (similar to how the A and B vectors looked in theprior problem - both starting at the origin). This is the normal way we write vectors since it allows us to add/subtract them easily. The former problem waswritten only to assist in visualizing the components of vectors.

=V 2 45

=V 3 180

=V 4 225

=V 5 270



Vector Addition and Subtraction

In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify acoordinate system in order to find the components of each arrow.) This problem gives you some practice with the components.

Let vectors , , and . Calculate the following, and express your answers as ordered triplets of valuesseparated by commas.

Part A

ANSWER:

Correct

Part B

ANSWER:

Correct

Part C

ANSWER:

Correct

Part D

ANSWER:

Correct

Part E

ANSWER:

Correct

= (1, 0,3)A = (2,5, 1)B = (3, 1, 1)C

= 3,-5,-4A B

= -5,4,0B C

= -6,4,3 + A B C

= -3,-2,-113 2A C

= -11,14,82 + 3 A B C



Part F

ANSWER:

Correct

Vector Magnitude and Direction Conceptual Question

A man out walking his dog makes one complete pass around a perfectly square city block. He starts at point A and walks clockwise around the block.

Let be the displacement vector from A to B, be the displacement vector from B to C, etc.

Part A

Which of the following vectors is equal to ?

Hint 1. Determining a vector

Recall that is a vector representing the displacement of the man and his dog as they walk from point A to point B. This vector has amagnitude equal to one block and a direction along the positive x axis.

Hint 2. Equal vectors

Two vectors are equal if they have the same magnitude and the same direction.

ANSWER:

= 17,-12,-62 3( )A B C

r AB r BC

r AB

r AB

only

only

only

All of the above

None of the above

r BCr CDr DA



CorrectRecall that, for vectors to be equal, they must have the same magnitude and direction.

Part B


ANSWER:

Correct

Part C


Hint 1. Determining the difference of two vectors

can be determined by adding the vector to the vector pointing opposite to . Thus looks like this:

Carefully perform the vector addition in each of the options and compare the resultant vectors to the one shown above.

ANSWER:

r AB

only

only

only

All of the above

None of the above

r BCr CDr DA

r AB r DA

r AB r DA r AB r DA r AB r DA

only

only

only

All of the above

None of the above

( + )r CD r DA+r AB r BCr BC r CD



Correct

Problem 3.6

Find x- and y-components of the following vectors.

Part A

Express your answers using two significant figures. Enter your answers numerically separated by a comma.

ANSWER:

Correct

Part B


ANSWER:

Correct

Part C


ANSWER:

Correct


For problem 3.37, you are asked to provide an answer with 2 significant figures. However it really wants the answer in three significant figures (this problemonly).

Problem 3.37

Jack and Jill ran up the hill at . The horizontal component of Jill's velocity vector was .

= (460m, 46 below positive x axis)r

, = 320,-330 rx ry m

= (530m/s, 36 above positive x axis)v

, = 430,310 vx vy m/s

= (5.6m/ , negative y direction)a s2

, = 0,-5.6 ax ay m/s2

3.0 m/s 2.5 m/s



Part A

What was the angle of the hill?

Give your answer to two significant figures.

ANSWER:

Correct

Part B

What was the vertical component of Jill's velocity?

Give your answer to two significant figures.

ANSWER:

Correct

Problem 3.8

Let ( 3.30 , 31 above the negative x-axis) and ( 25.1, 46 to the right of the negative y-axis).

Part A

Find the magnitude of .

ANSWER:

Correct

Part B

Find the x-component and the y-component .

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

ANSWER:

Correct

Part C

Find the magnitude of .

= 33.6

= 1.66 m/s vy

=C m =D

C

= 3.30 C m

C

, = -2.8,1.7 Cx Cy m

D



ANSWER:

Correct

Part D

Find the x-component and the y-component .

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

ANSWER:

Correct


For Problem 3.18, part A asks for the x-component while part B asks for the y-component. As is typical, you should provide your answers with 3 significantfigures unless told otherwise.

Problem 3.18

What are the x- and y-components of the velocity vector shown in the figure?

Part A

Express your answer with the appropriate units.

ANSWER:

Correct

Part B


ANSWER:

= 25.1D

D

, = 18,-17Dx Dy

= -86.6 vxms



Correct

Problem 3.34

A cannon tilted upward at 36.0 fires a cannonball with a speed of 100 .

Part A

What is the component of the cannonball's velocity parallel to the ground?


ANSWER:

Correct

Problem 3.14

Let , , and .

Part A

What is the component form of vector ?

ANSWER:

Correct

Part B

What is the magnitude of vector ?

ANSWER:

Correct

Part C

= 50.0 vy ms

m/s

= 80.9 v ms

= 6 3A ^ ^ = 3 + 6B ^ ^ = D A B

D

= 4 + 5D ^ ^

= 9 + 9D ^ ^

= 9 9D ^ ^

= 9 5D ^ ^

D

= 12.7D



What is the direction of vector ?

Express your answer using two significant figures.

ANSWER:

Correct

Problem 3.16

Let , , and .

Part A

What is the component form of vector ?

ANSWER:

Correct

Part B

What is the magnitude of vector

ANSWER:

Correct

Part C

What is the direction of vector

ANSWER:

Correct

Problem 3.24

D

= 45 below positive x-axis

= 6 4A ^ ^ = 4 + 6B ^ ^ = 5F A B

F

= 26 34F ^ ^

= 14 26F ^ ^

= 9 + 34F ^ ^

= 26 + 34F ^ ^

?F

= 42.8F

?F

= 52.6 below positive x-axis



Part A

What is the angle between vectors and in the figure?


ANSWER:

Correct

Part B

Use components to determine the magnitude of .

ANSWER:

Correct

Part C

Use components to determine the direction of .


ANSWER:

Correct


For problem 3.44, when asked for the angle note where they are asking you to take the angle from; counterclockwise from left (i.e. the negative x-axis).

Problem 3.44

The figure shows three ropes tied together in a knot. One of your friends pulls on a rope with 3.0 units of force and another pulls on a second rope with 5.0

E F

= 71.6

G = +E F

= 3.00G

G = +E F

= 90.0



units of force.

Part A

How hard must you pull on the third rope to keep the knot from moving?


ANSWER:

Correct

Part B

In what direction must you pull on the third rope to keep the knot from moving?


ANSWER:

Correct

Score Summary:Your score on this assignment is 99.6%.You received 37.85 out of a possible total of 38 points.

= 4.4 units of force F

= 83 counterclockwise from the left direction

Documents

Formal Homework Assignment 3