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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 1


How many significant figures does each of the following numbers have?

a)  53.2

b)  0.53

c)  5.320

d)  0.0532


Solution:

Part a

All nonzero digits are significant, therefore, 53.2 has 3 significant figures.

Part b

The number 0.53 can be written in scientific notation as 5.3×10-1. Therefore, 0.53 has 2 significant figures.

Part c

Trailing zeros are significant because they indicate increased precision. Therefore, 5.320 has 4 significant figures

Part d

The leading zeros are not significant but just locate the decimal point. Therefore, 0.0532 has only 3 significant figures.


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Chapter 1 Solution Guides to Physics for Scientists & Engineers 3rd Edition by Randall Knight


Conceptual Questions

Exercises and Problems

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Question 35

Question 36

Question 37

Question 38

Question 39

Question 40

Question 41

Question 42

Question 43

Question 44

Question 45

Question 46

Question 47

Question 48

Question 49

Question 50

Question 51

Question 52

Question 53

Question 54

Question 55

Question 56

Question 57


Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 2

Learning Goal:

To practice Tactics Box 1.2 Vector Subtraction.
Vector subtraction has some similarities to the subtraction of two scalar quantities. With numbers, subtraction is the same as the addition of a negative number. For example, 53 is the same as 5+(3). Similarly, A⃗ B⃗ =A⃗ +(B⃗ ). We can use the rules for vector addition and the fact that B⃗ is a vector opposite in direction to B⃗  to form rules for vector subtraction, as explained in this tactics box.
The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right.

To subtract B⃗  from A⃗  (Figure 1), perform these steps:

  1. Draw A⃗ .  The figure shows vector A that is directed upwards to the right.
  2. Draw B⃗  and place its tail at the tip of A⃗ .The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector minus B is on the right and is directed upwards to the left. The tip of vector A and the tail of vector minus B are at the same point.
  3. Draw an arrow from the tail of A⃗ to the tip of B⃗ . This is vector A⃗ B⃗ .     The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector minus B is on the right and is directed upwards to the left. The tip of vector A and the tail of vector minus B are at the same point. Vector A minus B starts at the tail of vector A and ends at the tip of vector minus B.
Part A
Draw vector C⃗ =A⃗ B⃗  by following the steps in the tactics box above. When drawing B⃗ , keep in mind that it has the same magnitude as B⃗  but opposite direction.
Part B
Draw vector F⃗ =D⃗ E⃗ by following the steps in the tactics box above. When drawing E⃗ , keep in mind that it has the same magnitude as E⃗  but opposite direction.
ANSWERS:
Part A
3
Part B
4

Arizona State University| PHY 121: Univ Physics I: Mechanics|Homework 1-1| Problem 1

Learning Goal:
To practice Tactics Box 1.1 Vector Addition.
Vector addition obeys rules that are different from those for the addition of two scalar quantities. When you add two vectors, their directions, as well as their magnitudes, must be taken into account. This Tactics Box explains how to add vectors graphically. 
To add B⃗  to A⃗  (Figure 1), perform these steps:
  1. Draw A⃗ .  The figure shows vector A that is directed upwards to the right.
  2. Place the tail of B⃗  at the tip of A⃗ .  The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right. The tip of vector A and the tail of vector B are at the same point.
  3. Draw an arrow from the tail of A⃗  to the tip of B⃗ . This is vector A⃗ +B⃗ .  The figure shows three vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right. The tip of vector A and the tail of vector B are at the same point. Vector A plus B starts at the tail of vector A and ends at the tip of vector B.
Part A
Create the vector R⃗ =A⃗ +B⃗  by following the steps in the Tactics Box above. When moving vector B⃗ , keep in mind that its direction should remain unchanged.
The figure shows two vectors. Vector A is on the left and is directed upwards to the right. Vector B is on the right and is directed downwards to the right.
Part B
Create the vector R⃗ =C⃗ +D⃗  by following the steps in the Tactics Box above. When moving vector D⃗ , keep in mind that its direction should remain unchanged. The location, orientation, and length of your vectors will be graded.
ANSWERS:
Part A
1
Part B
2

Grantham PHY220 Week 2 Assignment Problem 8

If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?

SOLUTION:

Draw the free-body diagram of the car

week 2 problem 8

Consider the vertical direction

\sum F_y=ma_y

F_N-mg=0

F_N=mg

Consider the motion in the horizontal direction

Solve for the acceleration of the car.

v^2=\left(v_0\right)^2+2a_x\Delta x

a_x=\frac{v^2-\left(v_0\right)^2}{2\Delta x}=\frac{0-50^2}{2\left(300\right)}=-4.17\:m/s^2

Solve for the coefficient of kinetic friction

\sum F_x=ma_x

-F_{fr}=ma_x

-\mu _kF_N=ma_x

\mu _k=\frac{ma_x}{-F_N}=\frac{m\:\left(-4.17\right)}{-m\left(9.80\right)}=\frac{4.17}{9.80}

\mu _k=0.43

Grantham PHY220 Week 2 Assignment Problem 7

A 7.93 kg box is pulled along a horizontal surface by a force F_p of 84.0 N applied at a 47^{\circ}  angle. If the coefficient of kinetic friction is 0.35, what is the acceleration of the box?

SOLUTION:

The free-body diagram of the box

week 2 problem 7

Solve for the normal force

\sum F_y=ma_y

F_N-mg+F_psin\left(47.0^{\circ} \right)=0

F_N-7.93\left(9.80\right)+84.0\:sin\left(47^{\circ} \right)=0

F_N=16.28\:N

Solve for the friction force

F_{fr}=\mu _kF_N=0.35\left(16.28\:N\right)=5.70\:N

Solve for the acceleration in the horizontal direction

\sum F_x=ma_x

F_p\:cos\:\left(47^{\circ} \right)-F_{fr}=7.93\left(a_x\right)

84.0\:N\cdot cos\:\left(47^{\circ} \right)-5.70\:N=7.93\:kg\:\cdot \left(a_x\right)

a_x=\frac{51.59\:N}{7.93\:kg}=6.51\:m/s^2

Therefore, the acceleration of the box is 6.51 m/s2 along the horizontal surface.

Grantham PHY220 Week 2 Assignment Problem 6

If the acceleration due to gravity on the Moon is 1/6 that what is on the Earth, what would a 100 kg man weight on the Moon? If a person tried to simulate this gravity in an elevator, how fast would it have to accelerate and in which direction?

SOLUTION:

The acceleration due to gravity on the moon is

g_m=\frac{1}{6}\left(9.80\:m/s^2\right)=1.63\:m/s^2

The weight of a 100-kg man on the moon is

W_m=mg_m=\left(100\:kg\right)\left(1.63\:m/s^2\right)=163.3\:N

If the elevator is accelerating upward then the acceleration would be greater. The person would be pushed toward the floor of the elevator making the weight increase. Therefore, the elevator must be going down to decrease the acceleration.

For a 100 kg man to experience a 163.3 N in an elevator,

F=ma

163.3\:N=100\:kg\:\left(9.80\:m/s^2-a_e\right)

9.80-a_e=\frac{163.3}{100}

a_e=9.80-\frac{163.3}{100}

a_e=8.167\:m/s^2

Therefore, the elevator should be accelerated at 8.167 m/s2 downward for a 100-kg man to simulate his weight just like his weight in the moon which has 1/6 of the Earth’s gravity acceleration.

Grantham PHY220 Week 2 Assignment Problem 4

A not so brilliant physics student wants to jump from a 3rd-floor apartment window to the swimming pool below. The problem is the base of the apartment is 8.00 meters from the pool’s edge. If the window is 20.0 meters high, how fast does the student have to be running horizontally to make it to the pool’s edge?

Solution:

Since the student will be running horizontally, there is no initial vertical velocity, v_{0_y}=0. We are also given \Delta x=8\:m, and \Delta y=-20\:m.

Consider the vertical component of the motion.

\Delta y=v_{0_y}t-\frac{1}{2}gt^2

-20=0-\frac{1}{2}\left(9.80\right)t^2

-20=-4.9t^2

t^2=\frac{20}{4.9}

t=\sqrt{\frac{20}{4.9}}

t=2.02\:s

Consider the horizontal component of the motion

\Delta x=v_{0_x}t

v_{0_x}=\frac{\Delta x}{t}

v_{0_x}=\frac{8}{2.02}

v_{0_x}=3.96\:m/s

Therefore, the student should be running 3.96 m/s horizontally to make it to the pool’s edge.

Grantham PHY220 Week 2 Problem 3

A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?

Solution:

Assuming that the gun was fired horizontally.

Consider the horizontal component of the motion.

v_{o_x}=1100\:m/s

\Delta x=V_{0_x}t=\left(1100\:m/s\right)\left(0.32\:s\right)=352\:m

Consider the vertical component.

v_{o_y}=0\:m/s

\Delta y=V_{0_y}t-\frac{1}{2}gt^2=0-\frac{1}{2}\left(9.80\:m/s^2\right)\left(0.32\:s\right)^2=-0.50

The negative sign of  \Delta y means that the bullet went downward.

Therefore, the bullet traveled 352 m horizontally and 0.50 m vertically downward in 0.32 seconds.