Tag Archives: Physics Solutions

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.

Advertisements

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

Conceptual Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Exercises and Problems

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

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

College Physics by Openstax Chapter 2 Problem 49

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

Solution:

The known values are a=-9.80\:\text{m/s}^2; v_o=15.0\:\text{m/s}; y=7.00\:\text{m}

The applicable formula is.

y=v_ot+\frac{1}{2}at^2

Using this formula, we can solve it in terms of time, t.

t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}

Substituting the known values, we have

\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a} \\
t & =\frac{-15.0\:\text{m/s}\pm \sqrt{\left(15.0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(7.00\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\
t&=\frac{-15.0\:\text{m/s}\pm 9.37\:\text{m/s}}{-9.80\:\text{m/s}^2}
\end{align*}

We have two values for time, t. These two values represent the times when the ball passes the tree branch.

 t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec \\
t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:sec

Therefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds.

t_2-t_1=2.49  \ \text{s} - 0.57 \ \text{s}=1.92 \ \text{s}  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 6-1: Odometer reading based on the number of wheel revolutions

Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions—it then calculates the distance traveled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometers should the odometer read?

Solution:

The formula for the total distance traveled is

\Delta s=\Delta \theta \times r

Therefore, the total distance traveled is

\begin{align*}
\Delta s & =\left(200000\:\text{rotations}\:\times \frac{2\pi \:\text{radian}}{1\:\text{rotation}}\right)\left(\frac{1.15\:\text{m}}{2}\right) \\
\Delta s & =722566.3103\:\text{m} \\
\Delta s & =722.6\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}
Advertisements
Advertisements

College Physics by Openstax Chapter 5 Problem 1

A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials, he quickly calculates the normal force. What is it?

Solution:

The formula for friction is

f=\mu _{k\:}N

When we solve for the normal force, N, in terms of the other variables, we have

N=\frac{f}{\mu _k}

The coefficient of kinetic friction is 0.04. Therefore, the normal force is

\begin{align*}
N & =\frac{f}{\mu _k} \\
N & =\frac{0.200\:\text{newton}}{0.04} \\
N & =5.00\:\text{newton} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Advertisements
Advertisements

College Physics by Openstax Chapter 4 Problem 1

A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s2. What is the net external force on him?

Solution:

So, we are given mass, m = 63.0 \ \text{kg} , and acceleration, a = 4.20 \ \text{m/s}^2.

The net force has a formula 

\text{F}=\text{m}a

Substituting the given values, we have

\begin{align*}
F & = \left( 63.0 \ \text{kg} \right)\left( 4.20 \ \text{m/s}^2 \right) \\
F & = 265 \  \text{kg}\cdot \text{m/s}^2 \\
F & = 265 \ \text{N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
Advertisements
Advertisements

College Physics by Openstax Chapter 3 Problem 1

Find the following for path A in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

\begin{align*}

\text{d} & =\left(3\times 120\ \text{m}\right)+\left(1\times 120\:\text{m}\right) \\
\text{d} & =480\:\text{m}  \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\

\end{align*}

Part B

The magnitude of the displacement is 

\begin{align*}

\text{s }& =\sqrt{\left( s_x \right)^{2\:}+\left( s_y \right)^2} \\
\text{s }& = \sqrt{\left(1\times 120\:\text{m}\right)^2+\left(3\times 120\:\text{m}\right)^2} \\
\text{s }& = 379\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

The direction is

\begin{align*}

 \theta & = \tan^{-1}\left(\frac{s_x}{s_y}\right) \\
\theta & = \tan^{-1}\left(\frac{1\times 120\:\text{m}}{3\times 120 \ \text{m}}\right) \\
\theta & =71.6^{\circ} ,\:\text{E of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}
Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 48

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?

Solution:

The known values are: y_0=2.20\:\text{m}; y=1.80\:\text{m}; v_0=11.0\:\text{m/s}; and a=-9.80\:\text{m/s}^2

We are going to use the formula

 \Delta y=v_0t+\frac{1}{2}at^2

Substituting the given values:

\begin{align*}
 \Delta y & =v_0t+\frac{1}{2}at^2 \\
1.80\:\text{m}-2.20\:\text{m} & =\left(11.0\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\
-0.40\:\text{m} & =\left(11.0\:\text{m/s}\right)t-\left(4.90\:\text{m/s}^2\right)t^2 \\
4.90t^2-11t-0.40 & =0
\end{align*}

Using the quadratic formula solve for t, we have

\begin{align*}
t & =\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\left(4.90\right)\left(-0.40\right)}}{2\left(4.90\right)} \\
\end{align*}
 t=2.28\:\text{sec}\:\:\:\:\:\text{or}\:\:\:\:\:\:t=-0.04 \ \text{sec}

We can discard the negative time, so

t=2.28\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 47

(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s.

(b) How long would it take to reach the ground if it is thrown straight down with the same speed?

Solution:

Part A

Refer to the figure below.

The known values are: t=2.35\:\text{s}; y=0\:\text{m}; v_0=+8.00\:\text{m/s}; and a=-9.8\:\text{m/s}^2

Based on the given values, the formula that we shall use is

y=y_0+v_0t+\frac{1}{2}at^2

Substituting the values, we have

\begin{align*}
y & =y_0+v_0t+\frac{1}{2}at^2 \\
0\: & =y_0+\left(8.00\:\text{m/s}\right)\left(2.35\:\text{s}\right)+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(2.35\:\text{s}\right)^2 \\
y_0 & =8.26\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Therefore, the cliff is 8.26 meters high.

Part B

Refer to the figure below

The knowns now are: y=0\:\text{m}; y_0=8.26\:\text{m}; v_0=-8.00\:\text{m/s}; and a=-9.80\:\text{m/s}^2

Based on the given values, we can use the formula 

y=y_0+v_0t+\frac{1}{2}at^2

Substituting the values, we have

\begin{align*}
y & =y_0+v_0t+\frac{1}{2}at^2 \\
0\:\text{m} & =8.26\:\text{m}+\left(-8.00\:\text{m/s}\right)t+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)t^2 \\
4.9 t^2+8t-8.26 & =0 \\
\end{align*}

Using the quadratic formula to solve for the value of t, we have

\begin{align*}
t &=\frac{-8\pm \sqrt{\left(8\right)^2-4\left(4.9\right)\left(-8.26\right)}}{2\left(4.9\right)} \\
t &=0.717\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
Advertisements
Advertisements

College Physics by Openstax Chapter 2 Problem 46

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool.

(a) How long are her feet in the air?

(b) What is her highest point above the board?

(c) What is her velocity when her feet hit the water?

Solution:

The known values are: y_0=1.80\:\text{m}, y=0\:\text{m}, a=-9.80\:\text{m/s}^2, v_0=4.00\:\text{m/s}.

Part A

Based from the knowns, the formula most applicable to solve for the time is \Delta y=v_0t+\frac{1}{2}at^2. If we rearrange the formula by solving for t, and substitute the given values, we have

\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2-2a\Delta y}}{a} \\
t & =\frac{-4.00\:\text{m/s}\pm \sqrt{\left(4.00\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.80\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\
t & =1.14\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We have the formula

\begin{align*}
\Delta y & =\frac{v^2-v_0^2}{2a} \\
\Delta y & =\frac{\left(0\:\text{m/s}\right)^2-\left(4.00\:\text{m/s}\right)^2}{2\left(-9.80\:\text{m/s}^2\right)} \\
\Delta y & =0.816\: \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

The formula to be used is

v^2=v_0^2+2a\Delta y

Substituting the given values

\begin{align*}
v^2 & =v_0^2+2a\Delta y \\
v & =\pm \sqrt{v_0^2+2a\Delta y} \\
v & =\pm \sqrt{\left(4.00\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(-1.80\:\text{m}\right)} \\
v & =\pm \sqrt{51.28\:\text{m}^2/\text{s}^2} \\
v &=\pm 7.16\:\text{m/s}
\end{align*}

Since the diver must be moving in the negative direction,

v=-7.16\:\text{m/s}  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements