Tag Archives: Physics PDF solutions

Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 8


Determine the signs (positive or negative) of the position, velocity, and acceleration for the particle in Figure Q1.8.

Figure Q1.8

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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 7


Determine the signs (positive or negative) of the position, velocity, and acceleration for the particle in Figure Q1.7.

Figure Q1.7

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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 6


Determine the signs (positive or negative) of the position, velocity, and acceleration for the particle in Figure Q1.6.

Figure Q1.6

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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 5


Does the object represented in Figure Q1.5 have a positive or negative value of \displaystyle a_y? Explain.

Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 5

Figure Q1.5


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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 4


Does the object represented in Figure Q1.4 have a positive or negative value of \displaystyle a_x? Explain.

Figure Q1.4 Physics for Scientists and Engineers 3rd Edition by Randall Knight

Figure Q1.4


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


Is the particle in FIGURE Q1.3 speeding up? Slowing down? Or can you tell? Explain.

Q1.3 Physics for Scientist and Engineers 3rdf Edition by Randall Knight

Figure Q1.3


Solution:

The diagram does not indicate any position in time that should have been represented by numbers on the dots. Without numbers on the dots we cannot tell if the particle in the figure is moving left or right, so we can’t tell if it is speeding up or slowing down. If the particle is moving to the right it is speeding up. If it is moving to the left it is slowing down. 


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Physics for Scientists and Engineers 3rd Edition by Randall Knight, Chapter 1 Conceptual Question 2


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

a)  310

b)  0.00310

c)  1.031

d)  3.10×105


Solution:

Part a

Zeroes before the decimal point merely locate the decimal point and are not significant. Thus, 310 has 2 significant figures

Part b

Trailing zeros after the decimal point are significant because they indicate increased precision. Therefore, 0.00310 has 3 significant figures

Part c

Zeroes between nonzero digits are significant. Therefore, 1.031 has 4 significant figures

Part d

Just like in part b, trailing zeros after the decimal point are significant because they indicate increased precision.Therefore, 3.10×105 has 3 significant figures.


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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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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)

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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*}

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