Tag Archives: Solution Manual for College Physics by Openstax

Problem 2-8: Motion of land mass around the San Andreas fault

Advertisements

PROBLEM:

Land west of the San Andreas fault in southern California is moving at an average velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is 590 km northwest, assuming the motion remains constant?

Advertisements

SOLUTION:

From the formula \overline{v}=\frac{\Delta x}{\Delta t}, we can solve for \Delta t as follows

\begin{align*}
\Delta t & =\frac{\Delta x}{\overline{v}} \\ \\
& =\frac{5.90 \times 10^{5}\ \text{m}}{6\ \text{cm/year}}\times \frac{100\ \text{cm}}{1\ \text{m}} \\ \\
& = 9.83 \times 10^{6}\ \text{years} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Therefore, it will take about 9.83 million years.

Advertisements
Advertisements

Problem 2-6: The average speed and average velocity of a spinning helicopter blade

Advertisements

PROBLEM:

A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation.

(a) Calculate the average speed of the blade tip in the helicopter’s frame of reference.

(b) What is its average velocity over one revolution?

Advertisements

SOLUTION:

Part A

The average speed of the blade tip is equal to the distance traveled divided by the time elapsed.

\begin{align*}
\text{speed} & =\frac{ \text{distance traveled}}{ \text{time elapsed}} \\ \\
& =\frac{2\pi \text{r}}{ \text{t}} \\ \\
& =\frac{2\pi \left(5.00\text{m}\right)}{60\:\text{s}}\times 100\:\text{rev} \\ \\
& =52.36\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

After one revolution, the tip of the blade is at the same position as it is originally. This means that the displacement is zero. Thus, the velocity is zero.

\text{v}=0\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-26: Uncertainty and percent uncertainty of a pound-mass (lbm) unit

Advertisements
Advertisements

PROBLEM:

When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1 lbm=0.4539 kg.

(a) If there is an uncertainty of 0.0001 kg in the pound-mass unit, what is its percent uncertainty?

(b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of 1 kg when converted to kilograms?

Advertisements
Advertisements

SOLUTION:

Part A

The percent uncertainty of the lbm is

\text{\%\:uncertainty}_{\text{lbm}}=\frac{0.0001\:\text{kg}}{0.4539\:\text{kg}}\times 100\%=0.022\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

For uncertainty of 1 kg, the corresponding lbm is

\begin{align*}
\text{lbm} & =\frac{\delta_{\text{lbm}}}{\text{\%\:uncertainty}_{\text{lbm}}}\times 100\% \\
\\
 & =\frac{1\:\text{kg}}{0.02\:\%}\times \frac{1\:\text{lbm}}{0.04539\:\text{kg}}\times 100\% \\
\\
& =11015.64\:\text{lbm} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) 
\end{align*}
Advertisements
Advertisements

Problem 1-20: Solving for the percent uncertainty of a given blood pressure of 120±2 mmHg

Advertisements
Advertisements

PROBLEM:

(a) A person’s blood pressure is measured to be 120 \pm 2 mmHg. What is its percent uncertainty?

(b) Assuming the same percent uncertainty, what is the uncertainty in a blood pressure measurement of 80 mmHg?

Advertisements
Advertisements

SOLUTION:

Part A

The percent uncertainty is computed as

\text{\% uncertainty}=\frac{2\:\text{mmHg}}{120\:\text{mmHg}}\times 100\%=1.7\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

The uncertainty in the blood pressure is

\delta _{bp}=\frac{1.7\:\%}{100\:\%}\times 80\:\text{mmHg}=1.3\:\text{mmHg} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-3: Converting 1.0 m/s to 3.6 km/h

Advertisements
Advertisements

PROBLEM:

Show that 1.0 m/s=3.6 km/h.

Hint: Show the explicit steps involved in converting 1.0 m/s=3.6 km/h.

Advertisements
Advertisements

SOLUTION:

We know that 1 hr = 3600 s, and 1 km = 1000 m. 

\begin{aligned}
1.0 \ \text{m/s} & = 1.0 \ \frac{\bcancel{\text{m}}}{ \bcancel{\text{s}}} \times \frac{3600 \ \bcancel{\text{s}}}{1 \ \text{hr}}\times \frac{1 \ \text{km}}{1000 \ \bcancel{\text{m}}} \\
\\
& =3.6 \ \text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{aligned}
Advertisements
Advertisements