Tag Archives: College Physics Solutions

Problem 1-28: Calculating the volume and its uncertainty of a car piston with dimensional uncertainties

Advertisements
Advertisements

PROBLEM:

A car engine moves a piston with a circular cross section of  7.500±0.002 cm diameter a distance of 3.250±0.001 cm  to compress the gas in the cylinder.

(a) By what amount is the gas decreased in volume in cubic centimeters?

(b) Find the uncertainty in this volume.

Advertisements
Advertisements

SOLUTION:

Part A

The average volume is

\begin{align*}
V & =\pi r^2h \\
& =\pi \left(\frac{7.5\:\text{cm}}{2}\right)^2\left(3.25\:\text{cm}\right) \\
& =143.5806\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part B

Solve for the percent uncertainties of each dimension

\begin{align*}
\%\:unc_r & =\frac{0.002\:\text{cm}}{7.500\:\text{cm}}\times 100\%=0.027\% \\
\%\:unc_h & =\frac{0.001\:\text{cm}}{3.25\:\text{cm}}\times 100\%=0.031\% \\
\end{align*}

The percent uncertainty in the volume is the combined effect of the uncertainties of the dimensions

\text{\%\:unc}_{vol}=0.027\%+0.031\%=0.058\%

The uncertainty in the volume is

 \delta _{vol}=\frac{0.058}{100}\times 143.5806=0.083\:\text{cm}^3  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-27: Calculating the area and its uncertainty of a room with a given dimensional uncertainties

Advertisements
Advertisements

PROBLEM:

The length and width of a rectangular room are measured to be 3.955 ±0.005 m and 3.050 ± 0.005 m . Calculate the area of the room and its uncertainty in square meters.

Advertisements
Advertisements

SOLUTION:

The average area of the room is

\begin{align*}
A & =l\times w \\
& =3.955\:\text{m}\times 3.050\:\text{m} \\
& =12.06\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Compute for the percent uncertainties of each dimension.

\begin{align*}
\text{\%\:unc}_{width} & =\frac{0.005\:\text{m}}{3.050\:\text{m}}\times 100\%=0.1639\% \\
\text{\%\:unc}_{length} & =\frac{0.005\:\text{m}}{3.955\:\text{m}}\times 100\%=0.1264\:\%
\end{align*}

The percent uncertainty in the area is the combined effect of the uncertainties of the length and width.

\text{\%\:unc}_{area}=0.1639\%+0.1264\%=0.2903\%

The uncertainty in the area is

\delta _{area}=\frac{0.2903\:\%}{100\:\%}\times 12.06\:\text{m}^2=0.035\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Therefore, the area is

A=12.06\pm 0.035\:\text{m}^2 \ \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 1-25: Solving for the uncertainty of the volume of a box with dimensional uncertainties

Advertisements
Advertisements

PROBLEM:

The sides of a small rectangular box are measured to be 1.80±0.01 cm, 2.05±0.02 cm, and 3.1±0.1 cm long. Calculate its volume and uncertainty in cubic centimeters.

Advertisements
Advertisements

SOLUTION:

The average volume of the box is

\begin{align*}
\text{Volume} & =l\times w\times h \\
& =\left(1.80\:\text{cm}\right)\left(2.05\:\text{cm}\right)\left(3.1\:\text{cm}\right) \\
&= 11.4\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

The percent uncertainty for each of the dimensions:

\begin{align*}
1.80\pm 0.01\:\text{cm}\:\rightarrow & \:\frac{0.01\:\text{cm}}{1.80\:\text{cm}}\times 100\%=0.556\% \\
\\
2.05\pm 0.02\:\text{cm}\:\rightarrow & \:\frac{0.02\:\text{cm}}{2.05\:\text{cm}}\times 100\%=0.976\% \\
\\
3.1\pm 0.1\:\text{cm}\:\rightarrow &\:\frac{0.1\:\text{cm}}{3.1\:\text{cm}}\times 100\%=3.226\% \\
\end{align*}

The percent uncertainty in the volume of the box is calculated by adding the percent uncertainties of the dimensions.

\begin{align*}
\%\:\text{uncertainty}_{\text{volume}} & =0.556\%+0.976\%+3.226\% \\
& =4.758\%
\end{align*}

The uncertainty of the volume is

\begin{align*}
\delta _{\text{volume}} & =0.04758\times 11.4\:\text{cm}^3 \\
& =0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the volume is

\text{Volume}=11.4\pm 0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-23: Calculating for the time it takes a marathon runner to finish 26.22 miles given a speed of 9.5 mi/h

Advertisements
Advertisements

PROBLEM:

If a marathon runner averages 9.5 mi/h, how long does it take him or her to run a 26.22-mi marathon?

Advertisements
Advertisements

SOLUTION:

The relationship of the velocity (or speed), time, and distance is

v=\frac{d}{t}

So, by rearranging the equation, we can equate time to

t=\frac{d}{v}

By direct substitution

t=\frac{26.22\:\text{mi}}{9.5\:\text{mi/hr}}=2.76\:\text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Advertisements
Advertisements

Problem 1-22: Solving for the area of a circle with a given diameter

Advertisements
Advertisements

PROBLEM:

What is the area of a circle 3.102 cm in diameter?

Advertisements
Advertisements

SOLUTION:

The area of a circle can be computed using the formula below when the radius is given.

A=\pi r^2

We also know that the radius is half the diameter, so the area can be calculated using the formula,

A=\pi \left(\frac{d}{2}\right)^2

So, by direct substitution

A=\pi \left(\frac{3.102\:\text{cm}}{2}\right)^2=7.557\:\text{cm}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The area of the circle is 7.557 square centimeters.

Advertisements
Advertisements

Problem 1-21: Counting heart rate with uncertainties in number of beats and time

Advertisements
Advertisements

PROBLEM:

A person measures his or her heart rate by counting the number of beats in 30 s. If 40±1  beats are counted in 30±0.5 s, what is the heart rate and its uncertainty in beats per minute?

Advertisements
Advertisements

SOLUTION:

In order to compute for the heart rate in beats per minute, we need to solve for the base. The base is

A=\frac{40\:\text{beats}}{30\:\text{sec}\:}\times \frac{60\:\text{sec}}{1\:\text{min}}=80\:\text{beats/min}

Then we compute for the percent uncertainty by combining the uncertainties of the number of beats and time. That is

\begin{align*}
\text{\%\:uncertainty} & =\left( \frac{1\:\text{beat}}{40\:\text{beats}}\times 100\% \right)+ \left(\frac{0.5\:\text{s}}{30.0\:\text{s}}\times 100\% \right)\\
&=2.5\%+1.7\% \\
& =4.2\% \\

\end{align*}

Based on this percent uncertainty, we compute for the tolerance

\begin{align*}
\delta _A & =\frac{\text{\%\:uncertainty}}{100\:\%}\times A \\
& = \frac{4.2 \%}{100 \%} \times 80 \ \text{beats/min} \\
& =3.4\:\text{beats/min}\\
\end{align*}

Therefore, the heart rate is

\displaystyle 80\pm 3\:\text{beats/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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 1-19: Calculating the percent uncertainty and range of speed with the same percent uncertainty

Advertisements
Advertisements

PROBLEM:

(a) If your speedometer has an uncertainty of 2.0 km/h at a speed of 90 km/h, what is the percent uncertainty?

(b) If it has the same percent uncertainty when it reads 60 km/ h, what is the range of speeds you could be going?

Advertisements
Advertisements

SOLUTION:

Part a

The percent uncertainty is computed as

\text{\% uncertainty}=\frac{2.0\ \text{km/hr}}{90\:\text{km/hr}}\times 100\%=2.2\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part b

The tolerance of the velocity is

\delta _v=\frac{2.2\:\%}{100\:\%}\times 60\:\text{km/hr}=1\:\text{km/hr}

Therefore, the range of the velocity is 60±1km/h, or that is 59 to 61 km/h. \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Advertisements
Advertisements

Problem 1-18: Significant figures, uncertainty, and accuracy of the numbers 99 and 100

Advertisements
Advertisements

PROBLEM:

(a) How many significant figures are in the numbers 99 and 100?

(b) If the uncertainty in each number is 1, what is the percent uncertainty in each?

(c) Which is a more meaningful way to express the accuracy of these two numbers, significant figures or percent uncertainties?

Advertisements
Advertisements

SOLUTION:

Part A

99 has 2 significant figures

100 has 3 significant figures

Part B

\begin{align*}
\frac{1}{99}\times 100\% & =1.0\:\%  \\ 
\frac{1}{100}\times 100\% & =1.00\%
\end{align*}

Part C

Based on the results of parts a and b, the percent uncertainties are more meaningful.

Advertisements
Advertisements