Tag Archives: College Physics Solutions

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

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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.

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SOLUTION:

The average area of the room is

A=l×w=3.955 m×3.050 m=12.06 m2  (Answer)\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.

% uncwidth=0.005 m3.050 m×100%=0.1639%% unclength=0.005 m3.955 m×100%=0.1264 %\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.

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

The uncertainty in the area is

δarea=0.2903 %100 %×12.06 m2=0.035 m2  (Answer)\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±0.035 m2  (Answer)A=12.06\pm 0.035\:\text{m}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-26: Uncertainty and percent uncertainty of a pound-mass (lbm) unit

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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?

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SOLUTION:

Part A

The percent uncertainty of the lbm is

% uncertaintylbm=0.0001 kg0.4539 kg×100%=0.022%  (Answer)\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

lbm=δlbm% uncertaintylbm×100%=1 kg0.02 %×1 lbm0.04539 kg×100%=11015.64 lbm  (Answer)\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*}
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Problem 1-25: Solving for the uncertainty of the volume of a box with dimensional uncertainties

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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.

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SOLUTION:

The average volume of the box is

Volume=l×w×h=(1.80 cm)(2.05 cm)(3.1 cm)=11.4 cm3  (Answer)\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:

1.80±0.01 cm → 0.01 cm1.80 cm×100%=0.556%2.05±0.02 cm → 0.02 cm2.05 cm×100%=0.976%3.1±0.1 cm → 0.1 cm3.1 cm×100%=3.226%\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.

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

The uncertainty of the volume is

δvolume=0.04758×11.4 cm3=0.54 cm3  (Answer)\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

Volume=11.4±0.54 cm3  (Answer)\text{Volume}=11.4\pm 0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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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

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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?

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SOLUTION:

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

v=dtv=\frac{d}{t}

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

t=dvt=\frac{d}{v}

By direct substitution

t=26.22 mi9.5 mi/hr=2.76 hours  (Answer)t=\frac{26.22\:\text{mi}}{9.5\:\text{mi/hr}}=2.76\:\text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-22: Solving for the area of a circle with a given diameter

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PROBLEM:

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

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SOLUTION:

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

A=πr2A=\pi r^2

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

A=π(d2)2A=\pi \left(\frac{d}{2}\right)^2

So, by direct substitution

A=π(3.102 cm2)2=7.557 cm2  (Answer)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.

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

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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?

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SOLUTION:

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

A=40 beats30 sec ×60 sec1 min=80 beats/minA=\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

% uncertainty=(1 beat40 beats×100%)+(0.5 s30.0 s×100%)=2.5%+1.7%=4.2%\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

δA=% uncertainty100 %×A=4.2%100%×80 beats/min=3.4 beats/min\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

80±3 beats/min  (Answer)\displaystyle 80\pm 3\:\text{beats/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-20: Solving for the percent uncertainty of a given blood pressure of 120±2 mmHg

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PROBLEM:

(a) A person’s blood pressure is measured to be 120±2120 \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?

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SOLUTION:

Part A

The percent uncertainty is computed as

% uncertainty=2 mmHg120 mmHg×100%=1.7%  (Answer)\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

δbp=1.7 %100 %×80 mmHg=1.3 mmHg  (Answer)\delta _{bp}=\frac{1.7\:\%}{100\:\%}\times 80\:\text{mmHg}=1.3\:\text{mmHg} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-19: Calculating the percent uncertainty and range of speed with the same percent uncertainty

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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?

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SOLUTION:

Part a

The percent uncertainty is computed as

% uncertainty=2.0 km/hr90 km/hr×100%=2.2%  (Answer)\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

δv=2.2 %100 %×60 km/hr=1 km/hr\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.   (Answer) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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

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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?

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SOLUTION:

Part A

99 has 2 significant figures

100 has 3 significant figures

Part B

199×100%=1.0 %1100×100%=1.00%\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.

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Problem 1-17: Stating the correct significant figures in a given calculation

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PROBLEM:

State how many significant figures are proper in the results of the following calculations:

(a) (106.7)(98.2)(46.210)(1.01)  (b) (18.7)2  (c) (1.60×10−19)(3712)\begin{align*} \left(a\right)\:\frac{\left(106.7\right)\left(98.2\right)}{\left(46.210\right)\left(1.01\right)} \ \qquad \ \left(b\right)\:\left(18.7\right)^2 \ \qquad \ \left(c\right)\:\left(1.60\times 10^{-19}\right)\left(3712\right) \end{align*}
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SOLUTION:

Part A

The answer is limited by 98.2 and 1.01. They are both 3 significant figures. So the result should be 3 significant figures.

Part B

The answer is limited by 18.7. The answer should be 3 significant figures. So the result should be 3 significant figures.

Part C

The answer is limited by 1.60 which is 3 significant figures. So the result should be 3 significant figures.

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