Tag Archives: Precision

Solution Guide to College Physics by Openstax Chapter 1 Banner

Chapter 1: Introduction: The Nature of Science and Physics

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Physical Quantities and Units

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Accuracy, Precision, and Significant Figures

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

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Approximation

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

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

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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.

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SOLUTION:

Part A

The average volume is

V=πr2h=π(7.5 cm2)2(3.25 cm)=143.5806 cm3  (Answer)\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

% uncr=0.002 cm7.500 cm×100%=0.027%% unch=0.001 cm3.25 cm×100%=0.031%\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

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

The uncertainty in the volume is

δvol=0.058100×143.5806=0.083 cm3  (Answer) \delta _{vol}=\frac{0.058}{100}\times 143.5806=0.083\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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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 1-24: Calculating uncertainties in distance, time and speed of a marathon runner

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PROBLEM:

A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time.

(a) Calculate the percent uncertainty in the distance.

(b) Calculate the uncertainty in the elapsed time.

(c) What is the average speed in meters per second?

(d) What is the uncertainty in the average speed?

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SOLUTION:

Part A

The percent uncertainty in the distance is

% uncertaintydistance=25 m42.188 km×1 km1000 m×100%=0.0593%  (Answer)\begin{align*} \text{\%\:uncertainty}_{\text{distance}} & =\frac{25\:\text{m}}{42.188\:\text{km}}\times \frac{1\:\text{km}}{1000\:\text{m}}\times 100\% \\ & =0.0593\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part B

The uncertainty in time is

% uncertaintytime=1 s9012 s×100%=0.0111%  (Answer)\begin{align*} \text{\%\:uncertainty}_{\text{time}} & =\frac{1\:\text{s}}{9012\:\text{s}}\times 100\% \\ & =0.0111\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part C

The average speed is

average speed=42.188 km9012 s×1000 m1 km=4.681 m/s  (Answer)\begin{align*} \text{average speed} & =\frac{42.188\:\text{km}}{9012\:\text{s}}\times \frac{1000\:\text{m}}{1\:\text{km}} \\ & = 4.681\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

Part D

The percent uncertainty in the speed is the combination of uncertainties of distance and time.

% uncertaintyspeed=% uncertaintydistance+% uncertaintytime=0.0593%+0.0111%=0.0704%\begin{align*} \text{\%\:uncertainty}_{\text{speed}} & =\text{\%\:uncertainty}_{\text{distance}}+\text{\%\:uncertainty}_{\text{time}} \\ & =0.0593\%+0.0111\% \\ & =0.0704\% \\ \end{align*}

Therefore, the uncertainty in the speed is

δspeed=0.0704%100%×4.681 m/s=0.003 m/s  (Answer)\begin{align*} \delta _{speed} & =\frac{0.0704\%}{100\%}\times 4.681\:\text{m/s} \\ & = 0.003\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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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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