Tag Archives: uncertainty

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

\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)
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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

\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)
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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

\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*}
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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=\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.

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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=\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)
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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 \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

\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)
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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

\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 1-14: Solving for the percent uncertainty of 130±5 beats/min

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PROBLEM:

An infant’s pulse rate is measured to be 130±5 beats/min. What is the percent uncertainty in this measurement?

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SOLUTION:

The percent uncertainty can be calculated as

\begin{align*}
\%\:\text{uncertainty}\:& =\frac{\delta _A}{A}\times 100\% \\
&  =\frac{5\:\text{beats/min}}{130\:\text{beats/min}}\times 100\% \\
& =3.85\:\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Therefore, the percent uncertainty is 3.85%.

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Problem 1-13: Computing for the range of possible speeds given 90 km/h and 5% uncertainty

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PROBLEM:

(a) A car speedometer has a  5.0% uncertainty. What is the range of possible speeds when it reads 90 km/h?

(b) Convert this range to miles per hour. (1 km=0.6214 mi)

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SOLUTION:

Part A

The uncertainty in the velocity of the car is computed as

\begin{align*}
\delta _v & =\frac{5.0\:\%}{100\:\%}\times 90.0\:\text{km/hr} \\
& = 4.5\:\text{km/hr}
\end{align*}

Therefore, the range of the possible speeds is 

\begin{align*}
\text{Range} & =90.0\:\pm 4.5\:\text{km/hr} \\
\text{Range} & :85.8\:\text{km/hr}\:-\:94.5\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

So the range of the possible speeds is 85.5 km/hr to 94.5 km/hr.

Part B

Convert the range to mi/h

For 85.5 km/hr

\begin{align*}
85.5\:\text{km/hr}=\left(85\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=53.13\:\text{mi/hr}
\end{align*}

For 94.5 km/hr

94.5\:\text{km/hr}=\left(94.5\:\frac{\text{km}}{\text{hr}}\right)\left(\frac{0.6214\:\text{mi}}{1\:\text{km}}\right)=58.72\:\text{mi/hr}

Therefore, the range can be represented as 53.13 mi/hr to 58.72 mi/hr.

\text{Range} : 53.13 \ \text{mi/hr} - 58.72 \ \text{mi/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-12: Solving for the percent uncertainty of a 20-m tape that is off by 0.50 cm

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PROBLEM:

A good-quality measuring tape can be off by 0.50 cm over a distance of 20 m. What is its percent uncertainty?

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SOLUTION:

The percent uncertainty can be calculated by the uncertainty divided by the total measure, but the units should be the same. That is,

\begin{align*}
\%\:\text{uncertainty} & =\frac{\delta _l}{l}\times 100\% \\
\\
& =\frac{0.5\:\text{cm}}{20\:\text{m}\left(\frac{100\:\text{cm}}{1\:\text{m}}\right)}\times 100\% \\
\\
& =\frac{0.5\:\text{cm}}{2000\:\text{cm}}\times 100\% \\
\\
&=0.025\:\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the percent uncertainty is 0.025%.

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