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.

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)

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?

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)

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?

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)

PROBLEM:

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

\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*}

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.