Physical Quantities and Units
Accuracy, Precision, and Significant Figures
Approximation
PROBLEM:
SOLUTION:
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*}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)PROBLEM:
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)PROBLEM:
SOLUTION:
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)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*}PROBLEM:
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:
SOLUTION:
Part A
The percent uncertainty in the distance is
\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
\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
\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.
\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
\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*}PROBLEM:
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:
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)^2So, 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.
PROBLEM:
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)
PROBLEM:
SOLUTION:
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)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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