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问题 1.

(a) Suppose that a person has an average heart rate of 72.0 beats $/ \mathrm{min}$. How many beats does he or she have in $2.0 y$ ? (b) In $2.00 y$ ? (c) In $2.000 y$ ?


Solution (a) To calculate the number of beats she has in 2.0 years, we need to multiply 72.0 beats/minute by 2.0 years and use conversion factors to cancel the units of time:
$$
\frac{72.0 \text { beats }}{1 \mathrm{~min}} \times \frac{60.0 \mathrm{~min}}{1.00 \mathrm{~h}} \times \frac{24.0 \mathrm{~h}}{1.00 \mathrm{~d}} \times \frac{365.25 \mathrm{~d}}{1.00 \mathrm{y}} \times 2.0 \mathrm{y}=7.5738 \times 10^7 \text { beats }
$$
Since there are only 2 significant figures in 2.0 years, we must report the answer with 2 significant figures: $7.6 \times 10^7 \text { beats. }$
(b) Since we now have 3 significant figures in 2.00 years, we now report the answer with 3 significant figures: $7.57 \times 10^7$ beats.
(c) Even though we now have 4 significant figures in 2.000 years, the 72.0 beats/minute only has 3 significant figures, so we must report the answer with 3 significant figures: $7.57 \times 10^7$ beats.

问题 2. A person measures his or her heart rate by counting the number of beats in $30 \mathrm{~s}$. If $40 \pm 1$ beats are counted in $30.0 \pm 0.5 \mathrm{~s}$, what is the heart rate and its uncertainty in beats per minute?


Solution To calculate the heart rate, we need to divide the number of beats by the time and convert to beats per minute.

$$
\frac{\text { beats }}{\text { minute }}=\frac{40 \text { beats }}{30.0 \mathrm{~s}} \times \frac{60.0 \mathrm{~s}}{1.00 \mathrm{~min}}=80 \text { beats } / \mathrm{min}
$$
To calculate the uncertainty, we use the method of adding percents.
$$
\% \text { unc }=\frac{1 \text { beat }}{40 \text { beats }} \times 100 \%+\frac{0.5 \mathrm{~s}}{30.0 \mathrm{~s}} \times 100 \%=2.5 \%+1.7 \%=4.2 \%=4 \%
$$
Then calculating the uncertainty in beats per minute:
$\delta A=\frac{\% \text { unc }}{100 \%} \times A=\frac{4.2 \%}{100 \%} \times 80$ beats $/ \mathrm{min}=3.3$ beats $/ \mathrm{min}=3$ beats $/ \mathrm{min}$
Notice that while doing calculations, we keep one EXTRA digit, and round to the correct number of significant figures only at the end.
So, the heart rate is $80 \pm 3$ beats $/ \mathrm{min}$.

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