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Q15PE

Expert-verifiedFound in: Page 11

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) Suppose that a person has an average heart rate of ${\mathbf{72}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{beats}}{\mathbf{/}}{\mathbf{min}}$** **beats/ min. How many beats does he or she have in 2.0 y ? (b) In 2.00 y ? (c) In 2.000 y ?**

- In 2.0 years , there are only 2 significant figures. Hence, the number of beats is $7.6\times {10}^{7}$ beats.
- In 2.0 years , since there are 3 significant figures. Hence, the number of beats is $7.57\times {10}^{7}\mathrm{beats}$.
- There are 4 significant figures in 2,000 years, the $72.0\mathrm{beats}/\mathrm{min}$ only has 3 significant figures, so report the answer with 3 significant figures i.e., $7.573\times {10}^{7}beats$ .

**A conversion factor that determines how many units are equal to a unit.**

**Given Data:**

Consider the given data as below.

Average heart rate is 72.0 beats/min .

- Multiply 72.0 beats/min by 2.0 years for calculating the number of beats she has in 2.0 years and hence after that use conversion factors to cancel the units of time.

$\mathrm{Numbers}\mathrm{of}\mathrm{beats}=\frac{72\mathrm{beats}}{1\mathrm{min}}\times \frac{60\mathrm{min}}{1\mathrm{hr}}\times \frac{24\mathrm{hr}}{1\mathrm{day}}\times \frac{365\mathrm{days}}{1\mathrm{yr}}\times 2\mathrm{years}\phantom{\rule{0ex}{0ex}}=7.57382\times {10}^{7}\mathrm{beats}$

In 2.0 years , there are only 2 significant figures. Hence, the number of beats is $7.6\times {10}^{7}\mathrm{beats}$.

(b) In 2.0 years , since there are 3 significant figures. Hence, the number of beats is $7.57\times {10}^{7}\mathrm{beats}$.

(c) Here, there are 4 significant figure in $2,000\mathrm{years}$ , the $72.0\mathrm{beats}/\mathrm{min}$ only has 3 significant figures, so report the answer with 3 significant figures i.e., $7.573\times {10}^{7}\mathrm{beats}$.

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