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Expert-verified Found in: Page 11 ### College Physics (Urone)

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 ?

1. In 2.0 years , there are only 2 significant figures. Hence, the number of beats is $7.6×{10}^{7}$ beats.
2. In 2.0 years , since there are 3 significant figures. Hence, the number of beats is $7.57×{10}^{7}\mathrm{beats}$.
3. 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., $7.573×{10}^{7}beats$ .
See the step by step solution

## Step 1: Definition of conversion factor

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 .

## Step 2: Calculating total beats

1. 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.

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

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

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

(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., $7.573×{10}^{7}\mathrm{beats}$. ### Want to see more solutions like these? 