Question

(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 ?

Short answer

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

Step-by-step solution

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 .

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.

$$\begin{gathered} \mathrm{Numbers}\mathrm{of}\mathrm{beats}=\frac{72\mathrm{beats}}{1\min }\times \frac{60\min }{1\mathrm{hr}}\times \frac{24\mathrm{hr}}{1\mathrm{day}}\times \frac{365\mathrm{days}}{1\mathrm{yr}}\times 2\mathrm{years} \\ =7.57382\times {10}^7\mathrm{beats} \end{gathered}$$

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 $72.0\mathrm{beats}/\min$ only has 3 significant figures, so report the answer with 3 significant figures i.e., $7.573\times {10}^7\mathrm{beats}$.