Q15PE - College Physics (Urone)

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 ?

Step-by-Step Solution

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 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.

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

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.,  $7.573\times {10}^7 \mathrm{beats}$.