Question

(a) Suppose that a person has an average heart rate of 72.0 beats/min. How many beats does he or she have in 2.0 y

Answer 1

Answer:

$7.56\times 10^7\ \text{beats}$

Step-by-step explanation:

It is given that,

A person has an average heart rate of 72.0 beats/min.

We need to find how many beats does he or she have in 2.0 years, t = 2 Y

Converting years to min.

$t=2\text{year}\times \dfrac{365\ \text{days}}{\text{year}}\times 24\dfrac{h}{\text{day}}\times \dfrac{60\text{min}}{\text{hour}}\\\\t=1051200\ \text{min}$

So, no of beats in 2 years is given by :

$b=72\ \text{beats/min}\times 1051200\ \text{min}\\\\b=7.56\times 10^7\ \text{beats}$

Hence, the required answer is $7.56\times 10^7\ \text{beats}$.