College Physics Problem 1.15

(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?
(b) In 2.00 y?
(c) In 2.000 y?

SOLUTION:

This is a problem on rounding off. Take note that 72.0 has three significant figures, 2.0 has two significant figures, 2.00 has three significant figures, and 2.000 has four significant figures.

In 2 years, the number of beats of an average person is

frac{72.0  beats}{1  min}times frac{60.0  min}{1.00  h}times frac{24.0  hr}{1.00  day}times frac{365.25  days}{1  year}=7.5738times10^{7}  beats

For item letter a), the answer is limited by 2.0, which is 2 significant figures. The answer is 7.6times10^{7}  beats.

For item letter b), the answer is limited by 2.00, which is 3 significant figures. The answer is 7.57times 10^{7}  beats.

For item letter c), the answer is limited by 72.0, which is 3 significant figures. The answer is 7.57times 10^{7}  beats.

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