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:

Part a

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

\displaystyle \left(\frac{72.0\:beats}{1\:min}\right)\left(\frac{60.0\:min}{1.00\:hr}\right)\left(\frac{24.0\:hr}{1.00\:day}\right)\left(\frac{365.25\:days}{1\:year}\right)

\displaystyle =7.5738\times 10^7\:beats

For item letter a), the answer is limited by 2.0, which is 2 significant figures. The answer is 7.6\times 10^7\:beats.

Part b

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

Part c

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