College Physics Problem 1.17


State how many significant figures are proper in the results of the following calculations:

\displaystyle \left(a\right)\:\frac{\left(106.7\right)\left(98.2\right)}{\left(46.210\right)\left(1.01\right)}

\displaystyle \left(b\right)\:\left(18.7\right)^2

\displaystyle \left(c\right)\:\left(1.60\times 10^{-19}\right)\left(3712\right)


Solution:

Part a

The answer is limited by 98.2 and 1.01. They are both 3 significant figures. So the result should be 3 significant figures.

Part b

The answer is limited by 18.7. The answer should be 3 significant figures. So the result should be 3 significant figures.

Part c

The answer is limited by 1.60 which is 3 significant figures. So the result should be 3 significant figures.