Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 4 - Section 4.6 - Multiplication and Division of Measurements - Exercise - Page 188: 24


$$7\bar{0}\ kg \ m/s^2$$

Work Step by Step

We have $$ \frac{(19 \ kg)(3.0 \ m/s)^2 }{ 2.46\ \ m}=69.512\ kg \ m/s^2 .$$ Now, round this product to two significant digits, which is the accuracy of the least accurate measurement. That is, $$69.512\ kg \ m/s^2=7\bar{0}\ kg \ m/s^2$$
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