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: 23


$$3\bar{0}0\ V^2/\Omega$$

Work Step by Step

We have $$ \frac{(120\ V)^2 }{ 47.6\ \Omega}=302.52\ V^2/\Omega .$$ Now, round this product to two significant digits, which is the accuracy of the least accurate measurement. That is, $$302.52\ V^2/\Omega=3\bar{0}0\ V^2/\Omega$$
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