Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 18 - Electric Currents - Problems - Page 521: 6

Answer

$2.7\times10^5C$

Work Step by Step

The ampere-hour is a unit that measures charge. $$(75A\cdot h)(\frac{1C/s}{A})(\frac{3600s}{1h})=2.7\times10^5C$$
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