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 - General Problems - Page 524: 77

Answer

a) 7.37hp b) 224km

Work Step by Step

a) $P=\frac{F\Delta x}{\Delta t}=Fv=(440N)(45km/h)(\frac{1000m\cdot1h}{3600s \cdot1km})=5500W$ $5500W\frac{hp}{746W}=7.37hp$ b) $\Delta t=\frac{\Delta PE}{P}=\frac{\Delta QV}{P}$ $\Delta x=v\frac{\Delta QV}{P}=(12.5m/s)\frac{(95A\times3600s)(12V)(24)}{5500W}=2.24\times10^5m=224km$
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