Electrical Engineering: Principles & Applications (6th Edition)

Published by Prentice Hall
ISBN 10: 0133116646
ISBN 13: 978-0-13311-664-9

Chapter 1 - 1.2 - Problems - Circuits, Currents, and Voltages - Page 35: P1.10

Answer

a) 17.632 km b) 587.877 m/s c) 400 times smaller than Gasoline

Work Step by Step

Power = Voltage $\times$ current = energy per unit time So total energy = Voltage $\times$ current $\times$ total time = 12 $\times$ 5 $\times$ (24 $\times$ 3600) = 5184000 J Energy needed to raise mass to height h = mgh h = $\frac{5184000}{9.8 \times 30}$ = 17.632 km Kinetic energy = $\frac{1}{2}mv^{2}$ Therefore, v = $\sqrt \frac{2E}{m}$ = $\sqrt \frac{2\times 5184000}{30}$ = 587.877 m/s Energy contained per unit mass in battery = energy/mass = 5184000/30 = 172800 J/kg, which is 400 times smaller than that of Gasoline
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