Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 16 - Section 16.4 - Exercises and Problems - Page 309: 22

Answer

Please see the work below.

Work Step by Step

We know that $E=(420Kcal)(\frac{4184J}{1 Kcal})=1757280J$ The power is given as $P=\frac{E}{t}$ This can be rearranged as: $t=\frac{E}{P}$ We plug in the known values to obtain: $t=\frac{1757280}{200}=8786.4s=2.44h$ Now we can find the distance as: $d=vt$ We plug in the known values to obtain: $d=(3)(2.44)$ $d=7.3km$
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