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 - Exercises and Problems - Page 310: 46

Answer

Please see the work below.

Work Step by Step

We know that $E=mgh$ We plug in the known values to obtain: $E=(60)(9.8)(1700)$ $E=999600J$ $E=(999600J)(\frac{1Kcal}{4184})$ $E=239Kcal$
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