Essential University Physics: Volume 1 (4th Edition)

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

Chapter 1 - Exercises and Problems - Page 12: 14

Answer

Please see the work below.

Work Step by Step

We know that the mass of the water is $m=14Eg$ $m=14\times 10^{18}$ $m=14\times 10^{18}\frac{1Kg}{1000g}$ $m=14\times 10^{15}Kg$
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