Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 6 - Exercises and Problems - Page 106: 60


Please see the work below.

Work Step by Step

We know that $KE=\frac{1}{2}mv^2$ This can be rearranged as: $m=\frac{2KE}{v^2}$ We plug in the known values to obtain: $m=\frac{2\times 500\times 4.18\times 10^{12}}{19\times 10^3}=11.58\times 10^6Kg$
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