College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 26 - Problems - Page 1011: 42


$E = 9.0\times 10^{13}~N$

Work Step by Step

We can find the amount of energy that is released: $E = mc^2$ $E = (1.0\times 10^{-3}~kg)(3.0\times 10^8~m/s)^2$ $E = 9.0\times 10^{13}~N$
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