College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1012: 45


$E = 2.0\times 10^{47}~J$

Work Step by Step

We can find the energy released by the explosion: $E = mc^2$ $E = (0.80)(1.4)(m_{sun})~c^2$ $E = (0.80)(1.4)(1.989\times 10^{30}~kg)(3.0\times 10^8~m/s)^2$ $E = 2.0\times 10^{47}~J$
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