Answer
$E = 2.0\times 10^{47}~J$
Work Step by Step
We can find the energy released by the explosion:
$E = mc^2$
$E = (0.80)(1.4)(m_{sun})~c^2$
$E = (0.80)(1.4)(1.989\times 10^{30}~kg)(3.0\times 10^8~m/s)^2$
$E = 2.0\times 10^{47}~J$
College Physics (4th Edition)
by Giambattista, Alan; Richardson, Betty; Richardson, Robert
Published by
McGraw-Hill Education
ISBN 10:
0073512141
ISBN 13:
978-0-07351-214-3
Chapter 26 - Problems - Page 1012: 45
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