Answer
$E = 1.08\times 10^{42}~J$
Work Step by Step
We can find the energy that would be released:
$E = (2~M_E)~c^2$
$E = (2)~(5.98\times 10^{24}~kg)~(3.0\times 10^8~m/s)^2$
$E = 1.08\times 10^{42}~J$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 44 - Quarks, Leptons, and the Big Bang - Problems - Page 1365: 47
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