ch43-p014 - 14. (a) Using the result of Problem 43-2, the...

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14. (a) Using the result of Problem 43-2, the TNT equivalent is (. . . 250 26 10 4 4 10 44 28 4 kg)(4.54 10 MeV / kg) MeV /10 ton ton kton. 26 6 × × = (b) Assuming that this is a fairly inefficiently designed bomb, then much of the remaining 92.5 kg is probably “wasted” and was included perhaps to make sure the bomb did not “fizzle.” There is also an argument for having more than just the critical mass based on
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This note was uploaded on 05/19/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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