p07_053.pdf

(70 KB) Pobierz

Chapter 7 - 7.53

53. (a) We set up the ratio

1km = E

1 / 3

50km

1megaton

10 5 megatons of TNT.

(b) We note that 15 kilotons is equivalent to 0 . 015 megatons. Dividing the result from part (a) by

0 . 013 yields about ten million bombs.

≈

and ﬁnd E =50 3