## University Physics with Modern Physics (14th Edition)

(a) $KE = 1.0\times 10^{16}~J$ (b) The amount of energy delivered to the ground by the meteor was equivalent to about 2.4 megaton bombs.
(a) $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(1.4\times 10^8~kg)(1.2\times 10^4~m/s)^2$ $KE = 1.0\times 10^{16}~J$ (b) We can find the amount of energy released by a 1.0-megaton bomb. $E = (4.184\times 10^9~J)(1.0\times 10^6)$ $E = 4.184\times 10^{15}~J$ We can consider the energy of the meteor compared to the bomb. $\frac{1.0\times 10^{16}~J}{4.184\times 10^{15}~J} = 2.4$ The amount of energy delivered to the ground by the meteor was equivalent to about 2.4 megaton bombs.