## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 228: 20

#### Answer

The kinetic energy of the meteoroid is $5.76\times 10^6~J$ The kinetic energy of the car is $4.63\times 10^5~J$ The kinetic energy of the meteoroid is greater than the kinetic energy of the car by a factor of 12.4

#### Work Step by Step

We can find the kinetic energy of the meteoroid: $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(0.0050~kg)(48,000~m/s)^2$ $KE = 5.76\times 10^6~J$ The kinetic energy of the meteoroid is $5.76\times 10^6~J$ We can find the kinetic energy of the car: $KE = \frac{1}{2}mv^2$ $KE = \frac{1}{2}(1100~kg)(29~m/s)^2$ $KE = 4.63\times 10^5~J$ The kinetic energy of the car is $4.63\times 10^5~J$ We can compare the kinetic energy of the meteoroid to the kinetic energy of the car: $\frac{5.76\times 10^6~J}{4.63\times 10^5~J} = 12.4$ The kinetic energy of the meteoroid is greater than the kinetic energy of the car by a factor of 12.4.

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