The correct answer is 1,102.5 J
KE = 1/2 * 0.005 * (21000)^2
KE = 1/2 * 0.005 * 441000000
KE = 2205000
KE = 1,102.5 J
So, the kinetic energy of the micrometeor when it enters Earth's atmosphere is 1,102.5 J.
A micrometeor has a mass of 0.005 grams. When it enters Earth’s atmosphere, it travels at 21,000 meters per second. What is its kinetic energy when it enters Earth’s atmosphere?
KE=12mv2
(1 point)
Responses
1,102.5 J
1,102.5 J
2,205 J
2,205 J
0.0525 J
0.0525 J
1,102,500 J
1 answer