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?
(1 point)
KE =
1
2mv
2
0.0525 J
2,205 J
1,102,500 J
1,102.5 J
To calculate the kinetic energy of the micrometeor, we can use the formula:
KE = 1/2 * m * v^2
Where:
KE = kinetic energy
m = mass of the micrometeor (0.005 grams = 0.005 * 10^-3 kg)
v = velocity of the micrometeor (21,000 m/s)
Plugging in the values:
KE = 1/2 * 0.005 * 10^-3 * (21,000)^2
KE = 1/2 * 0.005 * 10^-3 * 441,000,000
KE = 0.0525 J
Therefore, the kinetic energy of the micrometeor when it enters Earth's atmosphere is 0.0525 J.