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
K
E
=
1
2
m
v
2
(1 point)
Responses
1,102.5 J
1,102.5 J
0.0525 J
0.0525 J
1,102,500 J
1,102,500 J
2,205 J
2,205 J
To calculate the kinetic energy of the micrometeor, we can use the formula KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass of the micrometeor, and v is its velocity.
Given:
Mass (m) = 0.005 grams = 0.005 * 10^-3 kilograms = 5 * 10^-6 kilograms
Velocity (v) = 21,000 meters per second
Plugging in these values into the formula, we get:
KE = 1/2 * (5 * 10^-6) * (21000)^2
= 1/2 * 5 * 10^-6 * 441,000,000
= 1102.5 Joules
Therefore, the kinetic energy of the micrometeor when it enters Earth's atmosphere is 1,102.5 Joules.