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
2,205 J
2,205 J
0.0525 J
0.0525 J
1,102.5 J
1,102.5 J
1,102,500 J
To calculate the kinetic energy of the micrometeor, we can use the formula KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
Given:
m = 0.005 grams (converting to kilograms: 0.005 grams = 0.005/1000 = 0.000005 kg)
v = 21,000 meters per second
Plugging these values into the formula:
KE = (1/2)(0.000005 kg)(21000 m/s)^2
KE = 0.000005 kg * 441000000 m^2/s^2
KE = 2205 J
Therefore, the kinetic energy of the micrometeor when it enters Earth's atmosphere is 2,205 J.