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
1,102.5 J
1,102.5 J
0.0525 J
0.0525 J
1,102,500 J
To calculate the kinetic energy, we can use the formula:
KE = 1/2 * m * v^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
Plugging in the given values:
KE = 1/2 * 0.005 grams * (21,000 meters/second)^2
Now we need to convert the mass to kilograms because the SI unit for mass is kg:
0.005 grams = 0.005 * 0.001 kilograms = 0.000005 kilograms
Substituting this value back into the equation:
KE = 1/2 * 0.000005 kilograms * (21,000 meters/second)^2
Now we can calculate it:
KE = 1/2 * 0.000005 * (21,000)^2
= 1/2 * 0.000005 * 441,000,000
= 0.0011025 joules or 1,102.5 J
Therefore, the kinetic energy of the micrometeor when it enters Earth's atmosphere is 1,102.5 J.