a micrometeor has a mass of 0.005 when it enters earth atompshere it travels at 21,000 meters per second. what its kinetic energy when it enters earths atmosphere
2,205J
0.0525J
1,102,500J
1,102.5 J
I want to say its D am i correct?
Btw it was D
1. Kinetic energy is transferred from Terry's foot to the soccer ball.
2. A baseball bat strikes a baseball.
3. 173.28J
4. Kinetic energy is transferred from the racket to the ball.
5. 1,102.5 J
To calculate the kinetic energy of the micrometeor when it enters Earth's atmosphere, you can use the formula:
Kinetic Energy = (1/2) * mass * velocity^2
Given that the mass of the micrometeor is 0.005 kg and its velocity is 21,000 m/s, let's substitute these values into the formula:
Kinetic Energy = (1/2) * 0.005 kg * (21,000 m/s)^2
Now let's calculate the result:
Kinetic Energy = (1/2) * 0.005 kg * (441,000,000 m^2/s^2)
Kinetic Energy = 1,102,500 J
So the correct answer is C) 1,102,500 J.
To calculate the kinetic energy of the micrometeor when it enters Earth's atmosphere, you can use the formula:
Kinetic Energy (KE) = 1/2 * mass * velocity^2
Given that the mass of the micrometeor is 0.005 kg (you didn't provide the unit, but I'll assume it's kilograms) and the velocity is 21,000 m/s, we can substitute these values into the formula:
KE = 1/2 * 0.005 kg * (21,000 m/s)^2
Calculating this will give us:
KE ≈ 1/2 * 0.005 kg * 441,000,000 m^2/s^2
KE ≈ 0.0025 kg * 441,000,000 m^2/s^2
KE ≈ 1,102,500 J
Therefore, the correct option is C: 1,102,500 J.