Question

energy transfer and kinetic energy quick check
a micrometeor has a mass of 0.005 grams. when it enters earths atmosphere, it travels at 21,000 meters per second. what is its kinetic energy when it enters earths atmosphere?
ke = \frac{1}{2}mv^{2}
(1 point)
2,205 j
1,102,500 j
0.0525 j
1,102.5 j

Explanation:

Step1: Convert mass to kilograms

Given mass \(m = 0.005\space g\). Since \(1\space kg=1000\space g\), then \(m=\frac{0.005}{1000}=5\times10^{-6}\space kg\).

Step2: Substitute values into kinetic - energy formula

The formula for kinetic energy is \(KE=\frac{1}{2}mv^{2}\), with \(m = 5\times 10^{-6}\space kg\) and \(v = 21000\space m/s\).

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Answer:

1,102.5 J