To calculate the kinetic energy (KE) of the micrometeor, we can use the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m \) is the mass in kilograms,
- \( v \) is the velocity in meters per second.
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Convert the mass from grams to kilograms: \[ 0.005 \text{ grams} = 0.005 \div 1000 = 0.000005 \text{ kg} \]
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Use the velocity \( v = 21,000 \text{ m/s} \).
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Substitute \( m \) and \( v \) into the kinetic energy formula: \[ KE = \frac{1}{2} (0.000005 , \text{kg}) (21,000 , \text{m/s})^2 \]
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Calculate \( v^2 \): \[ (21,000 , \text{m/s})^2 = 441,000,000 , \text{m}^2/\text{s}^2 \]
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Now plug \( v^2 \) into the formula: \[ KE = \frac{1}{2} (0.000005 , \text{kg}) (441,000,000 , \text{m}^2/\text{s}^2) \]
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Calculate \( KE \): \[ KE = \frac{1}{2} \times 0.000005 \times 441,000,000 = 1,102.5 , \text{J} \]
So the kinetic energy of the micrometeor when it enters Earth’s atmosphere is 1,102.5 J.
Thus, the answer is:
1,102.5 J