To calculate the kinetic energy (KE) of the micrometeor when it enters Earth's atmosphere, we can use the kinetic energy formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m \) is the mass (in kilograms),
- \( v \) is the velocity (in meters per second).
First, we need to convert the mass from grams to kilograms:
\[ m = 0.005 \text{ grams} = 0.005 \times 10^{-3} \text{ kg} = 0.000005 \text{ kg} \]
Now, we can substitute the values into the kinetic energy formula:
\[ v = 21,000 \text{ m/s} \]
Now we calculate \( KE \):
\[ KE = \frac{1}{2} \times 0.000005 \text{ kg} \times (21,000 \text{ m/s})^2 \]
Calculating \( v^2 \):
\[ (21,000)^2 = 441,000,000 \text{ m}^2/\text{s}^2 \]
Now substitute it back into the kinetic energy equation:
\[ KE = \frac{1}{2} \times 0.000005 \times 441,000,000 \] \[ KE = 0.0000025 \times 441,000,000 \] \[ KE = 1,102.5 \text{ J} \]
Thus, the kinetic energy of the micrometeor when it enters Earth's atmosphere is:
1,102.5 J
The correct response is: 1,102.5 J
1 answer