Final answer:
To find the momentum of rain in one second, first find the mass by calculating the volume (1 cm depth over 10 km²) and then multiply by water's density.
Divide this mass by 3,600 to get mass per second, then multiply by the terminal velocity (10 m/s) to find momentum. The calculated momentum does not match any provided options, suggesting a need to review the calculations or question.
Step-by-step explanation:
To calculate the momentum of the rain that falls in one second, we must first find the mass of rain that falls over the 10 km² area in one hour.
With 1 cm of rain falling over this area, we can express this volume as 0.01 m × 10,000,000 m² (since 10 km² = 10,000,000 m²). This gives us a volume of 100,000 m³. The density of water is approximately 1,000 kg/m³, so the mass of rain falling in one hour would be 100,000 m³ × 1,000 kg/m³ = 100,000,000 kg.
We then convert the quantity to a per-second basis since 1 hour = 3,600 seconds. This gives us a per-second mass of 100,000,000 kg / 3,600 s = 27,777.78 kg.
Now, assuming each raindrop falls at a terminal velocity of 10 m/s, the momentum p (mass × velocity) of the rain per second can be calculated as:
p = m × v = 27,777.78 kg × 10 m/s = 277,777.8 kg·m/s.
However, none of the provided options match this calculation, implying a possible error in the setup or the options provided.
It would be best to recheck the question parameters and calculations.