Question

if your front lawn is 22.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snowflakes every minute, how much snow, in kilograms, accumulates on your lawn per hour? assume an average snowflake has a mass of 1.80 mg.

Explanation:

Answer:

First, find the area of the lawn: \( A = 22.0\,\text{ft} \times 20.0\,\text{ft} = 440\,\text{ft}^2 \).
Snowflakes per minute: \( 440\,\text{ft}^2 \times 1050\,\text{snowflakes/ft}^2 = 462000\,\text{snowflakes/min} \).
Snowflakes per hour: \( 462000\,\text{snowflakes/min} \times 60\,\text{min/hour} = 27720000\,\text{snowflakes/hour} \).
Mass of one snowflake: \( 1.80\,\text{mg} = 1.80 \times 10^{-6}\,\text{kg} \).
Total mass per hour: \( 27720000 \times 1.80 \times 10^{-6}\,\text{kg} = 49.896\,\text{kg} \approx 50.0\,\text{kg} \) (or more precisely \( 49.9\,\text{kg} \) if calculated with more precision, but following significant figures: 22.0, 20.0, 1050 (maybe 3 sig figs), 1.80 (3 sig figs) → result should have 3 sig figs: \( 49.9\,\text{kg} \) or \( 5.00 \times 10^1\,\text{kg} \), but let's compute exactly:

\( 22.0 \times 20.0 = 440 \)
\( 440 \times 1050 = 462000 \)
\( 462000 \times 60 = 27720000 \)
\( 27720000 \times 1.80 \times 10^{-3}\,\text{g} = 27720000 \times 1.80 \times 10^{-6}\,\text{kg} \)
\( 27720000 \times 1.80 = 49896000 \)
\( 49896000 \times 10^{-6} = 49.896\,\text{kg} \approx 49.9\,\text{kg} \) (or 50.0 kg with three significant figures).

So the final answer is \(\boxed{49.9}\) (or \(\boxed{50.0}\) depending on rounding, but 49.9 is more accurate from the calculation).