Question

If your front lawn is 24.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snow flakes every minute, how much snow (in kilograms) accumulates on your lawn per hour? assume an average snow flake has a mass of 1.70 mg.

Answer 1

First find the area of the front lawn:
Area = length * width
Area = 24.0ft * 20.0ft
Area = 480ft^2

Next, find the weight in kilograms of the snow per square foot per hour per ft^2:

`weight\ per\ ft^(2)=(1050\ snowflakes)/(minute)\\\\weight\ per\ ft^(2)=(1050\ snowflakes)/(minute)*(1.70\ milligrams)/(snowflake)*(1\gram)/(1000\milligrams)*(1\kilogram)/(1000\grams)* (60\minutes)/(1\hour)\\\\weight\ per\ ft^(2)=(1050*1.70)/(1000*1000*60)*(kg)/(hour)=2.975*10^(-5)\ (kg)/(hour)\\\\weight=2.975*10^(-5)\ (kg)/(hour*ft^(2))`

Now multiply the area of the lawn by the weight of the snow per hour per ft^2:

`weight\ of\ snow=lawn\ area*weight\ per\ hour\ per ft^(2)\\\\weight=480\ ft^(2)*2.975*10^(-5)\ (kg)/(hour*ft^(2))\\\\weight=0.01428\ (kg)/(hour)`



Thus your answer is .01428 kg.