Question

If your front lawn is 19.0 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 2.20 mg.

Answer 1

Answer:

$S [kg]=52.668\frac{kg}{hour}$

Explanation:

1) Since the question is stating how many snowflakes are accumulated in each square foot (ft²), we should know the area (A) of our front lawn, in ft² units. For that, we multiply its width by its length:

$A = (19.0 ft)*(20.0 ft) = 380 ft^{2}$

So we have an area of 380 ft².

2) Now, if each square foot of lawn accumulates 1050 new snowflakes every minute, how many new snowflakes per minute (SPM) accumulate on the entire area? To find that out, we want to multiply 1050 by the number of square feet we have in total:

$SPM = (1050 \frac{snowflakes}{minute*ft^{2} } )*(380 ft^{2} ) = 399 000\frac{snowflakes}{minute}$

Which means that each minute, our front lawn accumulates 399 000 new snowflakes.

3) We should now know how many snowflakes accumulate per hour (SPH). To do this, we simply multiply SPM by 60, since each hour has 60 minutes:

$SPH = (399 000\frac{snowflakes}{minute})*(60\frac{minutes}{hour})=23940000\frac{snowflakes}{hour}$

If we use scientific notation, we can then establish that our front lawn accumulates $2.394*10^{7}$ new snowflakes every hour.

4) Now we want to know how much snow (S), in kilograms, accumulates on our lawn per hour. Since we already have the number of snowflakes that accumulate in one hour, we only have to multiply that number (SPH) by the mass of an average snowflake, which is 2.20mg.

$S= (2.394*10^{7}\frac{snowflakes}{hour})*(2.20\frac{mg}{snowflake})=5.2668*10^{7}\frac{mg}{hour}$

Finally, to convert the mass from mg to kg, we simply have to divide it by one million (1000000):

$S=(\frac{5.2668*10^{7}\frac{mg}{hour}}{1000000\frac{mg}{kg} })=52.668\frac{kg}{hour}$