Answer:
62.64 kg/hr
Explanation:
Please consider the complete question.
If your front lawn is 16.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1150 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 2.00 mg.
Let us find area of the front lawn by multiplying length and width of the lawn as:
![\text{Area of front lawn}=\text{18 ft}* 20\text{ ft}](https://img.qammunity.org/2021/formulas/mathematics/high-school/b3bd47aamojlc10buq3s4u7drh306u82lh.png)
![\text{Area of front lawn}=360\text{ ft}^2](https://img.qammunity.org/2021/formulas/mathematics/high-school/vqdsdmeqks5w4vsgkfowbyz8ymgye58si1.png)
Let us find amount of snowflakes per minute.
![\text{Snowflakes}=360\text{ ft}^2* \frac{\text{1450 flakes}}{\text{ft}^2}=522000\text{ flakes}](https://img.qammunity.org/2021/formulas/mathematics/high-school/2x1tbosj2enjwxten151dnkm6no0v5j0bz.png)
Snowflakes per hour:
![522000* 60 \text{ flakes}=31,320,000\text{ flakes}](https://img.qammunity.org/2021/formulas/mathematics/high-school/5swgs69ortsty1g1pm636kgbtwo7kmgsn9.png)
![\text{Weight of snow}=31,320,000\text{ flakes}*\frac{\text{ 2 mg}}{\text{Snowflake}}](https://img.qammunity.org/2021/formulas/mathematics/high-school/h6tehtojifeno68l0vk8595r1c06ugbn0z.png)
![\text{Weight of snow}=62,640,000\text{ mg}](https://img.qammunity.org/2021/formulas/mathematics/high-school/vj3t7scprnz3nikldtjk0tmnmzwnw98bod.png)
1 kg = 1000,000 mg
![\text{Weight of snow}=62,640,000\text{ mg}* \frac{\text{1 kg}}{\text{1,000,000 mg}}=62.64\text{ kg}](https://img.qammunity.org/2021/formulas/mathematics/high-school/wjfyw9td3s1rqxv0mdg0krr96a7y5nk0ip.png)
Therefore, 62.64 kg of snow accumulates on your lawn per hour.