If your front lawn is 25.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1250 new snowflakes every minute, how much snow, in kilograms, accumulates on your lawn per hour

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Question:

If your front lawn is 25.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1250 new snowflakes 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.

Given Information:  

Area of lawn = 25 ft into 20 ft = 500 ft²

Rate of snowflakes = 1250 per minute

Mass of snowflakes = 1.70 mg = 0.0017 g = 0.0000017 kg

Required Information:  

Mass of snowflakes per hour = ?  

Mass of snowflakes = 63.75 kg

Explanation:  

Total snowflakes = Rate of snowflakes*Area of lawn

Total snowflakes = 1250*500

Total snowflakes = 625000/min

Multiply by 60 minutes to convert into hour

Total snowflakes = 625000*60

Total snowflakes = 37500000/hour

Weight of snowflakes = Total snowflakes*Mass of each snowflake

Weight of snowflakes = 37500000*0.0000017

Weight of snowflakes = 63.75 kg

Therefore, there will be 63.75 kg of snowflakes in an area of 500 ft².

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s