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Gnesinka [82]
3 years ago
15

If your front lawn is 18.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snow flakes every min

ute, how much snow (in kilograms) accumulates on your lawn per hour? assume an average snow flake has a mass of 2.00 mg
Physics
1 answer:
Ivanshal [37]3 years ago
8 0
<span>First, we need to determine the entire area of your front line by multiplying its length times its width.
18.0*20.0 = 360.0 square feet
We can use the rate of accumulation of snow, combined with this figure, to determine how much snow accumulates on your lawn per minute.
360.0 sq ft * 1050 flakes/min/sq ft = 378,000 flakes/min
We can then use the mass of a snowflake to calculate total snow accumulation per minute.
378,000 flakes/min * 2.00 mg/flake = 756,000 mg/min
Finally, we can use this number to determine accumulation per hour.
756,000 mg/min * 60 min/hr =
45,360,000 mg/hr</span>
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A uniform electric field is created by two parallel plates separated by a distance of 0.04 m. What is the magnitude of the elect
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Complete question:

A uniform electric field is created by two parallel plates separated by a

distance of 0.04 m. What is the magnitude of the electric field established

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Answer:

The magnitude of the electric field established between the plates is 2,000 V/m

Explanation:

Given;

distance between two parallel plates, d = 0.04 m

potential between first and second plate, = +40V and -40V respectively

The magnitude of the electric field established between the plates is calculated as;

E = ΔV / d

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d is the distance between the plates

ΔV = V₁ -V₂

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ΔV = 40 + 40

ΔV = 80 V

E = ΔV / d

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2 years ago
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Answer:

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C ( V/ V₂ - V₂ /V₂) = C₂

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