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Aloiza [94]
3 years ago
10

Atmospheric conditions near a mountain range are such that a cloud at an altitude of 2.00 km contains 3.20 ✕ 10^7 kg of water va

por. How long (in years) would it take for a 2.96 kW pump to raise this amount of water (at a constant speed) from Earth's surface to the altitude of the cloud?
Physics
1 answer:
kykrilka [37]3 years ago
3 0

Answer:

time required is 6.72 years

Explanation:

Given data

mass m = 3.20 ✕ 10^7 kg

height h = 2.00 km = 2 × 10^3 m

power p = 2.96 kW  =2.96 × 10^3 J/s

to find out

time period

solution

we know work is mass × gravity force × height

and power is work / time

so we say that power =  mass  gravity force × height / time

now put all value and find time period

power =  mass × gravity force × height / time

2.96 × 10^3 =  3.20 ✕ 10^7  × 9.81× 2 × 10^3  / time

time =  62.784 × 10^10 / 2.96 × 10^3

time = 21.21081081 × 10^7 sec

time = 58.91891892 × 10^3 hours

time = 6.72 years

so time required is 6.72 years

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(b) a = 22.87 m/s²

Explanation:

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∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

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F₁ : Horizontal force

F₂ : acting at an angle of θ to the horizontal,

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FN : Normal force : perpendicular to the direction the surface

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θ = 30° angle θ of F₂ with respect to the horizontal

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We apply the formula (1)

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FN = -6.2+30.38

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70.904 = ( 3.1)*a

a = (70.904 ) / ( 3.1)

a = 22.87 m/s²

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