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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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Answer:

d. The large pot of water and small cup of water have the same temperature, but the large pot of water has higher thermal energy.

Explanation:

Temperature is a measure of the average kinetic energy of individual molecules. While internal energy refers to the total kinetic energy of the molecules within the object. Since in this case we have the same amount of average kinetic energy, then the large pot of water and small cup of water have the same temperature. While the large pot of water has higher thermal energy, since has more water particles than the small cup.

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3 years ago
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imagine you are going on a rid in a spacecraft next to earth. Your trip takes one whole year. Describe earth's tilt in the north
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An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,
mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

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2 years ago
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Answer:

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Explanation:

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