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zheka24 [161]
3 years ago
11

A circular swimming pool has a diameter of 18 m. The circular side of the pool is 3 m high, and the depth of the water is 1.5 m.

(The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m 3 kg/m^3.)How much work (in Joules) is required to: i. pump all of the water over the side? ii. pump all of the water out of an outlet 2m over the side?
Physics
1 answer:
Dmitrij [34]3 years ago
8 0

Answer:

A) 11223450 J

B) 18705750 J

Explanation:

Work done = force x distance moved.

Force of fluid at bottom of the pool = pghA

Where

P = density of water = 1000 kg/m3

g = acceleration due to gravity = 9.8 m/s2

h = depth of water = 1.5 m

A = area of pool of diameter d = 18 m

A = (¶d^2)/4 = (3.142 x 18^2)/4 = 254.5 m2

Total wall height of pool = 3 m

Force on pool bottom due to water is

F = 1000 x 9.8 x 1.5 x 254.5 = 3741150 N

A) work (in Joules) required to pump all of the water over the side

Height of wall H = 3 m

Work = F x H = 3741150 x 3 = 11223450 J

B) work (in Joules) required to pump all of the water out of an outlet 2m over the side?

Total height will be 2 + 3 = 5 m

Work = F x H = 3741150 x 5 =

18705750 J

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1. What is the value of the acceleration that the car experiences? 2. What is the value of the change in velocity that the car e
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Answer:

All the answers are solved and explained below.

Explanation:

Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.

Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).

Data Given:

m = 1000 kg

F = 2000N

t = 10s

Q1 Solution:

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F = ma

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a = 2000/ 1000

a = 2 m/s^{2}

Q2: Solution:

Change in velocity = Δv = ?

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a = Δv/t

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Δv = 20 m/s

Q3: Solution:

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I = 20000 Ns

Q4: Solution:

Change in Momentum = Δp = ?

Δp = mΔv

Δp = 1000 x 20

Δp = 20000 Kgm/s

Q5: Solution:

Final velocity of the car at the end of 10 seconds = vf = ?

Δp = m x Δv

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Δp = 1000 x (vf - 0 )

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vf = 20000/1000

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Q6: Solution:

Change in momentum the car experiences as it continues at this velocity?

Δp = ?

Δp = mΔv

Δp = m x (0)

Δp = 0

Q7: Solution:

Impulse = Change in momentum

Impulse = Δp

Implulse = 0

Q8: Solution:

Change in momentum = Δp = mΔv

Δp = m(vf-vi)

Δp = 1000 x (0-20)

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Q9: Solution:

Impulse = Δp

Impulse = -20000 Ns

Q10: Solution:

Impulse = ?

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F = impulse/t

F = -20000/4s

F = -5000 N

Q11: Solution:

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a = F/m

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