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Fynjy0 [20]
3 years ago
12

Water flows over a section of Niagara Falls at the rate of 1.1 × 106 kg/s and falls 50.0 m. How much power is generated by the f

alling water? The acceleration of gravity if 9.81 m/s 2 . Answer in units of W.
Physics
1 answer:
Kryger [21]3 years ago
7 0
<h3>Answer:</h3>

5.395 × 10^8 Watts

<h3>Explanation:</h3>

<u>We are given;</u>

  • Rate of flow is 1.1 × 10^6 kg/s
  • Distance is 50.0 m
  • Gravitational acceleration is 9.8 m/s²

We are required to calculate the power that is generated by the falling water

  • Power is the rate of work done
  • It is given by dividing the energy or work done by time
  • Power = Work done ÷ time

But; work done = Force × distance

Therefore;

Power = (F × d) ÷ time

The rate is 1.1 × 10^ 6 Kg/s

But, 1 kg = 9.81 N

Therefore, the rate is equivalent to 1.079 × 10^7 N/s

Thus,

Power = Rate (N/s) × distance

           = 1.079 × 10^7 N/s × 50.0 m

           = 5.395 × 10^8 Watts

The power generated from the falling water is 5.395 × 10^8 Watts

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Answer:

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remember that angles are in radians

        w t₁ = 1.875 rad

x = 0.3 A

        0.3 A = A cos w t₂

        w t₂ = cos -1 (0.3)

         w t₂ = 1,266 rad

         

Now let's calculate the time of a complete period

x= -A

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        w t₃ = π rad

this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period

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The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

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and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

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The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

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2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

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