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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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The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa
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m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

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Explanation:

For this case we know the following info:

c_{pA} > c_{pB}

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

T_{iA}= T_{iB}

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

Q = m c_p \Delta T

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

Q_A = Q_B

And if we replace the formula for sensible heat we got:

m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B

And if we replace for the change of the temperature we got:

m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})

And since T_{iA}= T_{iB}= T we have this:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

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4 0
3 years ago
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This means that t1 is smaller than t2, ergo, it takes longer with wind

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