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liraira [26]
3 years ago
13

A power plant uses 365 GJ/hr of energy from a boiler and extracts 18 GJ/hr as work from a turbine and rejects the remaining ener

gy to a large cold resevoir of air (the atmosphere). What is the rate of heat rejection to the atmosphere for this plant? Give your answer in GJ/hr
Physics
1 answer:
ziro4ka [17]3 years ago
5 0

Answer:

The rate of heat rejection will be 347 GJ/hr

Explanation:

We have given that power plant uses energy of 365 GJ/hr

So energy uses Q_B=365GJ/hr

Work done from turbine W_T=18GJ/hr

We have to find the rate of heat rejection to the atmosphere Q_C

We know that Q_B=W_T+Q_C

365=18+Q_C

Q_C=347GJ/hr

So the rate of heat rejection will be 347 GJ/hr

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Explanation:

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                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
4 years ago
Click the links to open the resources below. These resources will help you complete the assignment. Once you have created your f
Paul [167]

Answer:

Explanation:

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3 years ago
One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given dista
Jlenok [28]

Answer:

(a) 0.94 m/s²

(b) g (planet) = 0.096g

Explanation:

(a)

From Newton's equation of motion,

S = ut + 1/2gt²......................... equation 1

Making g the subject of equation 1

g =( S - ut)/t² ........................ equation 2

Where  s = distance ( m), u = initial velocity (m/s), t = time (s), g = acceleration due to gravity (m/s²)

From the question, S = 12.02 m, t = 3.58 s, u= 0 ( at rest),

Substituting these values in equation 2

g = {12.02 -(0×3.58)}/3.58²

g = (12.02)/12.82

g = 0.94 m/s²

∴ The acceleration due to gravity on the planet = 0.94 m/s²

(b) g (planet)/g (earth) = 0.94/9.80

     g (planet) = 0.096 g (earth).

The acceleration due to gravity of the planet in terms of the earth g  is

g (planet) = 0.096g

5 0
3 years ago
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3 0
3 years ago
What is the instantaneous velocity of the hummingbird at t=1s?
Liono4ka [1.6K]

The distance - time graph of the humming bird is missing, so i have attached it.

Answer:

Instantaneous velocity = 0.5 m/s

Explanation:

From the attached graph, at time t = 1 s, the corresponding distance is 0.5 m.

Instantaneous velocity is the velocity at that point.

Thus;

Instantaneous velocity = 0.5/1

Instantaneous velocity = 0.5 m/s

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3 years ago
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