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noname [10]
3 years ago
11

An electrical power plant manages to transfer 88 percent of the heat produced in the burning of fossil fuel to convert water to

steam. Of the heat carried by the steam, 40 percent is converted to the mechanical energy of the spinning turbine. Which best describes the overall efficiency of the heat-to-work conversion in the plant
Physics
1 answer:
Shtirlitz [24]3 years ago
6 0

An electrical power plant manages to send 88% of the heat produced in the burning of fossil

fuel into the water-to-steam conversion. Of the heat carried by the steam, 40% is converted to

the mechanical energy of the spinning turbine. Which of the following choices best describes

the overall efficiency of the heat-to-work conversion in the plant (as a percentage)?

a. greater than 88%

b. 64%

c. less than 40%

d. 40%

Answer:

Option C is correct

Explanation:

This is the final heat transfer into mechanical energy / total energy production by fossil fuel

Let x = total heat produced by fossil fuel

-88% of x was used to convert water to steam so therefore

.88 *x = 0.88x

- 40% of heat in steam was converted to final mechanical work so therefore

0.4 * 0.88x = 0.352x

So therefore:

Overall efficiency = (0.352x/ x ) * 100 = 35.2%

So therefore option C is correct

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A 4 kg mass is in free fall. What is the velocity of the mass after 11 seconds
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5 0
3 years ago
A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.
Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

    k x = m g

k = \dfrac{mg}{x}

k = \dfrac{6.4 \times 9.8}{0.28}

      k = 224 N/m

b) f = \dfrac{\omega}{2\pi}

    \omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}

  f =0.94\ Hz

c)  v_b = -v cos \omega t

    v_b = -5.1 \times cos (5.92 \times 0.42)

    v_b = 4.04\ m/s

d)  a_{max} = v \omega

    a_{max} = 4.04 \times 5.92

    a_{max} =23.94\ m/s^2

e)  Y =- A sin (\omega t)

    A = \dfrac{v}{\omega}

    A = \dfrac{4.04}{5.92}

        A = 0.682 m

    Y =- 0.682 \times sin (5.92 \times 0.42)

    Y =- 0.42

Force =m \omega^2 |Y|

          =6.4 \times 5.92^2\times 0.42

F = 94.20 N

4 0
3 years ago
How much time will it take to perform 440 joule of work at a rare of 11 w?​
kifflom [539]

Answer:

40sec

Explanation:

Data

Work = 440 J

Power= 11watt

time = ?

Power = work done/time

===> time = work done/power

= 440/11

= 40sec

7 0
2 years ago
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