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maks197457 [2]
2 years ago
13

If steam enters a turbine at 600K and is exhausted at 400K, calculate the efficiency of the engine.

Physics
2 answers:
DochEvi [55]2 years ago
8 0
The formular for efficiency = change in temprature/ initial temprature
change in temprature = 600 - 400 = 200
efficiency = 200/600 = 0.333 = 33.3%'
ikadub [295]2 years ago
3 0

The correct answer to the question is 33.3 % i.e the efficiency of the engine is 33.3 %.

CALCULATION:

The steam enters a turbine at 600 K.

Hence, the temperature of the source T = 600 K.

The steam is exhausted at a temperature 400 K.

Hence, the temperature of the sink T' = 400 K

We are asked to calculate the efficiency of the engine.

The efficiency of a heat engine whose source is at temperature T K and sink at T' is calculated as -

                        efficiency \eta\ =\ 1-\frac{T'}{T}

                                             =\ 1-\frac{400\ K}{600\ K}

                                             =\ 1-\frac{2}{3}

                                             =\ \frac{1}{3}

                                             =\ 33.3\%             [ans]

Hence, the efficiency of the engine is 33.3 %.


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Which of the following statements accurately describes the atmospheric patterns that influence local weather?
timurjin [86]

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4 0
1 year ago
A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f
sasho [114]

We use the kinematic equations,

v=u+at                                          (A)

S= ut + \frac{1}{2} at^2                  (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given,  u=0, a=0.21 \ m/s^2 and s= 280 m.

Substituting these values in equation (B), we get

280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s.

Therefore from equation (A),

v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s

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8 0
2 years ago
PLEASE ANSWER ACCURATELY DO NOT GUESS PLEASE AND THANK YOU
krek1111 [17]
Hello! I can help you with this!

1. Providence- Rhode Island
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4 0
2 years ago
Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23
Sloan [31]

Answer:

The ratio is   \frac{T_1}{T_2}  = 3.965

Explanation:

From the question we are told that

   The  radius of Phobos orbit is  R_2 =  9380 km

    The radius  of Deimos orbit is  R_1  =  23500 \  km

Generally from Kepler's third law

    T^2 =  \frac{ 4 *  \pi^2 *  R^3}{G * M  }

Here M is the mass of Mars which is constant

        G is the gravitational  constant

So we see that \frac{ 4 *  \pi^2  }{G * M  } =  constant

   

    T^2 = R^3   *  constant      

=>  [\frac{T_1}{T_2} ]^2 =  [\frac{R_1}{R_2} ]^3

Here T_1 is the period of Deimos

and  T_1 is the period of  Phobos

So

      [\frac{T_1}{T_2} ] =  [\frac{R_1}{R_2} ]^{\frac{3}{2}}

=>    \frac{T_1}{T_2}  =  [\frac{23500 }{9380} ]^{\frac{3}{2}}]

=>    \frac{T_1}{T_2}  = 3.965

   

8 0
2 years ago
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