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Elza [17]
3 years ago
10

A Carnot engine has a power output of 150 kW. The engine operates between two reservoirs at 20.0°C and 500°

Physics
2 answers:
Nookie1986 [14]3 years ago
8 0
Efficiency η of a Carnot engine is defined to be: 
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
<span>Tc = Th - 75 = 266 K </span>

<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>
rjkz [21]3 years ago
5 0

Answer:

8.8 × 10³³J

Explanation:

efficiency of an engine η  = net work output ÷ heat input = 1 - \frac{T_{c} }{T_{H} }

T_{H} = 500 + 273 =773K

T_{C} = 20 + 273 =293K

n_{c} = 1 - \frac{293}{773}

  =0.62

power output= work output = 150kW

power output per hour = 150 × 1000×3600

                                     =5.4×10⁸J

recall, net work output ÷ heat input = 1 - \frac{T_{c} }{T_{H} }

∴ 0.62= 5.4×10⁸ ÷ heat input

heat input= 5.4×10⁸÷0.62

=8.81×10⁸J

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3 years ago
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A 74 kg man holding a 13 kg box rides on a skateboard at a speed of 11m/s. He throws the box behind him,giving it velocity of 6
Ivahew [28]

Answer:

After throwing the object the, the velocity of the man is 13.98 m/s

Explanation:

Given:

Let,

mass of man, m1 = 74 kg

mass of box, m2 = 13 kg

Initial velocity u = 11 m/s (initially both are together hence initial velocity will be same  for both)

Final velocity of man = v1

Final velocity of box = v2 = -6 m/s (the velocity is recoil velocity therefore it is negative)

To Find:

Final velocity of man,after throwing the object = v1 = ?

Solution:

Recoil velocity:

It is the backward velocity experienced.

Here recoil velocity is the backward velocity experience while throwing the box behind.Hence the velocity of the box is negative 6 m/s.

The recoil velocity is the result of conservation of linear momentum of the system. Therefore we will follow the law of conservation of momentum.

Law of conservation of momentum :

Total momentum of an isolated system before collision is always equal to total momentum after collision

\textrm{total momentum before collision}=\textrm{total momentum after collision}\\m_{1}\times u+ m_{2}\times u=m_{1}\times v_{1}+m_{2}\times v_{2}

substituting the values which are given above we get

74\times 11 + 13\times 11 = 74\times v_{1} +13\times -6\\ 957 = 74\times v_{1} -78\\v_{1}=\frac{1035}{74} \\v_{1}=13.98\ m/s

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3 0
3 years ago
A coil consists of 200 turns of wire. Each turn is a square of side d = 18 cm, and a uniform magnetic field directed perpendicul
vampirchik [111]

Answer:

Induced emf in the coil, \epsilon=4.05\ volts

Explanation:

Given that.

Number of turns in the coil, N = 200

Side of square, d = 18 cm = 0.18 m

The field changes linearly from 0 to 0.50 T in 0.80 s.

To find,

The magnitude of the induced emf in the coil while the field is changing.

Solution,

We know that due to change in the magnetic field, an emf gets induced in the coil. The formula of induced emf is given by :

\epsilon=\dfrac{d\phi}{dt}

\phi = magnetic flux

\epsilon=-\dfrac{d(NBA)}{dt}

A is the ares of square

\epsilon=AN\dfrac{d(B)}{dt}

\epsilon=AN\dfrac{B_f-B_i}{t}

\epsilon=(0.18)^2\times 200 \times \dfrac{0.5-0}{0.8}

\epsilon=4.05\ volts

So, the induced emf in the coil is 4.05 volts. Hence, this is the required solution.

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Rashid [163]
They seem to cancel each other out which is odd
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