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creativ13 [48]
3 years ago
7

Relate the output of energy from a heat engine to the energy put into the heat engine considering the second law of thermodynami

cs.
You're answer would be greatly appreciated.
Physics
1 answer:
Vinil7 [7]3 years ago
3 0

<u>Answer: </u>

<em>Considering the II law of thermodynamics</em>

<em>From the figure</em>

<em>Out put of energy: </em>

Heat supplied from the source/ reservoir  (Q₁) - Heat rejected to the surroundings from the system (Q) = Q₁ - Q₂. Also known as Net work done on the system.

<em>Input of energy: </em>

Amount of heat energy supplied to the system from the source (Q₁ ).

Efficiency (H.E) = η = Output÷ Input

                             η  = (Q₁ - Q₂) ÷ Q₁

                      OR η = Wnet ÷ Q₁ ;        since Wnet = (Q₁ - Q₂)


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What is black body radiation? Explain in detail.
tangare [24]

An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called black-body radiation

hope it helps

3 0
2 years ago
A 2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
krek1111 [17]

Answer:

a) Pb= 200 PA

b).work done= -3600 joules

c).3600joules

D).the system works under isothermal condition so no heat was transferred

Explanation:

2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.

a). PbVb= PaVa

Pb= (PaVa)/VB

Pb= (600*3)/9

Pb= 1800/9

Pb= 200 PA

b). work done= n(Pb-Pa)(Vb-Va)

Work done= 2*(200-600)(9-3)

Work done= -600(6)

Work done=- 3600 Pam³

work done= -3600 joules

C). Change in internal energy I the work done on the system

= 3600joules

D).the system works under isothermal condition so no heat was transferred

4 0
3 years ago
An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a c
Molodets [167]

Answer:

4x

Explanation:

Use v^{2} = u^{2} +2as to do the question.

For first instance,

0 = v^{2} +2ax -------------------( 1 )

for second instance,

0 = (2v)^{2} +2as-----------------( 2 )

So by (1) and (2),

s = 4x

3 0
3 years ago
Is 5m.what<br>In si system, the length of a body is<br>does it mean?​
lozanna [386]

Answer: Some conversions from one system of units to another need to be exact, without increasing or decreasing the precision of the first measurement. This is sometimes called soft conversion. It does not involve changing the physical configuration of the item being measured.

Explanation:

4 0
3 years ago
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
3 years ago
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