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o-na [289]
3 years ago
11

A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f

inal number of microstates is 0.599 times that of the initial number of microstates?
Physics
1 answer:
jonny [76]3 years ago
5 0

Answer:

<em>Entropy Change = 0.559 Times</em>

Explanation:

Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.

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Car A (mass 1100 kg) is stopped at a traffic light when it is rear­ended by car B(mass 1400 kg). Both cars then slide with locke
viva [34]

Answer:

Part a)

v_a = 3.94 m/s

Part b)

v_b = 3.35 m/s

Part C)

v_b = 6.44 m/s

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

a = -\frac{F_f}{m}

a = -\frac{\mu mg}{m}

a = - \mu g

now we will have

v_f^2 - v_i^2 = 2ad

0 - v_i^2 = 2(-\mu g)(6.1)

v_a = \sqrt{(2(0.13)(9.81)(6.1)}

v_a = 3.94 m/s

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

v_b = \sqrt{2(0.13)(9.81)(4.4)}

v_b = 3.35 m/s

Part C)

By momentum conservation we will have

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1400 v_b = 1100(3.94) + 1400(3.35)

v_b = 6.44 m/s

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0
3 years ago
Which transition metals have more in common than other
jeka94

Answer:

A. alkali metals

Explanation:

im not sure about that

8 0
3 years ago
Read 2 more answers
An air conditioner running with R-134a on a cycle executed under the saturationdome between the pressure limits of 0.8 MPa and 0
ohaa [14]

Answer:

The COP of the system is = 4.6

Explanation:

Given data

Higher pressure  = 1.8 M pa

Lower pressure = 0.12 M pa

Now we have to find out high & ow temperatures at these pressure limits.

Higher temperature corresponding to pressure 1.8 M pa

T_{H} = 62.9 °c = 335.9 K

Lower temperature corresponding to pressure 0.2 M pa

T_{L} = - 10.1 °c = 262.9 K

COP of the system is given by

COP = \frac{T_{L} }{T_{H} -T_{L}   }

COP = \frac{335.9}{335.9 -262.9}

COP = 4.6

Therefore the COP of the system is = 4.6

8 0
3 years ago
What are the characteristics of an ideal transformer ​
Vsevolod [243]

It is an imaginary transformer which has no core loss, no ohmic resistance and no leakage flux. The ideal transformer has the following important characteristic. The resistance of their primary and secondary winding becomes zero. The core of the ideal transformer has infinite permeability.

4 0
3 years ago
How is heat converted into work in a steam engine?
Dmitry_Shevchenko [17]
In an extremely simplified explanation,
What happens is that the steam (heat) that is generated by the burning of coal is used to rotate the motor of the steam engine.
8 0
3 years ago
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