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aleksandr82 [10.1K]
3 years ago
10

What is the entropy of a closed system in which 28 distinguishable grains of sand are distributed among 1000 distinguishable equ

al-sized compartments?
Physics
1 answer:
baherus [9]3 years ago
4 0

Answer:

193.4 J/K

Explanation:

For a closed system, the entropy is given as the natural logarithm of microstates.

The number of particles ( grains of sand ) N = 28

The number of boxes ( compartments ) M = 1000

Entropy is the logarithm of number of microstates

S=ln \psi

But the number of microstates is given by

\psi=M^{N} hence S=ln M^{N}=\ln (1000^{28})= \boxed{193.4 J/K}

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An airplane propeller is rotating at 1900 rpm (rev/min).
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Answer:

See explanation

Explanation:

We have to convert to angular velocity in rads-1 as follows;

Angular velocity in rad/s = 2π/60 × 1900 rpm = 199 rad/s

Given that

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Time taken = angle turned/angular velocity

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If we want to know the velocity that an object is traveling, we must know the
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Your friend says that if Newton’s third law is correct, no object would ever start moving. Here is his argument: You pull a sled
m_a_m_a [10]

Answer:

Explanation:

You pull a sled exerting a 50 N force on it , sled also exerts a force on you . These forces are action and reaction force , as per third law of Newton . These two forces are equal and  opposite . But they do not act on the same object so they do not cancel each other . They act on different objects , one on the sledge and the other on you . Due to force on sledge , sledge moves in the direction of force or towards you . You will start moving in opposite direction if frictional force of ground is nil or less .

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2 years ago
What quantity of heat must be removed from 20g<br>of water at 0°C to change it to ice at 0°C?​
seraphim [82]

The quantity of heat must be removed is 1600 cal or 1,6 kcal.

<h3>Explanation : </h3>

From the question we will know if the condition of ice is at the latent point. So, the heat level not affect the temperature, but it can change the object existence. So, for the formula we can use.

\boxed {\bold {Q = m \times L}}

If :

  • Q = heat of latent (cal or J )
  • m = mass of the thing (g or kg)
  • L = latent coefficient (cal/g or J/kg)
<h3>Steps : </h3>

If :

  • m = mass of water = 20 g => its easier if we use kal/g°C
  • L = latent coefficient = 80 cal/g

Q = ... ?

Answer :

Q = m \times L \\ Q = 20 \times 80 = 1600 \: cal

So, the quantity of heat must be removed is 1600 cal or 1,6 kcal.

<u>Subject : Physics </u>

<u>Subject : Physics Keyword : Heat of latent</u>

4 0
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