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klasskru [66]
3 years ago
15

Does this diagram illustrate the second law of thermodynamics? Why or why not

Physics
2 answers:
sergeinik [125]3 years ago
7 0

No, it does not because the diagram shows all the thermal energy added to the system being transformed into work. The second law of thermodynamics states that engines or machines cannot be 100 percent efficient because some thermal energy is released to the surroundings.

Savatey [412]3 years ago
5 0
<h2>Answer:</h2>

The diagram is not showing the second law of thermodynamics. It is the demonstration of 1st law of thermodynamics.

<h3>Explanation:</h3>

Second law of thermodynamics describes the entropy of the system increase with time, it does not decrease with time. It is constant for ideal systems.

While in first law of thermodynamics, it is stated that the energy of a system can not be lost but it is transferred from one form to other form.

And in this picture, it is shown that the energy released from heat source to cold sink is used in doing work.

Work and heat are forms of energy.

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Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the
Doss [256]

Answer:

v_{su} = 19.44 m/s

Explanation:

m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg

T=9.29x10^8\\r_{o}=1.43x10^{12}

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}

solve to r1

r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = \frac{m_{sa}*2*pi*r_1^2}{T}

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}

Since

L_{su}= m_{su}*v_{su}*r_1

then

v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}

v_{su} = 19.44 m/s

7 0
3 years ago
If the coefficient of kinetic friction between a 22 kg kg crate and the floor is 0.27, what horizontal force is required to move
alina1380 [7]

Answer:

58.27 N

Explanation:

the data we have is:

mass: m=22kg

coefficient of friction: \mu =0.27

and we also know the acceleration of gravity is g=9.81m/s^2

We need to do an analysis of horizontal and vertical forces acting on the object:

-------

Vertically the forces acting on the object:

  • Normal force N (acting up from the object)
  • weight: w=mg (acting down from)

so the sum of forces in the vertical axis "y" are:

F_{y}=N-w\\F_{y}=N-mg

from Newton's second Law we know that F=ma, so:

ma_{y}=N-mg

and since the object is not accelerating in the vertical direction (the movement is only horizontal) a_{y}=0, and:

0=N-mg\\N=mg

-----------

now let's analyze the horizontal forces

  • frictional force: f= \mu N and since N=mg  --> f=\mu mg
  • force to move the object: F

and the two forces just mentioned must be opposite, thus the sum of forces in the "x" axis is:

F=ma_{x}=F-f\\ma_{x}=F-\mu mg

and we are told that the crate moves at a steady speed, thus there is no acceleration: a_{x}=0

and we get:

0=F-\mu mg\\F=\mu mg

substituting known values:

F=(0.27)(22kg)(9.81m/s^2)\\F=58.27N

3 0
3 years ago
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Answer:

TO MEASURE THE ANGLES OF RAYS

Explanation:

6 0
3 years ago
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What was Marie's speed from point A to point B, in miles/minute?
sveticcg [70]

Answer:

40m

Explanation:

40m

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2 years ago
Three common elements that can reorient their electrons into magnetic domains and become magnetic are iron, nickel, and ________
ohaa [14]

Answer:

Cobalt

Explanation:

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2 years ago
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