2.27 mps repeating.
This is the last question ill ever answer here. Thanks for being the last.
Over lava!
I’m pretty sure that’s the answer cuz I haven’t seen the movie in a hot minute!
-YOUR WELCOME!! <3
Answer:
d. Not enough information is given to answer this question.
Explanation:
From first law of thermodynamics
Q= W + ΔU
Q=Heat ,W= Work , ΔU=Change in internal energy
If work done by the gas :
It means that W and Q both are positive
Q- W = ΔU
Ii Q > W ,then temperature of the gas will increase.
If Q< W ,Then temperature of the gas will decreases.
If work done on the gas:
Q positive but W will be negative
Q- W = ΔU
Q= W or Q>W or Q< W ,then temperature of the gas will increase.
There are three cases because they did not give any information about the work.That is why option d is correct.
Answer:
The work done is 5136.88 J.
Explanation:
Given that,
n = 1.90 mol
Temperature = 296 K
If the initial volume is V then the final volume will be V/3.
We need to calculate the work done
Using formula of work done
Put the value into the formula
The Work done on the system.
Hence, The work done is 5136.88 J.
Answer:
<em>d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.</em>
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Explanation:
Let us take the momentum of a photon unit as u
we know that the rate of change of momentum is proportional to the force exerted.
For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that
F = (u - 0)/t = u/t
for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,
F = u
For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to
F = (u - (-u))/t = 2u/t
just as the we did above, it becomes
F = 2u.
From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that <em>the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.</em>
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