Answer:
the answer is at the BOTTOM OF THEIR QUESTION
Explanation:
IT IS CORRECT BTW
Answer:
∆T = Mv^2Y/2Cp
Explanation:
Formula for Kinetic energy of the vessel = 1/2mv^2
Increase in internal energy Δu = nCVΔT
where n is the number of moles of the gas in vessel.
When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas
We say
1/2mv^2 = ∆u
1/2mv^2 = nCv∆T
Since n = m/M
1/2mv^2 = mCv∆T/M
Making ∆T subject of the formula we have
∆T = Mv^2/2Cv
Multiple the RHS by Cp/Cp
∆T = Mv^2/2Cv *Cp/Cp
Since Y = Cp/CV
∆T = Mv^2Y/2Cp k
Since CV = R/Y - 1
We could also have
∆T = Mv^2(Y - 1)/2R k
Answer:
The kinetic energy of the anti proton is 147.4 MeV.
Explanation:
Given that,
Energy = 2.12 GeV
Kinetic energy = 96.0 MeV
We need to calculate the kinetic energy of the anti proton
Using formula of energy
We know that,
So, 
Put the value into the formula
Hence, The kinetic energy of the anti proton is 147.4 MeV.
Answer: W = 1.5 J
Explanation: Work is the product of force and distance. It can be expressed in the following formula W = Fd.
W = Fd
= 5 N ( 0.3 M )
= 1.5 J
