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
Resultant force= (2*6^2)^(1/2)
=8.5m/s
answer is B.
Answer:
a). 87.5 mA or 
A
b). 1.78 
Explanation:
n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.
a).
A
b).
<span>Answer:
Therefore, x component: Tcos(24°) - f = 0 y component: N + Tsin(24°) - mg = 0 The two equations I get from this are: f = Tcos(24°) N = mg - Tsin(24°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24°) = 0.47 * (mg - Tsin(24°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24°) + sin(24°) * static coefficient) Then plugging in the values... T = 283.52.
Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/</span>
