Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg
It's velocity is not constant as direction is changing.
We know, velocity is speed with direction, so if direction is changing, velocity can't be constant, doesn't matter that speed is constant.
Hope this helps!
Answer:
Uh No thanks but make me brainiest!
Explanation:
Ok
........................
First we can say that since there is no external force on this system so momentum is always conserved.
now by the condition of elastic collision
![v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6](https://tex.z-dn.net/?f=v_%7B2f%7D%20-%20v_%7B1f%7D%20%3D%200.8%20-%200%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Enow%20add%20two%20equations%3C%2Fp%3E%3Cp%3E%5Btex%5D3%2Av_%7B2f%7D%20%3D%201.6)
also from above equation we have
So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.
