Chemical=B
Mechanical=C
Nuclear=A
Electromagnetic=D
Thermal=E
I think these are right but if not I’m sorry
Hope I helped :]
Answer:
The speed of the car when load is dropped in it is 17.19 m/s.
Explanation:
It is given that,
Mass of the railroad car, m₁ = 16000 kg
Speed of the railroad car, v₁ = 23 m/s
Mass of additional load, m₂ = 5400 kg
The additional load is dropped onto the car. Let v will be its speed. On applying the conservation of momentum as :
v = 17.19 m/s
So, the speed of the car when load is dropped in it is 17.19 m/s. Hence, this is the required solution.
Power = work done/time. Work done by gravity = mgh = 45x9.8x3.2 = 1411.2J. Therefore, power = 1411.2/2 = 705.6 ~ 706W.
Answer:
Explanation:
First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.
Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.
For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.
The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.
First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.
Answer:
a) 90 kJ
b) 230.26 kJ
Explanation:
The pressure at the first point 
= 10 bar —> 10 x 102 = 1020 kPa
The volume at the first point 
= 0.1 m^3
The pressure at the second point 
= 1 bar —> 1 x 102 = 102 kPa
The volume at the second point 
= 1 m^3
Process A.
constant volume V = C from point (1) to P = 10 bar.
Constant pressure P = C to the point (2).
Process B.
The relation of the process is PV = C
Required
For process A & B
(a) Sketch the process on P-V coordinates
(b) Evaluate the work W in kJ.
Assumption
Quasi-equilibrium process
Kinetic and potential effect can be ignored.
Solution
For process A.
V=C
There is no change in volume then
The work is defined by
║
V║limit 1--0.1
90 kJ
Process B
PV=C
By substituting with point (1) C = 10^2 x 1= 10^2
The work is defined by
║ ln(V)║limit 1--0.1
=230.26 kJ
