Answer:
work output is always less than work input - the ratio is less than 1.
Explanation:
This principle comes from the fact that a machine or system cannot produce more work than is supplied to it, because this would violate the energy conservation law (work is a type of mechanical energy).
In theoretical machines called "ideal machines" the input work is the same as the output work, but these machines are only theoretical because in real applications there is always some type of energy loss, either in heat produced by a machine or processes for its operation, for this reason the output work is always less than the input work.
Regarding the ratio work output to work input:
because work input WI is always greater than work output WO.
Out of the choices given, igniting the gas-air mixture supplies the heat for the hot reservoir in a car's engine. The correct answer is C.
Intensity:
Decibels
Amplitude:
Meters
Frequency: Hertz
<u>Explanation:</u>
The Wave is not visible to eyes and they can easily propagate through vacuum. the average power travelling at a given period of time in a space is the intensity. Decibels is the measure of intensity. it is measured in the decibel scale. The wave's strength and the intensity gives the amplitude of wave. It is measured using meters.
The wave's amplitude and the energy has a direct proportionality. The number occurrence of wave cycles per second refers to the frequency of wave. it is measured in hertz. it is also measured as the number of cycles that occurs per second.
Answer:to revise or edit
anything that can be made in the non draft one
Explanation:
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
