An aircraft engine takes in 9000 j of heat and discards 6400 j each cycle. (a) what is the mechanical work output of the engine
1 answer:
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Answer:
Explanation:
Given
Force P is acting upward
C is vertical contact Force
W is the weight of the crate
As P is unable to move the Block therefore Normal reaction keeps on acting on block
thus we can say that
P-W+C=0
P=W-C
Given:
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:
where
Ca = 3Cb (1)
where
Ca = heat capacity of object A
Cb = heat capacity f object B
Also,
Ta = 2Tb (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.
Let
Tf = final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.
Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb) (3)
Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb
Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb
Answer:
where
Answer:
Part a)
Part b)
Explanation:
Part a)
change in the energy due to decay of photon is given as
here we know that
now we have
Part b)
While electron return to its ground state it will emit a photon of energy 2/3rd of the total energy
so we have
now to find the wavelength we have