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Answer:
Explanation:
As the paint brush is dropped from the top of a tall ladder it started free falling
i.e. velocity of brush increases as it falls down due to the acceleration provided by gravity.
We know acceleration due to gravity is the attraction experienced by the object due to earth attraction
Value of acceleration due to gravity is constant i.e.
Therefore the correct choice is
velocity is increasing and acceleration is constant
Answer:
Thermal Efficiency, η = . . . . . . . . . . . . . . . . Eqn 1
where W₀ = Work Output = Q₁ - Q₀ =82500KW . . . . . . . . . . . . . . . Eqn 2
Q₁ = Heat Supplied/Input = mC(ΔT₁)
Q₁ = Heat Rejected/Output = mC(ΔT₀) . . . . . . . . . . . . . . . . . . . . . . . . Eqn 3
Note: From Carnot's theorem, for any engine working between these two temperatures (T₀/T₁), The maximum attainable efficiency is the Carnot efficiency given as follows;
Therefore, η = 1 - = 1 -
Remember, T₁ = 475K and T₀ = 308K
η = 1 - (308/475) = 1 - 0.648 = 0.352
Hence, the maximum efficiency at which this plant can operate = 35%
2. To determine the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.
Remember from Eqn 1, Q₁ = W/η,
Therefore, Q₁= 82500/0.35
Q₁=235,714KW,
So, from Eqn 2, Q₀ = 235714 - 82500
Q₀ = 153214KW (KJ/s) (Released Heat)
In t =24 hours, we can then use this to determine the minimum amount of heat rejected qₓ (KiloJoule), = Q₀t (Remember, you have to convert the time, t, unit to seconds)
= 153214 x t (KiloJoule)
qₓ = 153214 x 24 x 3600 (KiloJoule)
qₓ = 13238 MegaJoule
<h3>Therefore, the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours is 13238 MJoule</h3>