A heat engine takes in 840 kJ per cycle from a heat reservoir. Which is not a possible value of the engine's heat output per cyc
1 answer:
We have by the first law of thermodynamics tha energy is preserved, hence we cannot have over 840kJ per cycle. We have by the laws of thermodynamics (the 2nd one in specific) that the entropy of a system cannot increase. We cannot have an output of 840 kJ per cycle from a heat engine because then that would mean that the entropy would stay the same, while any heat engine increases it. Hence, any value 
is acceptable.
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The initial potential energy of the wagon containing gold boxes will enable
it roll down the hill when cut loose.
The Lone Ranger and Tonto have approximately <u>5.1 seconds</u>.
Reasons:
Mass wagon and gold = 166 kg
Location of the wagon = 77 meters up the hill
Slope of the hill = 8°
Location of the rangers = 41 meters from the canyon
Mass of Lone Ranger, m₁ = 65 kg
Mass of Tonto m₂ = 66 kg
Solution;
Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m
Potential energy = m·g·h
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912
Potential energy of wagon, P.E. ≈ 17457.0912 J
By energy conservation, P.E. = K.E.
Where;
v = The velocity of the wagon a the bottom of the cliff
Therefore;
Velocity of the wagon, v ≈ 14.5 m/s
Momentum = Mass, m × Velocity, v
Initial momentum of wagon = m·v
Final momentum of wagon and ranger = (m + m₁ + m₂)·v'
By conservation of momentum, we have;
m·v = (m + m₁ + m₂)·v'
Which gives;
The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s
The Lone Range and Tonto have approximately <u>5.1 seconds</u> to grab the
gold and jump out of the wagon before the wagon heads over the cliff.
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