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suter [353]
3 years ago
15

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

during one cycle? (b) what is the thermal efficiency of the engine?
Physics
1 answer:
charle [14.2K]3 years ago
4 0
Work done = heat in - heat out 
<span>= 9000 -6400 = 3600 J per cycle. </span>

<span>And the efficiency = work out / heat in </span>
<span>= 3600/9000 </span>
<span>= 0.4 or 40%

Sorry if this does not help.</span>
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A 0.54 kg air hockey puck is initially at rest. What will it's kinetic energy energy be after a net force of 0.56 N acts on it f
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Answer:

Kf = 470 mJ

Explanation:

  • According the work-energy theorem, the change in the kinetic energy of one object, is equal to the net work done on it.
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       W_{net} = F_{net} * \Delta X = 0.56 N * 0.84 m = 0.47 J = 470 mJ (1)

  • As we have already said, (1) is equal to the final kinetic energy of the puck:
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Uphill escape ramps are sometimes provided to the side of steep downhill highways for trucks with overheated brakes. For a simpl
o-na [289]

Answer:

404.4 m

Explanation:

Converting the initial speed from km/h to m/s then

140\times \frac {1000m}{3600s}=38.88888889 m/s \approx 38.89 m/s

The acceleration is resolved as shown in the figure hence

deceleration of the truck along the inclined plane will be

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Substituting g with 9.81 m/s^{2} then

a=-9.81 m/s^{2} sin 11^{\circ}=-1.871836245\approx -1.87 m/s^{2}

Using kinematic equation

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s=\frac {v^{2}-u^{2}}{2a} where v and u are final and initial velocities respectively

Substituting 0 for v, 38.89 m/s for u and -1.87 m/s^{2} then

s=\frac {0^{2}-38.89^{2}}{2\times -1.87}=404.3936096 m\approx 404.4 m

3 0
3 years ago
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