The height at which the mass will be lifted is; 3 meters
<h3>How to utilize efficiency of a machine?</h3>
Formula for efficiency is;
η = useful output energy/input energy
We are given
η = 60% = 0.6
Input energy = 4 KJ = 4000 J
Thus;
0.6 = useful output energy/4000
useful output energy = 0.6 * 4000
useful output energy = 2400 J
Work done in lifting mass(useful output energy) = force * distance moved
Useful output energy = 800 * h
where h is height to lift mass
Thus;
800h = 2400
h = 2400/800
h = 3 meters
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Answer:
Heat transfer = 2.617 Kw
Explanation:
Given:
T1 = 300 k
T2 = 440 k
h1 = 300.19 KJ/kg
h2 = 441.61 KJ/kg
Density = 1.225 kg/m²
Find:
Mass flow rate = 1.225 x [1.3/60]
Mass flow rate = 0.02654 kg/s
mh1 + mw = mh2 + Q
0.02654(300.19 + 240) = 0.02654(441.61) + Q
Q = 2.617 Kw
Heat transfer = 2.617 Kw
Answer:
I think it is( More expensive than immature technologies) I took engineering class and this question is still quite tricky.
Explanation:
