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:
the torque capacity is 30316.369 lb-in
Explanation:
Given data
OD = 9 in
ID = 7 in
coefficient of friction = 0.2
maximum pressure = 1.5 in-kip = 1500 lb
To find out
the torque capacity using the uniform-pressure assumption.
Solution
We know the the torque formula for uniform pressure theory is
torque = 2/3 ×
× coefficient of friction × maximum pressure ( R³ - r³ ) .....................................1
here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in
now put all these value R, r, coefficient of friction and maximum pressure in equation 1 and we will get here torque
torque = 2/3 ×
× 0.2 × 1500 ( 4.5³ - 3.5³ )
so the torque = 30316.369 lb-in
Answer:
0.008
Explanation:
From the question, the parameters given are:
Velocity V = 5 m/s
Pressure = 10 pa
But pressure = F/A
10 = F/A
F = 10A
Substitute all the parameters into the formula below
Coefficient of viscosity (η) = F × r /[AV]
Where
F = tangential force,
r = distance between layers,
A = Area, and
V = velocity
(η) = 10A × 0.004 /[A × 5]
The A will cancel out
(η) = 10 × 0.004 /[5]
(η) = 0.04 /5
(η) = 0.008
Therefore, the coefficient of viscosity of the fluid is 0.008
Answer:
d
Explanation:
is the because that's the amount of work in making machine can do producing heat