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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All of the above. Answer.
1.Only suitable for dc
2.more expensive than moving iron type
3. Easily damaged
Answer:
The inductance of the inductor is 0.051H
Explanation:
From Ohm's law;
V = IR .................. 1
The inductor has its internal resistance referred to as the inductive reactance, X
, which is the resistance to the flow of current through the inductor.
From equation 1;
V = IX
X = 
................ 2
Given that; V = 240V, f = 50Hz, 
= 
, I = 15A, so that;
From equation 2,
X= 
= 16Ω
To determine the inductance of the inductor,
X = 2fL
L = 
= 
= 0.05091
The inductance of the inductor is 0.051H.
Answer:
the rate of increase of radius is dR/dt = 0.804 m/hour = 80.4 cm/hour
Explanation:
the slick of oil can be modelled as a cylinder of radius R and thickness h, therefore the volume V is
V = πR² * h
thus
h = V / (πR²)
Considering that the volume of the slick remains constant, the rate of change of radius will be
dh/dt = V d[1/(πR²)]/dt
dh/dt = (V/π) (-2)/R³ *dR/dt
therefore
dR/dt = (-dh/dt)* (R³/2) * (π/V)
where dR/dt = rate of increase of the radius , (-dh/dt)= rate of decrease of thickness
when the radius is R=8 m , dR/dt is
dR/dt = (-dh/dt)* (R³/2) * (π/V) = 0.1 cm/hour *(8m)³/2 * π/1m³ *(1m/100 cm)= 0.804 m/hour = 80.4 cm/hour
