Since the track is friction less, block’s kinetic energy at the bottom of the track is equal to its potential energy at the top of the track.
PE = 81 * 9.8 * 3.8 = 3016.44 J
Work = 1/2 * 1888 * d^2
PE = Kinetic energy at the base.
1/2 * 1888 * d^2 = 3016.44
d = 1.78 approx 1.8
F = Ke = 1888 * 1.8 = 3398.4N
Answer:
The output power the weightlifter is 2916.67 W.
Explanation:
Given;
weight lifted, W = 700 N
height the weight is lifted, h = 2.5 m
time taken to lift the weight, t = 0.60 s
The output power the weightlifter is calculated as;
Power = Energy applied / time taken
Energy applied = weight lifted x height the weight is lifted
Energy applied = 700 x 2.5
Energy applied = 1750 J
Power = 1750 / 0.6
Power = 2916.67 J/s = 2916.67 W.
Therefore, the output power the weightlifter is 2916.67 W.
Answer:
When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.
During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to

where
m is the mass of the water

is the latent heat of evaporation.
For water, the latent heat of evaporation is

, while the mass of the water is

so, the amount of heat released in the process is