From a dam,water is pouring down at the rate of 150 kg/s, on the blades of a turbine.if a ball dropped from the same dam takes 2✓5 s to hit the turbine,then the power delivered to the turbine is approximately equal to
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Answered by
7
Hi friend,
Height of the dam=
S= ut + 1/2 g t²
Here initial velocity (u) = 0.
S= 0.5 × 10×(2√5)²
S= 100m.
power= total change inpotential energy
= ∆mgh
= 150×10×100
= 150000watt = 150kw.
Hope this helped you a little!!!
Height of the dam=
S= ut + 1/2 g t²
Here initial velocity (u) = 0.
S= 0.5 × 10×(2√5)²
S= 100m.
power= total change inpotential energy
= ∆mgh
= 150×10×100
= 150000watt = 150kw.
Hope this helped you a little!!!
Answered by
0
height of the dam s = 1/2 g t^2
s = 1/2 × 10×(2 sqrt(5))^2 = 100m.
Power in the water striking the turbines = m g h / time
P = 150 kg/s* 10 m/s^2* 100 m
= 150 kW
s = 1/2 × 10×(2 sqrt(5))^2 = 100m.
Power in the water striking the turbines = m g h / time
P = 150 kg/s* 10 m/s^2* 100 m
= 150 kW
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