Answer:
the required mass flow rate is 49484.37 kg/s
Explanation:
Given the data in the question;
we first determine the relation for mass flow rate of water that passes through the turbine;
so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;
= m ( Δ P.E )
so we substitute (gh) for ( Δ P.E );
= m (gh)
m = / gh
so we substitute our given values into the equation
m = 100 MW / ( 9.81 m/s²) × 206 m
m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m
m = 10 × 10⁷ / 2020.86
m = 49484.37 kg/s
Therefore, the required mass flow rate is 49484.37 kg/s
Answer:
Final Velocity = √(eV/m)
Explanation:
The Workdone, W, in accelerating a charge, 2e, through a potential difference, V is given as a product of the charge and the potential difference
W = (2e) × V = 2eV
And this work is equal to change in kinetic energy
W = Δ(kinetic energy) = ΔK.E
But since the charge starts from rest, initial velocity = 0 and initial kinetic energy = 0
ΔK.E = ½ × (mass) × (final velocity)²
(Velocity)² = (2×ΔK.E)/(mass)
Velocity = √[(2×ΔK.E)/(mass)]
ΔK.E = W = 2eV
mass = 4m
Final Velocity = √[(2×W)/(4m)]
Final Velocity = √[(2×2eV)/4m]
Final Velocity = √(4eV/4m)
Final Velocity = √(eV/m)
Hope this Helps!!!
VO2 max is considered to be the most valid measure<span> of </span>cardio respiratory fitness<span>. It </span>measures<span> the capacity of the heart, lungs, and blood to transport oxygen to the working muscles, and </span>measures<span> the utilization of oxygen by the muscles during exercise.</span>
Answer:
Explanation:
Since the roundabout is rotating with uniform velocity ,
input power = frictional power
frictional power = 2.5 kW
frictional torque x angular velocity = 2.5 kW
frictional torque x .47 = 2.5 kW
frictional torque = 2.5 / .47 kN .m
= 5.32 kN . m
= 5 kN.m
b )
When power is switched off , it will decelerate because of frictional torque .
