Answer:
80 J
Explanation:
PE = mgh
PE = (4 kg)(9.8 m/s^2)(2 m)
PE = 78.4 J and with sig figs, it would be 80 J
Answer:
2649600 Joules
Explanation:
Efficiency = 40%
m = Mass of air = 92000 kg
v = Velocity of wind = 12 m/s
Kinetic energy is given by
The kinetic energy of the wind is 6624000 Joules
The wind turbine extracts 40% of the kinetic energy of the wind
The energy extracted by the turbine every second is 2649600 Joules
You first find the mass and the volume of that object. Then you divide mass ÷ volume
Answer:
(a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Explanation:
Given that,
Power factor = 0.6
Power = 600 kVA
(a). We need to calculate the reactive power
Using formula of reactive power
...(I)
We need to calculate the
Using formula of
Put the value into the formula
Put the value of Φ in equation (I)
(b). We draw the power triangle
(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95
Using formula of reactive power
We need to calculate the difference between Q and Q'
Put the value into the formula
Hence, (a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.