Answer:
Magnetic fields can be used to make electricity
Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current. Electricity generators essentially convert kinetic energy (the energy of motion) into electrical energy
Explanation:
Answer:
at the beginning: 
when the plates are pulled apart: 
Explanation:
The capacitance of a parallel-plate capacitor is given by

where
k is the relative permittivity of the medium (for air, k=1, so we can omit it)
is the permittivity of free space
A is the area of the plates of the capacitor
d is the separation between the plates
In this problem, we have:
is the area of the plates
is the separation between the plates at the beginning
Substituting into the formula, we find

Later, the plates are pulled apart to
, so the capacitance becomes

A sound wave<span> in a steel rail </span>has<span> a </span>frequency of<span> 620 </span>Hz<span> and a </span>wavelength<span> of 10.5 ... Find the </span>speed<span> of </span>a wave<span> with a </span>wavelength of 5<span> m and a </span>frequency of<span> 68 </span>Hz<span>.</span>
Answer:
D. Because they are using space technology on a shirt so people can wear it on earth as well
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.