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.
M=2.45 because you multiply out the equation on the right and divide by 10
The distance between the two adjacent nodes = λ/2.
<h3>What is Wavelength?</h3>
A periodic wave's wavelength is its spatial period, or the length over which its form repeats. It is a property of both travelling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings. The spatial frequency is the reciprocal of wavelength. The Greek letter lambda (λ) is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.
The distance between the two adjacent nodes = λ/2.
for the standing wave ,the distance between any two adjacent nodes or antinodes is 1/2 λ.
to learn more about the wavelength go to - brainly.com/question/6297363
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Energy that transfers through the medium
Explanation:
Given that,
A girl pushes her little brother on his sled with a force of 200. N for 850. m
We need to find the work done by the force of friction acting on the sled. The force of friction is 200 N
Net force = 200 N - 200 N = 0
Work done = 0
If force of friction is 300 N.
Net force = 300 N - 200 N = 100 N
The work done by a force is given by :
W = Fd
Put values,
W = 100 × 850
W = 85000 J
Hence, this is the required solution.
