Answer:
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Explanation:
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The capacitive reactance is reduced by a factor of 2.
<h3>Calculation:</h3>
We know the capacitive reactance is given as,
where,
= capacitive reactance
f = frequency
C = capacitance
It is given that frequency is doubled, i.e.,
f' = 2f
To find,
=?
Therefore, the capacitive reactance is reduced by a factor of 2.
I understand the question you are looking for is this:
A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?
- The capacitive reactance is doubled.
- The capacitive reactance is traduced by a factor of 4.
- The capacitive reactance remains constant.
- The capacitive reactance is quadrupled.
- The capacitive reactance is reduced by a factor of 2.
Learn more about capacitive reactance here:
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The magnetizing current in a transformer is rich in 3rd harmonic. This is because harmonics are AC voltages and currents with frequencies that are generally higher.
A decrease in velocity is referred to as deceleration. If car is moving at 30 m/s and stop in 50 m .The value of deceleration is 11.56 ms−2.
<h3>How to calculate deceleration ?</h3>
While acceleration is motion in which an object's speed varies every second, deceleration is motion that causes an object to slow down.
We are aware that acceleration refers to an object's rate of increase in speed, and deceleration refers to an object's rate of decrease in speed. For instance, when we apply the brakes while driving, we benefit from the vehicle's ability to decelerate and slow down.
The Deceleration Formula is the final velocity minus the initial velocity, with a negative sign in the result because the velocity is decreasing, if starting velocity, final velocity, and time taken are given.
velocity of car = 30 m/s
car need to stop in 50m
Deceleration a = v^2 – u^2 / 2s
= 0^2 - 50^2 / 2*30
= 11.56
Deceleration of the care = 11.56 ms−2
To learn more about deceleration refer :
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Average speed =
(distance covered during some period of time)
divided by
(length of time to cover that distance).
