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.
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<span>v/2
This is an exercise in the conservation of momentum.
The collision specified is a non-elastic collision since the railroad cars didn't bounce away from each other. For the equations, I'll use the following variables.
r1 = momentum of railroad car 1
r2 = momentum of railroad car 2
x = velocity after collision
Prior to the collision, the momentum of the system was
r1 + r2
mv + m*0
So the total momentum is mv
After the collision, both cars move at the same velocity since it was non-elastic, so
r1 + r2
mx + mx
x(m + m)
x(2m)
And since the momentum has to match, we can set the equations equal to each other, so:
x(2m) = mv
x(2) = v
x = v/2
Therefore the speed immediately after collision was v/2</span>
Answer:
Part a)
Part b)
Part c)
Explanation:
Part a)
As we know that it is having constant torque so here the time taken by it to accelerate is given as
Part b)
angular displacement is given as
Part c)
As we know that the angular deceleration produced by the brakes is given as
now we have
As we know that
so we have
Answer:
Speed and direction affect pitch.
Ohms law says that V = IR
The amount of work done is P = IV or P = I^2R (using Ohm's law to substitute)
Therefore P = 64*15 = 960W
So the total energy released is Pt = 960*60 = 57600J