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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Answer: B. the isovolumetric process
Explanation:
In the graph given, the volume is constant throughout. It represents a constant volume process. Such processes are called the isovolumetric process or isochoric process.
<em>Hence, option B is the correct answer.</em>
Option A is incorrect because in an isobaric process, the pressure is constant.
Option C is incorrect because in an isothermal process, the temperature is constant.
Option D is incorrect because in an adiabatic process there is no heat transfer.
Answer:

Explanation:
Given:
- mass of solid cylinder,

- diameter of cylinder,

- mass of bucket of water,

<em>When the bucket is released to fall in the well, it fall under the acceleration due to gravity.</em>
We have formula for angular acceleration as:

where:
g = acceleration due to gravity
r = radius of the cylinder


Explanation:
Below is an attachment containing the solution.
The initial speed of car A is 15.18 m/s.
Momentum is defined as mass in motion. If there are two objects (the two objects in motion or only one object in motion and the other in stationary) that collide and no other forces work in the system, the law of momentum conservation applies in the system.
p=p'
pa+pb = pa'+pb'
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
- ma = mass of object A (kg) = 1,783 kg
- mb = mass of object B (kg) = 1,600 kg
- va = speed of object A before collides (m/s)
- va' = speed of object A after collides (m/s) = 8 m/s
- vb = speed of object B before collides (m/s) = 0 m/s
- vb' = speed of object B after collides (m/s) = 8 m/s
- p = momentum before collision (Ns)
- p' = momentum after collision (Ns)
(ma×va) + (mb×vb) = (ma×va') + (mb×vb')
(1,783×va) + (1,600×0) = (1,783×8) + (1,600×8)
(1,783×va) + 0 = 14,264+12,800
(1,783×va) = 27,064

va = 15.18 m/s
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