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:
CO2 emissions from fossil fuel burning should be minimized at all cost. The CO2 are gotten when the carbons from hydrocarbons react with air(oxygen). This gas erodes the ozone layer which makes the melting of ice caps faster due to increased amount of heat radiations on the earth. This is the only best and permanent solution to the reduction of the amount of heat rays on the earth which is a global problem.
Objects which reflects back the sunrays could also be inserted into the sea to prevent the melting of the ice caps.
Answer:
the acceleration of harry is equal to 66.126 m/s²
Explanation:
given,
harry is 35 m behind Draco
speed of Draco = 40 m/s
original speed of harry = 50 m/s
acceleration = ?
time taken by the Draco
t =
t = 1.875 s
distance covered by Harry
d = 35 + 175 = 210 m
to calculate the acceleration of harry


a × 3.516 × 0.5 = 116.25
a = 66.126 m/s²
hence, the acceleration of harry is equal to 66.126 m/s²
Answer:
the distance that the object is raised above its initial position is 5.625 m.
Explanation:
Given;
applied effort, E = 15 N
load lifted by the ideal pulley system, L = 16 N
distance moved by the effort, d₁ = 6 m
let the distance moved by the object = d₂
For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.
M.A = V.R

Therefore, the distance that the object is raised above its initial position is 5.625 m.
Answer:
8. acceleration =
= 1 unit .
9. acceleration =
= -1 unit.
10. acceleration =
= 0 units.
Explanation:
8. i) acceleration = velocity / time
ii) In this figure velocity = time
iii) therefore acceleration =
= 1 unit .
9. i) acceleration = velocity / time
ii) In this figure 4 = m + 5, therefore m = -1
therefore velocity = (-0.5
time) + 5
iii) therefore acceleration =
= -1 units.
10.) velocity is constant at 2
therefore acceleration =
= 0 units