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:
the cell is the smallest unit
Explanation:
a. Net force is mass times acceleration (Newton's second law).
∑F = ma
∑F = (5.0 kg) (2.0 m/s²)
∑F = 10 N
b. The net force is the sum of the individual forces.
10 N = F − 5 N
F = 15 N
c. Friction force here is mgμ.
mgμ = 5 N
(5.0 kg) (10 m/s) μ = 5 N
μ = 0.1
An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called black-body radiation
hope it helps
Answer:
A) 89.39 J
B) 30.39J
C) 23.8 J
Explanation:
We are given;
F = 30.2N
m = 3.5 kg
μ_k = 0.646
d = 2.96m
ΔEth (Block) = 35.2J
A) Work done by the applied force on the block-floor system is given as;
W = F•d
Thus, W = 30.2 x 2.96 = 89.39 J
B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;
ΔEth = μ_k•mgd
Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J
Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.
Thus,
ΔEth = ΔEth (Block) + ΔEth (floor)
Thus,
ΔEth (floor) = ΔEth - ΔEth (Block)
ΔEth (floor) = 65.59J - 35.2J = 30.39J
C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;
W = K + ΔEth
Therefore;
K = W - ΔEth
K = 89.39 - 65.59 = 23.8J