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:
Mass is measured using a scale. Volume is the amount of space matter takes up. Volume can be measured by multiplying the length, width, and height of an object, or by using a measuring cup and seeing how the measurement of water changes in a cup. Density is the amount of matter an object has in a certain space.
Answer:
λ = 6.97 *10⁻⁶ C/m
Explanation:
Conceptual Analysis:
The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:
E= 2k*(λ/r) Formula (1)
Where:
E: electric field .( N/C)
k: Coulomb electric constant. (N*m²/C²)
λ: linear charge density. (C/m)
r : distance from the charge line to the surface where E calculates (m)
Known data
E= 6.6 x 10⁴ N/C
r = 1.9 m
k= 8.99 *10⁹ N*m²/C²
Problem development
We replace data in the formula (1):
E= 2k*(λ/r)
6.6*10⁴ = 2*(8.99 *10⁹)*(λ/1.9)
(6.6*10⁴)*(1.9) = 2*(8.99 *10⁹) *λ
(12.54* 10⁴) / (17.98*10⁹) = λ
λ = 6.97 *10⁻⁶ C/m
Answer:
in accelerated motion
Explanation: tbh I just guessed if im wrong sorry