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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:
moves into a region of higher potential
Potential difference = 835 V
Ki = 835 eV
Explanation:
given data
initial speed = 400000 m/s
solution
when proton moves against a electric field so that it will move into higher potential region
and
we know Work done by electricfield W is express as
W = KE of proton K
so
q × V = 0.5 × m × v² ......................1
put here va
lue
1.6 × × V = 0.5 × 1.67 × × 400000²
Potential difference V = 1.336 × 10-16 / 1.6 × 10-19
Potential difference = 835 V
and
KE of proton in eV is express as
Ki = V numerical
Ki = 835 eV
Answer:
A.
Explanation:
It will stay in motion unless for example, someone stops it from going.