Answer:
M1 V1 = M1 V2 + M2 V3 conservation of momentum
V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact
V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s
Note: All speeds are in the same direction and have the same sign
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.
Learn more about capacitive reactance here:
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That would be an asteroid
Answer:
The height of the cliff is 90.60 meters.
Explanation:
It is given that,
Initial horizontal speed of the stone, u = 10 m/s
Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)
The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s
Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :



h = 90.60 meters
So, the height of the cliff is 90.60 meters. Hence, this is the required solution.
Answer:
2.55sec
Explanation:
time = distance/speed = (87 m)/(34 m/s) = 2.55sec