A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the
1 answer:
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Answer:
<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>
Explanation:
The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC
The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC
Since the negative ends are joined together the total charge on both capacity would be;
q =
q = 200 - 180
q = 20 mC
In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage
q = (4 x V) + (6 x V)
20 = 10 V
V = 2 V
For the final charge on 6.0 mF;
q = CV
q = 6.0 mF x 2 V
q = 12 mC
Therefore the final charge on the 6.0 mF capacitor would be 12 mC
Answer:
10.791 m/s
5.93505 m
Explanation:
m = Mass of ball
= Final velocity
= Initial velocity
= Final time
= Initial time
g = Acceleration due to gravity = 9.81 m/s²
From the momentum principle we have
Force
So,
The speed that the ball had just after it left the hand is 10.791 m/s
As the energy of the system is conserved
The maximum height above your hand reached by the ball is 5.93505 m