Answer:
I got the answar ...............................
Half the potential difference of the the1-µF
A circuit must have a capacitance of 2 F across a 1 kV potential difference for an electrical technician. He has access to a sizable number of 1F capacitors, each of which can sustain a potential difference of no more than 400 V. Please suggest a configuration that uses the fewest capacitors possible.
The 2-mu F capacitor has the following characteristics: none of the aforementioned; half the charge of the 1-mu F capacitor; twice the charge of the 1-mu F capacitor; and half the potential difference of the 1-mu F capacitor.
Q = C V, C = Capacitance of the capacitor gives the charge stored by a capacitor with an applied voltage V. V is the applied voltage.
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Answer:
h = 20 m
Explanation:
given.
height, h = 10 m
Potential energy at 10 m = 50 J
Kinetic energy at 10 m = 50 J
maximum height the ball will reach, H = ?
Total energy of the system
T E = 50 J + 50 J
T E = 100 J
now,
A h = 10 m
P E = m g h
50 = m g x 10
mg = 5 ..............(1)
at the top most Point the only Potential energy will be acting on the body.
now, TE = Potential energy
100 = m g h
5 h = 100
h = 20 m
hence, the maximum height reached by the ball is equal to 20 m.
<u>Explanation:</u>
Velocity of B₁ = 4.3m/s
Velocity of B₂ = -4.3m/s
For perfectly elastic collision:, momentum is conserved
where,
m₁ = mass of Ball 1
m₂ = mass of Ball 2
v₁ = initial velocity of Ball 1
v₂ = initial velocity of ball 2
v'₁ = final velocity of ball 1
v'₂ = final velocity of ball 2
The final velocity of the balls after head on elastic collision would be
Substituting the velocities in the equation
If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.
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