Answer:
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Explanation:
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All of them have the same potential energy <span />
Since there are no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, we would assume that momentum is conserved. This means that
total momentum of car before collision = total momentum of car after collision.
Recall,
momentum = mass x velocity
From the information given,
mass of car before collision = 2000
velocity of car before collision = 3
Thus,
total momentum of car before collision = 2000 x 3 = 6000
Also,
mass of sack = 500
mass of car and sack after collision = 500 + 2000 = 2500
velocity after collision = v
momentum after collision = 2500 x v = 2500v
Since momentum is conserved, then
6000 = 2500v
v = 6000/2500
v = 2.4
the speed of the flatcar is 2.4 m/s
Answer:
Explanation:
i )
When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2
So charged stored in it will remain unchanged .
ii )
Potential difference = charge / capacitance
in the first case potential difference = Q / C
in the second case potential difference = Q / 2C
So potential difference becomes half .
iii ) electric field = potential diff / plate separation
in the first case electric field = Q / (d x C )
in the second case electric field = 2 Q / (d x 2C)
= Q / (d x C )
So electric field remains unchanged .
iv)
energy stored in first case = Q² / 2C
In the second case energy stored = Q² / 2x2C
so energy stored becomes half .
