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bagirrra123 [75]
3 years ago
8

What is the difference between a battery losing charge (discharging) and a capacitor losing charge (discharging)?

Physics
1 answer:
aivan3 [116]3 years ago
6 0
The capacitor takes the charge from the battery but the electromotive force EMF of the battery remains same. But, during discharging, the internal resistance of the battery decreases because the charge moves out of the battery to the capacitor.
But, when the capacitor discharges its emf,  decreases with time.
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A person climbs out of swimming pool and stands in the open air.Explain why evaporation of water from the surface of the persons
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A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball
lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.  

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

M_{1} u_{1} +M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

Since initial velocity for M1 ball is zero, then

M_{2} u_{2}=M_{1}  v_{1}+M_{2}  v_{2}

and

M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+  M_{2}v_{2} ^{2}

So, on solving all the above equation, we get an equation for velocity and that is

\frac{2M_{2}u_{2} }{(M_{1}+M_{2}  }=final velocity of ball with mass 2.5 kg

v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is \frac{1}{2}*2.5*(4.67)^{2}  =27.09 J

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

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