You leave a 3kW heater on in your room. You put it on at 8am and leave it on until 4pm. If a unit (1kWh) costs 12.5p, how much w
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1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we call
the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
(1)
where
is the mass of ball A
is the mass of ball B
is the initial velocity of ball A
is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)
If we solve (1) to find, we find that the final velocity of the balls is
and the positive sign means the two balls are going to the right.
2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
where
is the final velocity of ball A
is the final velocity of ball B
If we solve simultaneously the two equations, we find:
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.
3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
and the total momentum after the collision is:
If we call
where
If we solve (1) to find
and the positive sign means the two balls are going to the right.
2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
where
If we solve simultaneously the two equations, we find:
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.
3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
and the total momentum after the collision is:
Answer:
Explanation:
Electric field due to a point charge Q at a point at distance d is given by the relation
E =
Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.
The charges are situated on the corners of a square in such a way that
equal charges of Q1 and Q3 are situated on the diametrically opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two charges will be zero.
On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero
Overall, net field due to all the four charges will be zero