Answer:-q
Explanation:
Given
Capacitor is charged to a battery and capacitor acquired a charge of q i.e.
+q on Positive Plate and -q on negative Plate.
If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by
where A=area of capacitor plate
d=Separation between plates
This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q
Newtons second law says that the acceleration of an object (produced by a net force) is directly proportional to that magnitude of the net force. E.g. F = ma
where F is the net force of an object, m is mass and a is acceleration.
For example, if an object had a large mass, there would have to be more force in order to move it than if it was lighter.
In a linear motion, if you pushed two objects, one slightly larger than the other, with the same force, the acceleration of the smaller object would be bigger than the larger one. So the motion (change in position over time), of the larger object would be seen as lesser than the smaller one (in a situation where both forces are equal).
Answer:
Thus induced emf is 0.0531 V
Solution:
As per the question:
Diameter of the loop,
Thus the radius of the loop, R = 0.048 m
Time in which the loop is removed, t = 0.15 s
Magnetic field, B = 1.10 T
Now,
The average induced emf, e is given by Lenz Law:
where
= magnetic flux =
where
A = cross sectional area
Also, we know that:
e = 0.0531 V
The sketch is shown in the figure, where I indicates the direction of the induced current.
Answer:
w= 62.75 J
Explanation:
Given that
Force vector F= 5 x i + 4 y j
Space vector or displacement vector d= 5.01 i
We know that work (w)
w=∫ F.ds
w= ∫(5 x i + 4 y j) .dx ( only object is moving in x- direction)
w= 2.5 x 5.01² J
w= 62.75 J