The centrifugal force working against the car would be the entire car's weight, so the centripetal force that is supplied to the car must be the car's tires traction, aka the friction of the car's tires and the surface.
Answer:
0.00158 A
Explanation:
A = Area of loop = 7.8 cm²
= Final magnetic field = 3.2 T
= Initial magnetic field = 0.5 T
R = Resitance = 
The emf is given by

The current is given by

The resulting induced current in the loop is 0.00158 A
Answer:
Rotating the coil faster
Increasing the number of coils
Using stronger magnets
<h2>
Answer:</h2>
4.2 C
<h2>
Explanation:</h2>
The charge (Q) moving is the product of the current(I) flowing through the torso of the person and the time taken (t) for the flow.
i.e
Q = I x t
Where;
I = current = 14.0A
t = time taken = 0.0300s
Substituting the values of I and t into the equation above gives
Q = 14.0 x 0.0300
Q = 4.2 C
Therefore quantity of charge moving is 4.2 C
Answer:
Explanation:
Given
Charge Q is uniformly spread over large non-conducting Elastic sheet
Electric field due to non-conducting Elastic sheet

where
surface charge density

for side 2d Electric Field is given by


