How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c
1 answer:
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Answer:
The required work done is
Explanation:
Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if '' be the coefficient of friction, then
where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.
Since the application of force by the movers does not create any acceleration to the block, we can write
So the work done (W) in moving the crate by a distance s = 10.6 m is
Answer:
a) 1.34*10^-8 W
b) 1.18*10^-5 H
c) 20mV
Explanation:
a) To find the average magnetic flux trough the inner solenoid you the following formula:
mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A
N: turns of the solenoid = 340
I: current of the inner solenoid = 0.100A
A: area of the inner solenoid = pi*r^2
r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m
You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:
the magnetic flux is 1.34*10^{-8}W
b) the mutual inductance is given by:
N1: turns of the outer solenoid = 22
N2: turns of the inner solenoid
A_2: area of the inner solenoid
l: length of the solenoids = 25.0cm=0.25m
by replacing all these values you obtain:
the mutual inductance is 1.18*10^{-5}H
c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:
by replacing you obtain:
the emf is 20mV