Answer:
self-inductance is 23.4 mH
0.145 seconds it be turned off
Explanation:
given data
length L = 42.2 cm
diameter d = 10 cm
no of loop N = 1000 loops
induced emf = 3 V
current I = 18.6 A
to find out
self-inductance and How fast it turned off emf
solution
we apply here formula of self-inductance that is
self-inductance = u×N²×A / L .......................1
put here value
self-inductance = 4π×1000²×(π×r² / 0.422)
self-inductance = 4π×1000²×(π×0.05² / 0.422)
self-inductance = 0.0234 H
so self-inductance is 23.4 mH
and
we know emf = Ldi /dt
so time = 0.0234 × 18.6
time = 0.145 seconds
about 1/10th as great by bending your legs
Time extended decreases acceleration value.
solve using Newton's 2nd and Uniform Acceleration Laws. plug in values for descending.
Vf = 0.0 m/s
Vi = 10 m/s
t = 1 s
Acceleration is given by
a = [Vf - Vi] / t
a = [ (0.0 m/s) - (10 m/s) ] / (1 s)
a = [ -10 m/s ] / (1 s)
a = -10 m/s^2
Solve the same thing, but with time 10x, so t = 10 s
a = [ (0.0 m/s) - (10 m/s) ] / (10 s)
a = [ -10 m/s ] / (10 s)
a = -1 m/s^2
Solve for first force and second force, using a mass of 100 kg
F = m * a
F = (100 kg) * (-10 m/s^2)
F = -1,000 N
F = (100 kg) * (-1 m/s^2)
F = -100 N
So divide long-force by short-force
(-100 N) / (-1,000 N) = 0.1 (which is 1/10)
Answer:
the answer to your question is c
Answer:
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