Answer:
a) E_{L} = -360 V
, b) t = 3 s
Explanation:
The electromotive force in an inductor is
= - L dI/ dt
in the exercise they give us the relation of i (t)
i (t) = 1.00 t² -6.00t
we carry out the derivative and substitute
E_{L} = - L (2.00 2t - 6.00 1)
a) the electromotive force at t = 1.00 s
E_{L} = - 90.0 (4.00 1 - 6.00)
E_{L} = -360 V
b) for t = 4 s
E_{L}= - 90 (2 4 2 - 6 4)
E_{L} = - 720 V
c) for the induced electromotive force to zero, the amount between paracentesis must be zero
(2.00 t2 - 6.00t) = 0
t (2.0 t-6.00) = 0
the solutions of this equation are
t = 0
2 t -6 = 0
t = 3 s
to have a different solution the trivial (all zero) we must total t = 3 s
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Answer:
Explanation: V = 3 cm/s = 0.03 m/s
BY THE FORmULA OF K.E
K.E = 1/2 mV^2
300 =1/2 m (0.03)^2
m = 300 x 2/0.0009
m= 666666.66kg