How fast can the 140 a current through a 0.200 h inductor be shut off if the induced emf cannot exceed 80.0 v?
1 answer:
Recall that to compute for the emf of a circuit given current and inductance, we must recall that
where I is the current (A), M is the mutual inductance (h), and t is the time (ms). Since the current must not exceed 80.0 V, we have
From this, we see that it must take at least 0.35 ms so it doesn't exceed 80 V.
Answer: 0.35 ms
where I is the current (A), M is the mutual inductance (h), and t is the time (ms). Since the current must not exceed 80.0 V, we have
From this, we see that it must take at least 0.35 ms so it doesn't exceed 80 V.
Answer: 0.35 ms
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Answer
given,
mass of the ball = 6.3 kg
speed of the ball = 10.4 m/s
angle made with horizontal = 43°
m_a = 1.8 kg v_a = 2.2 m/s
m_b = 1.6 kg v_b = 1.8 m/s
mass of third particle = 6.3 - 1.8 - 1.6
= 2.9 kg
u cos θ = 10.4 x cos 43° = 7.61 m/s
by using conservation momentum along x-axis
6.3 x 7.61 = 1.8 × (-2.2) + 0 + 1.6 × V₃ₓ
V₃ₓ = 32.44 m/s (toward right)
by using conservation momentum along y-axis
0 = 0 + 1.6 x 1.8 + 1.6 × V₃y
V₃y = -1.8 m/s (indicate downward)
velocity of the third particle
v = 32.49 m/s
θ = 3.176° (downward with horizontal)