If you adjusted the frequency to be lower than the resonant frequency, would the voltage have been leading ahead of or lagging b
ehind the current? Explain
1 answer:
Answer:
current will lead the voltage
Explanation:
From the bellow figure we can see the plot between the frequency and impedance
From the figure it is clear that when the frequency is less than the resonance frequency then the impedance decreases as the frequency increases it is possible only when impedance is capacitive in nature.
As the impedance is capacitive in nature so current will lead than the voltage
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Answer:
The answer is below
Explanation:
The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm
radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m
a) Initial angular velocity () = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s
θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
angular acceleration (α) is:
b)
c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
a) When it stops, the final angular velocity is 0. Hence:
θ = 323 rad