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pogonyaev
3 years ago
13

Before designing the filter, one must understand the relationship between the output voltage of the circuit and the frequency. F

or a series RLC circuit, how do the impedances of the circuit elements depend on the frequency?
a. The impedance magnitudes do not depend on frequency.
b. Both impedance magnitudes decrease when the frequency increases.
c. The magnitude of the impedance of the capacitor decreases when the frequency increases; the magnitude of the impedance of the inductor increases when the frequency increases.
d. The magnitude of the impedance of the capacitor increases when the frequency increases; the magnitude of the impedance of the inductor decreases when the frequency increases.
Physics
1 answer:
Kamila [148]3 years ago
5 0

Answer:

Option b. is correct

Explanation:

An RLC electrical circuit consists of constituent components: a resistor (R), an inductor (L), and a capacitor (C). A resistor, an inductor, and a capacitor are connected in series or parallel.

The impedances of the circuit elements depend on the frequency.

Both impedance magnitudes decrease when the frequency increases

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Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

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Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

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