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Zina [86]
3 years ago
8

A series RLC circuit has a resistance of 57.61 W, a capacitance of 13.13 mF, and an inductance of 196.03 mH. The circuit is conn

ected to a variable-frequency sourcewith a fixed rms output of 23.86 V. If the source frequency is set at the circuit’s resonance frequency, what is the magnitude of the rms voltage across each of the circuit elements
Physics
1 answer:
Firdavs [7]3 years ago
4 0

Answer:

voltage across = 1.6 V

Explanation:

given data

resistance R = 57.61 Ω

capacitance c = 13.13 mF = 13.13 ×10^{-3} F

inductance L = 196.03 mH = 0.19603 H

fixed rms output Vrms = 23.86 V

to find out

voltage across circuit

solution

we know resonant frequency that is

resonant frequency = 1 / ( 2π√(LC)

put the value

resonant frequency = 1 / ( 2π√(0.19603×13.13 ×10^{-3})

resonant frequency f = 3.1370 HZ

so current will be at this resonant is

current = Vrms / R

current =  23.86 / 57.61

current = 0.4141 A

and

so voltage across will be

voltage across = current / ( 2π f C )

voltage across = 0.4141 / ( 2π ( 3.1370) 13.13 ×10^{-3} )

voltage across = 1.6 V

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3 years ago
A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force)
katen-ka-za [31]

Answer:

Velocity will be equal to 7.31 m/sec

Explanation:

We have given mass of the student m = 61 kg

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Potential energy is equal to U=mgh=61\times 9.8\times 12.3=7352.94J

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So energy remained = 7352.94-5800 = 1552.94 J

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7 0
3 years ago
When a potential difference of 12 v is applied to a wire 6.8 m long and 0.35 mm in diameter the result is an electric current of
Nonamiya [84]
The total resistance is R = voltage / current

This resistance R = (L/A)r where:
r is the resistivity
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A is the cross-sectional area of the conductor.

thus r = RA/L

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8 0
3 years ago
(a) Find the voltage near a 10.0 cm diameter metal sphere that has 8.00 C of excess positive charge on it. (b) What is unreasona
antoniya [11.8K]

Answer:

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Part b)

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here we can assume the sphere is placed at vacuum so that there is no break down of air.

Explanation:

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As we know that the potential near the surface of metal sphere is given by the equation

V = \frac{kQ}{R}

here we have

Q = 8 C

R = 10.0 cm

now we have

V = \frac{(9\times 10^9)(8 C)}{0.10}

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Part b)

this is a large potential which can not be possible because at this high potential the air will break down and the charge on the sphere will decrease.

Part C)

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Can someone please help on this one?
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Please READ Newton's first law of motion, and you'll see how nicely it explains that fact.

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