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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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True or false: when hydrochloric acid is dissolved in water the reaction produces carbonand oxygen?
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3 years ago
Coherent light with wavelength 540 nm passes through narrow slits with a separation of 0.370 mm . At a distance from the slits w
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Answer:

∅=2021rad

Explanation:

Coherent light with wavelength 540 nm passes through narrow slits with a separation of 0.370 mm . At a distance from the slits which is large compared to their separation, what is the phase difference (in radians) i the light from the two slits at an angle of 28.0 ∘ from the centerline?

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using the formula below

∅=\frac{2\pi*d }{\beta } sin\alpha

the phase difference is given from the above

d= distance/separation of slits .00037m

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540*10^-9m

\alpha=angle of two slits  28.0 ∘

∅=(\frac{2\pi*0.00037 }{540*10^-9} sin 28

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there are large bright fringes from the centre line

3 0
3 years ago
HELP PLZ! A parachutist jumps out of an aeroplane and accelerates uniformly at 9.8 ms-2 for ten seconds. a. How far does she tra
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Answer:

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b.) 98m/s

Explanation:

Given that the

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Time = 10s

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U = 0

Distance covered = height H

The height can be calculated by using second equation of motion

H = Ut + 1/2gt^2

Substitute g and t into the formula

H = 1/2 × 9.8 × 10^2

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Therefore, she travels as far as 490 m

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V^2 = U^2 + 2gH

Substitute g and H into the formula.

Remember that U = 0

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4 0
4 years ago
A rocket has initail mass M begins to move from space , with an exhaust constant speed . Find the mass of the rocket while has t
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Answer:

m=\frac{m_{0}}{e}

Explanation:

Equation of the rocket is,

m\frac{dv}{dt} =F-v'\frac{dm}{dt}

Here, v' is the relative velocity of rocket.

In space F is zero.

So,

m\frac{dv}{dt} =-v'\frac{dm}{dt}\\dv=-v'\frac{dm}{m} \\v=-v'ln\frac{m}{m_{0} }

Now the momentum can be obtained by multiply by m on both sides.

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Therefore, the mass of the rocket while having maximum momentum is \frac{m_{0}}{e}

3 0
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