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Nadya [2.5K]
3 years ago
5

Reactance Frequency Dependence: Sketch a graph of the frequency dependence of a resistor, capacitor, and inductor. RLC Circuit R

esonance Frequency: Identify the relation between the capacitive reactance (Xc) and inductive reactance (XL) that will minimize the total impedance (Z) of an RLC circuit. Using this condition, derive the resonance frequency () of an RLC circuit.
Physics
1 answer:
jolli1 [7]3 years ago
8 0

Answer:

f=\frac{1}{2\pi \sqrt{LC}}

Explanation:

We know that impedance of a RLC circuit is given by Z=R+J(X_L-X_C)

So Z=\sqrt{R^2+(X_L-X_C)^2} here R is resistance X_L is inductive reactance and X_C is capacitive reactance

To minimize the impedance X_L-X_C should be zero we know that X_L=\omega L\ and \ X_C=\frac{1}{\omega C}

So \omega L-\frac{1}{\omega C}=0

\omega ^2=\frac{1}{LC}

\omega =\sqrt{\frac{1}{LC}}

We know that \omega =2\pi f

So \omega =2\pi f=\frac{1}{\sqrt{LC}}

f=\frac{1}{2\pi \sqrt{LC}}

Where f is resonance frequency  

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An x-ray beam of wavelength 1.4×10−10m makes an angle of 20° with a set of planes in a crystal(the Bragg angle)causing first ord
Gnom [1K]

Answer:

Second order line appears at 43.33° Bragg angle.

Explanation:

When there is a scattering of x- rays from the crystal lattice and interference occurs, this is known as Bragg's law.

The Bragg's diffraction equation is :

n\lambda=2d\sin\theta      .....(1)

Here n is order of constructive interference, λ is wavelength of x-ray beam, d is the inter spacing distance of lattice and θ is the Bragg's angle or scattering angle.

Given :

Wavelength, λ = 1.4 x 10⁻¹⁰ m

Bragg's angle, θ = 20°

Order of constructive interference, n =1

Substitute these value in equation (1).

1\times1.4\times10^{-10} =2d\sin20

d = 2.04 x 10⁻¹⁰ m

For second order constructive interference, let the Bragg's angle be θ₁.

Substitute 2 for n, 2.04 x 10⁻¹⁰ m for d and 1.4 x 10⁻¹⁰ m for λ in equation (1).

2\times1.4\times10^{-10} =2\times2.04\times10^{-10} \sin\theta_{1}

\sin\theta_{1} =0.68

<em>θ₁ </em>= 43.33°

4 0
2 years ago
A woman launches a boat from one shore of a straight river and wants to land at the point directly on the opposite shore. If the
I am Lyosha [343]

Answer:

If she stands on the North side of a river flowing to the East at 5 mph,

she must head towards the SouthWest to arrive on the South side of the river directly across from her starting point and we have

x^2 + 5^2 = 10^2 where x is her speed directly across the river

x = (75)^1/2 = 8.66 mph towards the South

sin theta = 5 / 10 = 1/2

She must angle the boat at 30 deg from straight South

4 0
2 years ago
You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of
Rzqust [24]

Answer:

<h2>3.36J</h2>

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

                                       = 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

4 0
2 years ago
A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of
PtichkaEL [24]

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box.  There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

8 0
3 years ago
Students in a science class were provided with this image and asked to identify which color of light would refract the most if w
djverab [1.8K]
Violet would refract the most as it has the shortest wavelength amongst the visible spectrum.
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