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Mazyrski [523]
4 years ago
14

Radio waves travel at a speed of 300,000,000 m/s. WFNX broadcasts radio waves at a

Physics
1 answer:
yKpoI14uk [10]4 years ago
3 0

Answer:

Wavelength, \lambda=2.94\ m

Explanation:

It is given that,

Speed of radio waves is v=3\times 10^8\ m/s

Frequency of radio waves is f = 101,700,000 Hz

We need to find the wavelength of WFNX’s radio waves. The relation between wavelength, frequency and speed of a wave is given by :

v=f\lambda

\lambda is wavelength

\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{3\times 10^8}{101,700,000}\\\\\lambda=2.94\ m

So, the wavelength of WFNX’s radio waves is 2.94 m.

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In a bi-prism experiment the eye-piece was placed at a distance 1.5m from the source. The distance between the virtual sources w
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Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

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Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

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The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoi
Novay_Z [31]

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L = [μ(0) * N² * A] / l

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A = cross sectional area of the solenoid

l = length of the solenoid

7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24

1.752*10^-3 = 4π*10^-7 * 202500 * A

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N*ΔΦ = ΔI*L

Substituting this in eqn 1, we have

emf = - ΔI*L / Δt

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