Physics
21.03.2022 01:05
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local AM radio station broadcasts at a frequency of 685.9 kHz. Calculate the wavelength

local AM radio station broadcasts at a frequency of 685.9 kHz. Calculate the wavelength at which they are broadcasting. Wavelength
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maritzahernandez32
maritzahernandez32
5,0(99 marks)

λ = 437 m.

Explanation:

Since a radio wave is a electromagnetic type of wave, it propagates approximately at the same speed of light in vacuum, 3*10⁸ m/s.As in any wave, there exist a fixed relationship between the propagation speed, the wavelength and the frequency of the radio wave, as follows:

       c = \lambda* f (1)

Replacing by the values of the speed and the frequency, and solving for the wavelength λ, we get:

       \lambda = \frac{c}{f} =\frac{3e8 m/s}{685.9e31/s} =437 m  (2)

MickeyAppleX
MickeyAppleX
4,7(47 marks)

Explanation:

0.632 m

Explanation:

let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.

According to the diagram,

d = L Sinθ + L Sinθ = 2 L Sinθ      .....(1)

Let T be the tension in the string.

Resolve the components of T.

T Sinθ  = k q1 q2 / d^2

T Sinθ = k q1 q2 / (2LSinθ)²     .....(2)

T Cosθ = mg    .....(3)

Dividing equation (2) by equation (3), we get

tanθ = k q1 q2 / (4 L² Sin²θ x mg)

tan θ Sin²θ = k q1 q2 / (4 L² m g)

For small value of θ, tan θ = Sin θ

So,

Sin³θ = k q1 q2 / (4 L² m g)

Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)

Sin³θ =  0.2523

Sinθ = 0.632

θ = 39.2 degree

So, the separation between the two charges, d = 2 x L x Sin θ

d = 2 x 0.5 x 0.632 = 0.632 m


Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. the

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