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Sources A and B emit long-range radio waves of wavelength 360 m, with the phase of the emission from A ahead of that from source B by 90°. The distance rA from A to a detector is greater than the corresponding distance rB from B by 140 m. What is the magnitude of the phase difference at the detector?

Respuesta :

gbenewaeternity

Answer:

The magnitude of the phase difference at the detector = 0.8692 rad

Explanation:

Given Data:

Wavelength (λ) = 360 m

Distance (x) = 140 m

The phase difference between the two phases can be calculated using the formula;

Ф = 2π/λ * x

   = (2*π/360) * 140

   =2.44 rad

Calculating the magnitude of the phase difference at the detector as;

ΔФ = 2.44 rad -90°(2π/360)

      = 2.44 - 1.5708

      = 0.8692 rad

 

chamberlainuket

Answer:

5.5859rad

Explanation:

2pi/lambda(140)

=2pi/360x140

= 2.44rad

Thus total phase difference

Pi + 2.44rad

3.142+2.44

= 5.5859rad

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