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°. T

Question

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?

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Dâu 2 months 2021-07-31T22:21:57+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-31T22:23:03+00:00

    Answer:

    5.5859rad

    Explanation:

    2pi/lambda(140)

    =2pi/360×140

    = 2.44rad

    Thus total phase difference

    Pi + 2.44rad

    3.142+2.44

    = 5.5859rad

    0
    2021-07-31T22:23:34+00:00

    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

     

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