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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 130 m. What is the magnitude of the phase difference at the detector?

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okpalawalter8

Answer:

[tex]X=165 \textdegree[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda =360m[/tex]

Angle [tex]\theta=90 \textdegree[/tex]

Distance variation [tex]d=130m[/tex]

Generally the equation for phase difference X is mathematically given by

 [tex]X=\theta+\frac{d}{\lambda}*100[/tex]

 [tex]X=90+\frac{150}{360}*100[/tex]

 [tex]X=165 \textdegree[/tex]

Therefore Phase difference X is given as

 [tex]X=165 \textdegree[/tex]

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