i really have some proplem with this question(adsbygoogle = window.adsbygoogle || []).push({});

Two antennas located at points A and B are broadcasting radio waves of frequency 95.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d= 9.30 m. An observer, P, is located on the x axis, a distance x= 60.0 m from antenna A, so that APB forms a right triangle with PB as hypotenuse. What is the phase difference between the waves arriving at P from antennas A and B?

1.426 rad

You are correct. Computer's answer now shown above.

Your receipt is 155-1772

Now observer P walks along the x axis toward antenna A. What is P's distance from A when he first observes fully destructive interference between the two waves?

If observer P continues walking until he reaches antenna A, at how many places along the x axis (including the place you found in the previous problem) will he detect minima in the radio signal, due to destructive interference?

i calculated the phase difference between the two antenas but i didnt know how to to calculate the first time that i will recieve a destructive interfernce i think that it is when the phase difference is pi*(2m+1)

this is how i calculted the first quesion first i calculated the distance BP which is 60.716m

then i used the equation (phase difference=2*pi*delta L/ג ) delta L=BP-AP and ג=c/frequency which is given

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# Two antenas (two waves wih the same frequency)

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