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dangina [55]
3 years ago
10

A sound-producing object is moving away from an observer. The sound the observer hears will have a frequency that actually being

produced by the object.
the same as

higher than

lower than

need ASAP!!!
Physics
2 answers:
kozerog [31]3 years ago
6 0
Lower than, this is due to the doplar effect
padilas [110]3 years ago
5 0

Answer:

Lower than

Explanation:

According to Doppler’s effect, the observer will not hear the same frequency of sound if there is relative motion between the source and observer. When source is moving away from the observer the frequency perceived by the observer will be less than actual frequency as given by the following formula:

                                   f1 = (V/V+Vs)*f  

where,

f1 = frequency heard by the observer  

f = actual frequency  

V = velocity of sounds waves  

Vs = velocity of source

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Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid
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Answer:

a

    z= 2.5 \ m

b

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Explanation:

From the question we are told that

     Their distance apart is  d =  5.00 \ m

      The  wavelength of each source wave \lambda =  6.0 \ m

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              z - (d-z) =  m \lambda

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

  so

        z - (5-z) =  0

=>     2 z - 5 =  0

=>     z= 2.5 \ m

Generally the path difference for destructive  interference is

           |z-(d-z)| = (2m + 1)\frac{\lambda}{2}

=>         |2z - d |= (0 + 1)\frac{\lambda}{2}

=>        |2z - d| =\frac{\lambda}{2}

substituting values

          |2z - 5| =\frac{6}{2}

=>      z  =  \frac{5 \pm 3}{2}

So  

      z =  \frac{5 + 3}{2}

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and

      z =  \frac{ 5 -3 }{2}

=>   z =  1 \ m

=>    z =  (1 \ m ,  4 \ m )

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