Answer:
Explanation:
Given
Distance between two loud speakers 
Distance of person from one speaker 
Distance of person from second speaker 
Path difference between the waves is given by

for destructive interference m=0 I.e.




frequency is given by

where 

For next frequency which will cause destructive interference is
i.e.
and 


frequency corresponding to this is

for 


Frequency corresponding to this wavelength

You can't. Velocity and acceleration measure two different things, so their units are incompatible. It's like asking, "How many meters does this book weigh?"
Maybe you mean "find" acceleration using given velocities, or a velocity function?
A because the girl in that instant is not moving up or down so
( up forces)=(down forces )
The up forces is the tension of the rope and down forces us mg -the gravitational force on the girl by the earth
C is correct. The work-force relation is given by W=F·d, where F is force vector, and d is the displacement vector. The dot is the dot product, which is a measure of how parallel the two vectors are. It can be restated as the product of two vector magnitudes times the cosine of the angle between them. Therefore work is a scalar, not a vector, since the dot product returns a scalar.
We will hear the sound of siren of frequency 1553.4606 Hz.
<h3>What is Doppler Effect?</h3>
The apparent change in wave frequency brought on by the movement of a wave source is known as the Doppler effect. When the wave source is coming closer and when it is moving away, the perceived frequency changes. The Doppler effect explains why we hear a passing siren's sound changing in pitch.
according to Dopplers Effect,
![f'=[\frac{v + v_{0} }{v - v_{s} } ]f](https://tex.z-dn.net/?f=f%27%3D%5B%5Cfrac%7Bv%20%2B%20v_%7B0%7D%20%7D%7Bv%20-%20v_%7Bs%7D%20%7D%20%5Df)
![f'= [\frac{700+68.1}{700-94.8} ]* 1224](https://tex.z-dn.net/?f=f%27%3D%20%5B%5Cfrac%7B700%2B68.1%7D%7B700-94.8%7D%20%5D%2A%201224)

the frequency would be 1553.4606 Hz.
to learn more about Doppler Effect go to - brainly.com/question/9165991
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