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

It would be 12hz because it
Answer:
No, because as distance increases, gravitional force decreases.
It has both magnitude and direction