Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be = 
..............1
put here value in equation 1
lowest frequency will be = 
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength = 
..............2
put here value
wavelength = 
wavelength = 0.08575 m
so distance = 
..............3
distance = 
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
If you mean S is the distance then it is true
Velocity = Distance / time
Answer:
the moon lacks a sizeable iron core
Explanation:
Answer
The intensity of a sound wave depends on the pressure of the wave,density of the medium and speed of sound in the medium. Higher density and higher sound speed both give a lower intensity. and may be it is because that sound wave is more characterize by wavelength than frequency..explanation
Explanation:
As decibel levels get higher, sound waves have greater intensity and sounds are louder. For every 10-decibel increase in the intensity of sound, loudness is 10 times greater. Intensity of sound results from two factors: the amplitude of the sound waves and how far they have traveled from the source of the sound.
Answer:
just search up a ven-diagram and then try to draw it or trace it then use it for ur question
Explanation:
