A 14gram ovarian tumor is treated using a sodium phosphate in which the phosphorus atoms are the radioactive phosphorus 32 isoto
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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