Answer :
Answer:
a) 1.1mm
b) 2.513kg/ms^2
Explanation:
You can use the formula for the calculation of the wavelength of a wave
f=1.40MHz=1.40*10^{6}Hz
1 )
[tex]\lambda=\frac{v}{f}=\frac{1540m/s}{1.40*10^{6}Hz}=1.1*10^{-3}m=1.1mm[/tex]
2)
The Young modulus can be computed by using the expression:
[tex]v=\sqrt{\frac{Y}{\rho}}\\\\Y=v^2\rho[/tex]
where Y is the Young modulus and p is the density of the material. Here, you have considered that the prostate gland can be taken as a vibrating membrane or string.
By replacing you obtain:
[tex]Y=(1540m/s)^2(1060kg/m^3)=2.513*10^9kg/ms^2[/tex]
hence, the Young modulus of the prostate glande is 2.513kg/ms^2
Answer:
1) 1.1 x [tex]10^{-3}[/tex] m
2)2.51 x[tex]10^{9}[/tex]Pa
Explanation:
Given:
Speed of sound in prostate 'V'= 1540m/s
frequency 'f' = 1.40 MHz = 1.40 x [tex]10^{6}[/tex]Hz
density of prostate'ρ' = 1060 kg/m3
1) As we know that the relationship of the speed of sound, its frequency, and wavelength is the same as for all waves
V= fλ
λ= V/f => 1540/1.40 x [tex]10^{6}[/tex]
λ= 1.1 x [tex]10^{-3}[/tex] m
Thus, the wavelength of the sonogram ultrasound is 1.1 x [tex]10^{-3}[/tex] m
2)The speed of sound in a solid the depends on the Young's modulus of the medium and the density
V=√Y/ρ.
V² = Y/ρ
Y= V² x ρ=> 1540² x 1060
Y= 2.51 x[tex]10^{9}[/tex]Pa
Thus, Young's modulus for the prostate gland is 2.51 x[tex]10^{9}[/tex]Pa