A diagnostic sonogram produces a picture of internal organs by passing ultrasound through the tissue. In one application, it is used to fid the size, location, and shape of the prostate in preparation for surgery or other treatment. The speed of sound in the prostate is 1540 m/s, and a diagnostic sonogram uses ultrasound of frequency 1.40 MHz. The density of the prostate is 1060 kg/m3.1) What is the wavelength of the sonogram ultrasound? 2)What is Youngâs modulus for the prostate gland?

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 )

2)

The Young modulus can be computed by using the expression:

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:

hence, the Young modulus of the prostate glande is 2.513kg/ms^2

1) 1.1 x m

2)2.51 xPa

Explanation:

Given:

Speed of sound in prostate 'V'= 1540m/s

frequency 'f'  = 1.40 MHz = 1.40 x 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

λ= 1.1 x m

Thus, the wavelength of the sonogram ultrasound is 1.1 x 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 xPa

Thus, Young's modulus for the prostate gland is 2.51 xPa

the answer is b ) square root of 5

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west