Answer:
a) f ’’ = f₀
, b) Δf = 2 f₀ 
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
f ’’ = fo
f ’’ = fo 
leave the linear term
f ’’ = f₀ + f₀ 2
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ 
If i remeber correctly when dealing with real world cordinate systems as you rotate around clockwise you move in a positive direction. but all the examples i have done said north was 0 degrees, so i may be wrong
Explanation:
In a longitudinal wave, the crest and trough of a transverse wave correspond respectively to the compression, and the rarefaction. A compression is when the particles in the medium through which the wave is traveling are closer together than in its natural state, that is, when their density is greatest.
Answer:
a) C = (5-4D)/3
Explanation:
Given the simultaneous equation
3C+4D = 5 .... 1
2C+5D = 2 .... 2
In order to isolate one of the variables, we will make one of the variables in any of the equation.
Using equation 1:
3C+4D = 5
Make C the subject of the formula:
Subtract 4D from both sides of the equation.
3C+4D-4D= 5-4D
3C = 5-4D
Divide both sides by 3:
3C/3 = (5-4D)/3
C = (5-4D)/3
Hence the expression that would result from the process is C = (5-4D)/3