Suppose that the ultrasound source placed on the mother's abdomen produces sound at a frequency 2 MHz (a megahertz is 10^610 ​6 ​​ Hz). Sound travels through tissue at roughly the same speed as in water (v\approx 1500v≈1500m/s). Find the maximum change in frequency between the sound that is emitted by the device and the sound that is observed at the wall of the baby's heart. Treat the heart wall as a moving observer. Hint: you will need to use your answer from part (a). Give your answer as a positive number in Hz.

the maximum frequency observed is 2.0044 10⁶ Hz

Explanation:

This is a Doppler effect exercise. Where the emitter is still and the receiver is mobile, therefore the expression that describes the process is

          f ’=

the + sign is used when the observer approaches the source

typical speeds of a baby's heart stop are around 200 m / min

let's reduce to SI units

        v₀ = 200 m / min (1 min / 60 s) = 3.33 m / s

let's calculate

         f ’= 2 10⁶ ()  

         f ’= 2.0044 10⁶ Hz

         f ’= 1,9956 10⁶ Hz

therefore the maximum frequency observed is 2.0044 10⁶ Hz

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

prolly a beacuse the momentuem the car has increases while going down hill