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Pie
3 years ago
14

An ultrasound unit is being used to measure a patient’s heartbeat by combining the emitted 2.0 MHz signal with the sound waves r

eflected from the moving tissue of one point on the heart. The beat frequency between the two signals has a maximum value of 520 Hz. What is the maximum speed of the heart tissue?
Physics
1 answer:
Rus_ich [418]3 years ago
6 0

Answer: v= 0.2m/s

Explanation:

from Doppler effect

f = (2v/c) * fo

where:

f = the max beat frequency between two signals

fo = the emitted signal frequency from the organ.

v= The max speed of the organ tissue

c= 1540m/s (speed of Ultrasound wave in human tissue)

substituting the values, we have

520 Hz/ 2000000Hz = 2v/1540m/s

simplifying,

2v = (520 x 1540)/2000000

2v = 0.4004

v = 0.4/2 = 0.2 m/s

The max speed of the heart tissue is 0.2m/s

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Scientists measure the time between the arrival of an earthquake's __P____ and ___S____ waves to help determine the distance between the recording seismograph and the earthquake epicenter.

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P- (compressional) and S- (shear) waves produced in earthquakes travel at different speeds. P waves are faster than S waves and hence will be detected first by a seismograph after an earthquake. The further away a seismograph is from the epicenter of an earthquake,  the longer the time difference between the two (2) waves will be.

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A basketball player jumped straight up to grab a rebound. If she was in the air for 0.80 second, how high did she jump?
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4 0
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The wave length of violet light rounded to the nearest nanometer is a __ nm
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6 0
3 years ago
if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and
Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62

\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99

||v|| = 85

θ = 335°

\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03

\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92

Resultant

R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15

\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0
3 years ago
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