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3 years ago

Which of the following is not an application of Doppler technology?

2 answers:

8 0

The correct answer to the question above is The third Option: C; ultrasound imaging of the liver. The ultrasound imaging of the liver is definitely not an application of Doppler technology.

Hope this helps! :)

Reil [10]3 years ago

4 0

Answer:

ultrasound imaging of the liver.

Explanation:

Liver ultrasound is not an application of the Doppler effect. The use of ultrasound is more common in medicine, for example, during pregnancy.

Another application in medicine is with the movement of blood. It is a flowmeter which allows the blood flow to be discovered through the Doppler effect, since a sound wave affects the red blood cells that are in motion inside the vessel and collect the echo signal.

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An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass

vladimir2022 [97]

The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N
the slow masses that must be quicker are the pulley, ring, and the rolling sphere.
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

6 0

2 years ago

An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e

VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

$r = \frac{8.26}{2} = 4.13m$

And the distance 'd' is

$d = lsin\theta$

$d = 1.14 sin 16.2\°$

$d = 0.318m$

Through the free-body diagram the tension components are given by

$Tcos\theta = mg$

$Tsin\theta = \frac{mv^2}{R}$

Here we can watch that,

$R = r+d$

Dividing both expression we have that,

$tan\theta = \frac{v^2}{Rg}$

Replacing the values,

$tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}$

$v = 4.83371m/s$

PART B) Using the vertical component we can find the tension,

$Tcos\theta = mg$

$T = \frac{mg}{cos\theta}$

$T = \frac{(10+26.2)(9.8)}{cos(16.2)}$

$T = 369.42N$

6 0

3 years ago

Which statement is TRUE concerning the particle arrangement of a solid

Anni [7]

It's packed together

8 0

3 years ago

Read 2 more answers

g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,

Stolb23 [73]

Answer:

$523269.9\ \text{N/m}$

Explanation:

q = Charge

r = Distance

$q_1=25\ \text{C}$

$r_1=3000\ \text{m}$

$q_2=40\ \text{C}$

$r_2=850\ \text{m}$

The electric field is given by

$E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}$

The electric field at the aircraft is $523269.9\ \text{N/m}$

4 0

3 years ago

1. Synthesize Information You push your

RSB [31]

Answer:separate

Explanation:

6 0

2 years ago