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3 years ago

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A patient has an ongoing history of cancer. She has a tumor in the abdominal region, and has been undergoing treatment for it. T

here may be other tumors and a potential blockage in the surrounding area that need to be investigated. The imaging technique that might provide the most information in this case is . Joe has ongoing issues with his throat and feels some sort of blockage or abnormality as he swallows. The doctor decides to use X-ray imaging to visualize Joe’s internal anatomy as he swallows to help determine the nature of the problem. will be used for this procedure.
answer: 1. a CT scan
2. Fluoroscopy

1 answer:

Vadim26 [7]3 years ago

6 0

Answer:

the answer is at the BOTTOM OF THEIR QUESTION

Explanation:

IT IS CORRECT BTW

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A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

An experiment is conducted such that an applied force is exerted on a 5kg object as it travels across a horizontal surface in wh

Katen [24]

If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.

The graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.

A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m

For more information on work done, visit

brainly.com/subject/physics

5 0

3 years ago

Read 2 more answers

The vertical height of an ocean wave from a crest to a trough is 10 meters. What is the amplitude of the ocean wave?

natima [27]

The amplitude of the wave would be 5 meters.

6 0

3 years ago

Read 2 more answers

9. A Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table. How long does it take to hit the floor if no one sneezes? Wh

Paha777 [63]

By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s

Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.

Height h = 1.0 m

As the ball rolls off the table, it will be fallen under gravity. Where

g = 9.8 m/ $s^{2}$

Initial vertical velocity $u_{y}$ = 0

Initial horizontal velocity $u_{x}$ = 5 m/s

Considering the vertical distance, the formula to use to calculate the time will be;

h = ut + 1/2g $t^{2}$

1 = 0 + 1/2 x 9.8 $t^{2}$

1 = 4.9 $t^{2}$

$t^{2}$ = 1/4.9

t = $\sqrt{0.204}$

t = 0.45 seconds

It takes 0.45 seconds to hit the floor if no one sneezes.

To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.

Vertical component

$V_{y}$ = $U_{y}$ + gt

$V_{y}$ = 0 + 9.8(0.45)

$V_{y}$ = 4.41 m/s

Horizontal component

$V_{x}$ = $u_{x}$ + at

but a = 0

$V_{x}$ = 5 m/s

Final velocity V = $\sqrt{5^{2} + 4.41^{2} }$

V = 6.67 m/s

Therefore, it will hit the floor at a velocity of 6.7 m/s

Learn more here: brainly.com/question/5063616

8 0

3 years ago

In a video game, a flying coconut moves at a constant velocity of 20 meters/second. The coconut hits an obstacle and moves in th

qwelly [4]

Change in velocity = Final velocity - Initial velocity

Initial velocity = 20 m/s
Final velocity = 10 m/s

Change in velocity = 10 - 20 = -10m/s

Correct option B

Hope This Helps You!

4 0

4 years ago