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3 years ago

Which imaging technique uses X-rays to make cross-sectional images of the body?

Physics

2 answers:

Crank3 years ago

7 0

Answer: Computed tomography (CT) is an imaging tool that combines x-rays with computer technology to produce a more detailed, cross-sectional image of your body. A CT scan lets your doctor see the size, shape, and position of structures that are deep inside your body, such as organs, tissues, or tumors.

So, the answer must be a CT scan. I may be wrong.

PilotLPTM [1.2K]3 years ago

5 0

Answer:

B. CT SCAN

Explanation:

Edge 2021

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the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i

Naily [24]

Answer:

(a) $\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b) P = 0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

$P = \sigma AT^4$

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

$\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}$

(b)

Now, for 90% radiator blackbody at 2000 K:

$P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4$

7 0

3 years ago

The increase of kinetic energy at the square of the speed of your vehicle has a major influence on all motor vehicles in three p

Viktor [21]

This implies that stopping distance and impact force grow as a function of speed. The best ways to improve manoeuvrability and lessen crash severity are to drive at an appropriate pace and to slow down as soon as you spot dangers in front of you.

Keep in mind that stopping distance increases with speed; at 50 mph, it is four times longer than at 25 mph, and at 75 mph, the force of impact is nine times greater.

<h3>What is the impact of speed on kinetic energy ?</h3>

When your car expends or absorbs energy to speed up or slow down, you may feel a pull or a jolt, called impulse. Impulse increases as the energy or force increases, and increases as the duration of the force decreases. You'll feel a harder jolt if you speed up or slow down suddenly.

Consider: coming to a stop from 60 mph in ten seconds doesn't hurt you or your vehicle because the force of this event is spread out over a long time. But if you hit a wall and come to a stop in just half a second, you'll feel twenty times the impulse, causing severe damage.

Learn more about Kinetic energy here:

brainly.com/question/25959744

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5 0

2 years ago

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece

Brilliant_brown [7]

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

$v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m$

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

8 0

3 years ago

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitt

Tamiku [17]

Answer:

7 m .

Explanation:

For destructive interference

Path difference = odd multiple of λ /2

Wave length of sound from each of A and B.

= speed / frequency

λ = 334 / 172 = 2 m

λ/2 = 1 m

If I am 1 m away from B , the path difference will be

8 - 1 = 7 m which is odd multiple of 1 or λ /2

So path difference becomes odd multiple of λ /2.

This is the condition of destructive interference.

So one meter is the closest distance which I can remain at so that i can hear destructive interference.

8 0

4 years ago

Read 2 more answers

. If she

I am Lyosha [343]

Answer:

9 meters

Explanation:

Given:

Mass of Avi is, $m=40\ kg$

Spring constant is, $k=176,400\ N/m$

Compression in the spring is, $x=20\ cm=0.20\ m$

Let the maximum height reached be 'h' m.

Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.

So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.

Decrease in elastic potential energy is given as:

$EPE=\frac{1}{2}kx^2\\EPE=\frac{1}{2}\times 176400\times (0.20)^2\\EPE=88200\times 0.04=3528\ J$

Now, increase in gravitational potential energy is given as:

$GPE=mgh=40\times 9.8\times h=392h$

Now, increase in gravitational potential energy is equal decrease in elastic potential energy. Therefore,

$392h=3528\\\\h=\frac{3528}{392}\\\\h=9\ m$

Therefore, Avi will reach a maximum height of 9 meters.

6 0

3 years ago