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Elena-2011 [213]
3 years ago
11

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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What creates an ionic bond?
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A volcano erupts spewing ash into the air and sending lava flowing down the side of the mountain. Looking at the image explain h
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A lateral eruptions or lateral blast is a volcanic eruption which is directed laterally from a volcano rather than upwards from the summit. Lateral eruptions are caused by the outward expansion of flanks due to rising magma. Breaking occurs at the flanks of volcanoes making it easier for magma to flow outward.

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3 years ago
Yolanda is studying two waves. The first wave has an amplitude of 2 m, and the second has an amplitude of 3 m. Which statement a
Anna71 [15]

Answer:

a) Not Accurate

b) Not Accurate

c) Accurate

d) Accurate

Explanation:

Part a

Not Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m

Part b

Not Accurate, because constructive interference would lead to minimum possible magnitude of > 2 m

Part c

Accurate, because destructive interference would lead to maximum possible magnitude of < 3 m by varying the phase difference between two waves she can achieve the desired results.

Part d  

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8 0
3 years ago
A 60-kg rollerblader rolls 10 m down a 30? incline. When she reaches the level floor at the bottom, she applies the brakes. The
VARVARA [1.3K]

Answer:

s = 20 m

Explanation:

given,

mass of the roller blader = 60 Kg

length = 10 m

inclines at = 30°

coefficient of friction = 0.25

using conservation of energy

\dfrac{1}{2}mu^2 = m g d sin \theta

u^2 = 2 g d sin30^0

u= \sqrt{2\times 9.8 \times 10 sin30^0}

u = 9.89 m/s

Using second law of motion  

ma =μ mg

a = μ g

a = 0.25 x 9.8

a = 2.45 m/s²

Using third equation of motion ,  

v² - u² = 2 a s

0² - 9.89² = 2 x 2.45 x s

s = 20 m

the distance moved before stopping is 20 m

3 0
3 years ago
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