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Iteru [2.4K]
4 years ago
5

Which are methods of reducing exposure to ionizing radiation? Check all that apply.

Physics
2 answers:
gulaghasi [49]4 years ago
8 0
<h2>Answer:Answer: using a lead apron during medical X-rays, using a short burst of X-rays to take an image, taking an image of the smallest possible area</h2>

Ionizing radiation is a type of energy released by atoms in the form of electromagnetic radiation (gamma rays or X-rays) or particles (alpha and beta particles or neutrons).

This is a product of the spontaneous disintegration of atoms, where the surplus energy emitted is a form of ionizing radiation.

In this context, people are exposed daily to ionizing radiation, both of natural origin (for example <u>radon</u> emanating from rocks and soil or <u>cosmic rays </u>that go through the Earth) or human, as for example by the generation of <u>nuclear energy</u> or for the <u>medical use of radiation</u> for diagnostic or therapeutic purposes.

In the latter case, the most common artificial sources of ionizing radiation are medical devices, such as X-ray machines.

In this regard, it should be noted that X-rays are dangerous, because they contain enough energy to trigger mutations in organs and tissues and cause cancer (<u>the damage caused depends on the dose received or absorbed</u>). That is why people who undergo X-rays should be exposed to a brief burst of this radiation to capture the image on a photographic film, as well as avoid taking very large areas of the body and the staff that works with them must be protected wearing a lead apron and be placed behind a thick concrete wall as these materials prevent these rays from passing.

Dominik [7]4 years ago
5 0

Answer: (1,4,5)

Explanation:

1, 4, 5, are the answers on edg

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A car drives 40 miles on local roads at 20 mph, and 180 miles on the highway at 60 mph, what is the average speed of the entire
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The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0
nirvana33 [79]

(a) 7.25\cdot 10^{-8}m

Wien's displacement law is summarized by the equation

\lambda = \frac{b}{T}

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\lambda is the peak wavelength

b=2.898\cdot 10^{-3} m \cdot K is Wien's displacement constant

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T = 40,000 K

Therefore, its peak wavelength is

\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m

(b) Ultraviolet

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gamma rays  

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ultraviolet  380 nm - 1 nm

visible light  750 nm - 380 nm

infrared  25 \mu m - 750 nm

microwaves  1 mm - 25 \mu m

radio waves  > 1 mm

The peak wavelength of this star is

\lambda=7.25\cdot 10^{-8}m=72.5 nm

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(c) No

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The two telescopes, thanks to several instruments, are able to detect  much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

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3 years ago
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