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babymother [125]
3 years ago
11

Why do X-Ray technicians stand behind a shield when giving X-Rays to patients?

Physics
2 answers:
scoray [572]3 years ago
7 0
Lol I just learned this, here's the short answer:
Large doses of X-Rays are known to cause cell damage and cancer, so
<span>X-Ray technicians have to stand behind the shield. Here's the lengthy answer:
</span> This is because the patient would only be exposed to the small dose of radiation once, which is hardly harmful. However, radiographers deal with this all day every day, which could mean that the small doses could add up. This is why they stand behind a lead screen; the small exposure to the X-Rays aren't harmful alone but frequent exposure could begin to damage your cells and lead to potential genetic mutation. If a patient is often in radiography (a long-stay patient for example), they begin to wear lead aprons whilst being X-rayed in order to prevent the risk of harm. Hope I answered correctly.



Mashcka [7]3 years ago
7 0

Answer:

because large doses of X-Rays are known to cause cell damage and cancers

Explanation:^^

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A dog walking to the right with a speed of 1.5 m s 1.5 s m ​ 1, point, 5, start fraction, start text, m, end text, divided by, s
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Answer:

b.) Length

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3 years ago
Which equation would you use to calculate how much work is done in pushing a rock to the edge of the cliff?
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d

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4 0
2 years ago
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
3 years ago
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