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3 years ago

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

2 answers:

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
X-Ray technicians have to stand behind the shield. Here's the lengthy answer:
 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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Why are electromagnets used in metal scrap yards PLEASE HELP!!!!!!

Alinara [238K]

To help pick to up the metal and place it in a incinerator

4 0

3 years ago

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounce

Ierofanga [76]

Answer:

2.67 m

Explanation:

k = Spring constant = 1.5 N/m

g = Acceleration due to gravity = 9.81 m/s²

l = Unstretched length

Frequency of SHM motion is given by

$f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Frequency of pendulum is given by

$f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$

Given in the question

$f_p=\dfrac{1}{2}f_s$

$\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\\\Rightarrow \sqrt{\dfrac{g}{l}}=\dfrac{1}{2}\sqrt{\dfrac{k}{m}}\\\Rightarrow \dfrac{g}{l}=\dfrac{1}{4}\dfrac{k}{m}\\\Rightarrow l=\dfrac{4gm}{k}\\\Rightarrow l=\dfrac{4\times 9.81\times \dfrac{1}{9.81}}{1.5}\\\Rightarrow l=2.67\ m$

The unstretched length of the spring is 2.67 m

6 0

3 years ago

Not in book

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$