Given what we know, we can confirm that doubling the distance between you and a source of radiation decreases your exposure by 75%.
<h3>How is distance related to radiation exposure?</h3>
- As expected, increasing the distance from the source of the radiation will reduce its negative effects.
- Counter-intuitively however, doubling the distance does not reduce by half, but rather reduces its effects by 3/4th.
- This is due to the fact that the radiation effects from the source are inversely proportional to the square of the distance.
- This causes the changes to be far greater than expected.
Therefore, given that the radiation is proportional to the square of the distance, instead of being of a more direct relation, we can confirm that when doubling the distance between yourself and the source of the radiation, you can reduce its effects by 3/4 or 75%.
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<span>Presuming that the gas behaves like an ideal gas:
Volume in cubic meter V= 59.6x 0.001= 0.0596 mÂł
n=3.01 mol
Temperature in Kelvin T=44.9+273.15=318.05 K
R= 8.413 J/ mol K
The ideal gas law:
PxV= nxRxT
Substituting the give information from the question into the ideal gas law:
P= 3.01 mol x 8.413 J/mol k x 318.05 K / 0.0596= 133544.392 Pa
1351.3458 Pa / 101325 Pa/ atm = 1.3179806 atm</span>
Carbon-12 is described as "Stable". I just had this question on my homework :P Hope it helped!
The answer is Fuse, the Fuse has a low melting point
"Dispersion forces" is the one intermolecular force among the following choices given in the question that <span>explains why iodine (I2) is a solid at room temperature. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that the answer has helped you.</span>