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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Answer:
By decreasing the surface area exposed to the air. Vertical body position instead of horizontal.
Explanation:
So the terminal velocity formula is as follows.
v = sqrt[(2*m*g)/(p*A*c)]
m = mass
g = 9.81 m/s^2
p = density of air
A = surface area of object
c = Drag coefficient
So the only thing you can change in mid-air is surface area.
... directly over the Tropic of Capricorn, about 23.5 degrees south of the equator. The date is around December 22 or 23.
Answer: 20.2 m/s
Explanation:
From the question above, we have the following data;
M1 = 800kg
M2 = 1200kg
V1 = 13m/s
V2 = 25m/s
U (common velocity) =?
M1V1 + M2V2 = (M1 + M2). U
(800*13) + (1200*25) = (800+1200) * U
10400 + 30000 = 2000u
40400 = 2000u
U = 40400 / 2000
U = 20.2 m/s