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:
132.64m/s
Explanation:
Averagevelocity= distance/time
From the question, the distance d = 825m
time t = 6.22 secs
Using the above formula
We have
average velocity = 825/6.22
= 132.64m/s
The mutual inductance of the two coils is 1.28*10^-3H.
To find the answer, we have to know about the mutual inductance.
<h3>How to find the
mutual inductance of the two coil?</h3>
- We have the expression for emf in terms of mutual inductance and rate of change of current as,
- From the question, it is clear that,
- Thus, the mutual inductance is,
Thus, we can conclude that, the mutual inductance of the two coils is 1.28*10^-3H.
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1. T<span>he fields of science and mathematics are so interdependent because:
</span><span>Scientists need to be able to take careful measurements.
</span><span>Scientists need to be able to make precise measurements.
2. </span>T<span>he following questions can use science as a method of inquiry:
</span><span>How can we protect crops from drought?
</span><span>Did dinosaurs exist?
P.S. </span>No probs man, I will answer the same question over and over again, as long as you are willing to give away so many points. (Would be a real shame if I'm wrong, tho, but let's hope for the best :))
Time taken by the boat to accelerate is 1 min 4 s.
<h3>How to calculate the time taken?</h3>
V - V = 2 a X ,
a = Vf2 - Vo2 / 2 X
= (26 m/s)2 -(13m/s)2 / 2(1250m)
= 507 m2/s2 / 2.5 m = 0.203 m/s2
Solve for time using the equation with acceleration known.
Vf = at + Vo ,
t = Vf - Vo / a = 26m/s -13m/s/ 0.203 m/s2
= 64 s
= 1 min 4 s
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