If the atmosphere were 50% more dense, sunlight would be much redder then it is now. As the atmosphere increase in density, more and more of the blue light would be scattered away in all directions, making the light that reaches the ground very red. Think of the color of a deep red sunset, but this would be the color even at noon.
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The magnitude of the electric field at the third vertex of the triangle is determined as zero.
<h3>Electric field at the third vertex of the triangle </h3>
The electric field at the third vertex of the equilateral triangle due to the other charges placed on the first and second vertices is calculated as follows;
E = E(13) + E(23)
E = (kq₁)/r² + (kq₂)/r²
where;
- q1 is positive charge
- q2 is negative charge
E = (kq₁)/r² - (kq₂)/r²
E = 0
Thus, the magnitude of the electric field at the third vertex of the triangle is determined as zero.
Learn more about electric field here: brainly.com/question/14372859
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Originally there must been
1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start
% = 2.25 / 150 = 1.5 % of 235 U left
I need more to answer this
