Answer:

Explanation:
The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

is the material's resistance, L is the legth and A is the cross-sectional area.
For the first and second coils, we have:

For the third and fourth coils, we have:

The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.
- The Universe was 5 percent its current size when light left objects now at redshift of <u>19</u>.
Reasons:
The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

Where;
a₀ = The current size of the Universe
a = The size of the early Universe = 5% of a
Therefore;


0.05 + 0.05·z = 1

- The redshift is of the observed light is, z = <u>19</u>
Learn more here:
brainly.com/question/14459434
brainly.com/question/3654558
A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North.