A supernova is a star that suddenly increases greatly in brightness because of a catastrophic explosion that ejects most of its mass.
Answer:
1) Current decreases; 2) Inverse proportionally; 3) 1[A]
Explanation:
1)
As we can see as the resistance increases the current decreases, if we take two points as an example, when the resistance is equal to 50 [ohms] the current is equal to 1[amp] and when the resistance is equal to 200 [ohms] the current tends to have a value below 0.5 [amp]. Thus demonstrating the decrease in current.
2)
Inverse proportionally, by definition we know that the law of ohm determines the voltage according to resistance and amperage. This is the voltage will be equal to the product of the voltage by the resistance.
![V=I*R\\V = voltage [volts]\\I = current[amp]\\R = resistance [ohms]](https://tex.z-dn.net/?f=V%3DI%2AR%5C%5CV%20%3D%20voltage%20%5Bvolts%5D%5C%5CI%20%3D%20current%5Bamp%5D%5C%5CR%20%3D%20resistance%20%5Bohms%5D)
where:

And whenever we have in a fractional number the denominator the variable we are interested in, we can say that this is inversely proportional to the value we are interested in determining. In this case, we can see from the two previous expressions that both the current and the resistance appear in the denominator, therefore they are inversely proportional to each other.
3)
If we place ourselves on the graph on the resistance axis, we see that at 50 [ohm] will correspond a current value equal to 1 [A].
Answer:
Sherpas do work that is much more meaningful than the work other climbers do.
Explanation:
Answer:
The new current in the straight wire is 4.98 A
Explanation:
Given;
initial magnetic force on the wire, F₁ = 0.017 N
initial current flowing on the straight wire, I₁ = 1.1 A
When the current in the wire is changed,
new magnetic force on the wire, F₂ = 0.077 N
the new current in the wire, I₂ = ?
Applying equation of magnetic force on conductor;
F₁ = I₁BLsinθ
F₂ = I₂BLsinθ
BLsinθ = F₁/I₁ = F₂/I₂
I₂ = (F₂I₁)/F₁
I₂ = (0.077 x 1.1) / 0.017
I₂ = 4.98 A
Therefore, the new current in the straight wire is 4.98 A