1950 g This is the answer due to the kilograms of lead being distributed
Answer:
4 A
Explanation:
The relationship between current, voltage and resistance in a circuit is given by Ohm's law:

where
V is the voltage
R is the resistance
I is the current
The equation can also be rewritten as

from which we see that the current is inversely proportional to the resistance, R.
In this problem, the initial current is I = 8 A. Then the resistance is doubled:
R ' = 2R
So the new current is

so the current is halved.
Answer:
1. Luminosity
2.Apparent brightness
Explanation:
There are two factors on which brightness of star appear to be in the sky
The two factors are
1. Luminosity
2.Apparent brightness
1.Luminosity :It is defined as the total energy emitted by the object in a given time.Luminosity vary with the distance of observer from the star.Luminosity is a intrinsic property which depends on the fundamental chemical composition and structure of the material.Luminosity is depends on the size of star.Lager the star luminosity will be more.
2.Apparent brightness: It is defined as how bright a star appears from an observer on the earth and the amount of starlight reaching the earth.if the distance is large then the brightness decreases.When the distance of star from us small then the brightness of star increases.Distance is inversely proportional to brightness of the star.
V=IR
Potential Difference (v)= Current (A) * Resistance (Ω)
As V increases, R also increases.