Explanation:
<u>Given</u><u>.</u> <u>Required</u><u> </u>
wo = 400N. MA=?
wi = 50N
<u>Solution</u>
MA = <u> </u><u>wo</u> = <u>400N</u> =8
wi. 50N
Explanation:
It is given that,
Let q₁ and q₂ are two small positively charged spheres such that,
.............(1)
Force of repulsion between the spheres, F = 1 N
Distance between spheres, d = 2 m
We need to find the charge on the sphere with the smaller charge. The force is given by :
............(2)
On solving the system of equation (1) and (2) using graph we get,
So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.
Answer:
The term “cold-blooded” implies that these animals are in a never-ending struggle to stay warm. That really isn't correct. A cold-blooded animal can warm up their blood by being in the sun for hours.
Let the rod be on the x-axes with endpoints -L/2 and L/2 and uniform charge density lambda (2.6nC/0.4m = 7.25 nC/m).
The point then lies on the y-axes at d = 0.03 m.
from symmetry, the field at that point will be ascending along the y-axes.
A charge element at position x on the rod has distance sqrt(x^2 + d^2) to the point.
Also, from the geometry, the component in the y-direction is d/sqrt(x^2+d^2) times the field strength.
All in all, the infinitesimal field strength from the charge between x and x+dx is:
dE = k lambda dx * 1/(x^2+d^2) * d/sqrt(x^2+d^2)
Therefore, upon integration,
E = k lambda d INTEGRAL{dx / (x^2 + d^2)^(3/2) } where x goes from -L/2 to L/2.
This gives:
E = k lambda L / (d sqrt((L/2)^2 + d^2) )
But lambda L = Q, the total charge on the rod, so
E = k Q / ( d * sqrt((L/2)^2 + d^2) )
To calculate force, use the formula force equals mass times acceleration, or F = m × a. Make sure that the mass measurement you're using is in kilograms and the acceleration is in meters over seconds squared. When you've solved the equation, the force will be measured in Newtons.
