Answer:
μk = (Vf - Vc)/(T×g)
Explanation:
Given
Vi = initial velocity of the car
Vf = final velocity of the car
T = Time of application of brakes
g = acceleration due to gravity (known constant)
Let the mass of the car be Mc
Assuming only kinetic frictional force acts on the car as the driver applies the brakes,
The n from Newtown's second law of motion.
Fk = Mc×a
Fk = μk×Mc×g
a = (Vf - Vc)/T
Equating both preceding equation.
μk×Mc×g = Mc × (Vf - Vc)/T
Mc cancels out.
μk = (Vf - Vc)/(T×g)
The current in the home is (AC) and the AC tends the flow near the surface of the cross section area the wire according to (skin effect) ,on the other hand , the current in a battery is (DC) and DC flow in all of the cross section area of the wire
<span>As it is descended from a vertical height h,
The lost Potential Energy = Mgh
The gained Kenetic Energy = (1/2)Mv^2; The rotational KE = (1/2)Jw^2
The angular speed w = speed/ Radius = v/R
So Rotational KE = (1/2)Jw^2 = (1/2)J(v/R)^2; J is moment of inertia
Now Mgh = (1/2)Mv^2 + (1/2)J(v/R)^2 => 2gh/v^2 = 1 + (J/MR^2)
As v = (5gh/4)^1/2, (J/MR^2) = 2gh/v^2 - 1 => (J/MR^2) = (8gh/5gh) - 1
so (J/MR^2) = 3/5 and therefore J = (3/5)MR^2.</span>
