What is the ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 125 km/h speed limit (about 78 mi/h), assuming everyone travels at the limit

An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let θ denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,

• the normal force with magnitude n

• the car's weight with magnitude w

and the net force points toward the center of the circle made by the turn, with centripetal acceleration

a = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²

Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,

∑ F (⟂) = N + W (⟂) = m a (⟂)

and

∑ F (//) = W (//) = m a (//)

Let the direction of N be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle θ with the banked curve, and W makes the same angle with the negative perpendicular axis, so that the equations above reduce to

N - m g cos(θ) = m a sin(θ)

and

m g sin(θ) = m a cos(θ)

The second equation is all we need at this point to find the ideal θ. The mass m cancels out, and we can solve for θ to get

tan(θ) = a/g ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615

→   θ ≈ 3.52°

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j