Consider a hybrid car that has 200 k j of kinetic energy at a certain speed. the car's regenerative braking is 43 % efficient ( η = 0.43 ) at converting kinetic energy ( e k = 1 / 2 m v 2 ) to energy stored in a battery. when the car slows down to half of its original speed using the regenerative braking, what is the energy, δ e , added to the car's battery?

600 kJ

Solution:

As per the question:

Kinetic energy of the car, = 200 kJ

      (1)

Now,

The speed of the car reduces to half of its initial speed

If the initial speed is 'v' m/s, then the

Also,

KE varies in proportion to the square of the velocity of the car.

Thus

        (2)

Thus from eqn (1) and (2):

Therefore,

the answer is to check all of them! they all apply well at least i think so!

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.