A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is low, the penny rides up and down without difficulty. If the frequency is steadily increased, there comes a point at which the penny leaves the surface. Part A At what point in the cycle does the penny first lose contact with the piston

in the downward movement of the movement when the constant is lost

Explanation:

When the coin is on the piston it has a relationship given by

       a = d²x / dt²

the piston position is

      x = A cos wt

      a = - A w² cos wt

the maximum acceleration is

     a = - A w²

When the piston raises the acceleration of gravity and that of the piston go in the same direction, when the piston descends they relate it is contrary to gravity, therefore when the frequency increases, the point where the acceleration of the piston is greater than gravity arrives and the coin loses contact.

The point where you lose contact is

      a = g

      g = A w²

In short, in the downward movement of the movement when the constant is lost

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

the total distance traveled by the car is

30km + 49km = 79 km .