What happens in terms of energy when a moving car hits a parked car, causing the parked car to move?

Physics

2 answers:

Helen [10]3 years ago

8 0

the answer to this question is c

meriva3 years ago

5 0

This is an example Newton's Third Law. All the kinectic energy from the moving car transferred the potential energy of the parked car. This potential is not much since the brakes are on (hopefully) and it's not in a non-moving position.

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A box weighing 43.2 N is pulled horizontally until it slides uniformly lat a constant

GREYUIT [131]

Your diagram should include four forces:

• the box's weight, pointing down (magnitude w = 43.2 N)

• the normal force, pointing up (mag. n)

• the applied force, pointing the direction in which the box is sliding (mag. p = 6.30 N, with p for "pull")

• the frictional force, pointing oppoiste the applied force (mag. f )

The box is moving at a constant speed, so it is inequilibrium and the net forces in both the vertical and horizontal directions sum to 0. By Newton's second law, we have

n + (-w) = 0

and

p + (-f ) = 0

So then the forces have magnitudes

w = 43.2 N

n = w = 43.2 N

p = 6.30 N

f = p = 6.30 N

5 0

3 years ago

Three balls of equal mass are fired simultaneously with equal speeds from the same height h above the ground. Ball 1 is fired st

pantera1 [17]

Answer:

d) v1 = v2 = v3

Explanation:

This can be answered using conservation of energy. We calculate the mechanical energy E=K+U (sum of kinetic and gravitational potential energies) at the original and final points, and impose they are equal.

At the original point we have, for the three balls:

$E_i=K_i+U_i=\frac{mv_i^2}{2}+mgh_i=\frac{mv^2}{2}+mgh$

At the final point we have, for the three balls:

$E_f=K_f+U_f=\frac{mv_f^2}{2}+mgh_f=\frac{mv_f^2}{2}$

Since we have $E_i=E_f$ , and $E_i$ is the same for all balls, then $E_f$ is the same for all balls, which means that $v_f$ , the final velocity, is the same for all balls.