Describe several uses of plastic, and explain why plastic is a good choice for these products

Kinetic energy is represented as

KE = (0.5) m v²

In each scenario, v = the velocity of the bottle set at  4 m/s

with m = 0.125 kg

KE = (0.5) m v² =  (0.5) (0.125) (4)² = 1 J

for m = 0.250 kg

KE = (0.5) m v² =  (0.5) (0.250) (4)² = 2 J

if m = 0.375 kg

KE = (0.5) m v² =  (0.5) (0.375) (4)² = 3 J

when m = 0.500 kg

KE = (0.5) m v² =  (0.5) (0.500) (4)² = 4 J

The answer is

-Small f and large D.

The explanation:

-A car jack acts as a machine, defined as an apparatus that aids individuals in exerting force more easily.

-Hence, by applying a small force to the jack, the height at which the car is elevated increases.

Machines are essential for people to amplify their strength; without them, lifting a car would be impossible.

Employing leverage or hydraulic principles, machines enhance your exerted force.

Utilizing a greater lever allows for extensive movement with minimal force, resulting in the opposite side moving shorter distances with an increased force.

Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) =

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

According to Newton's second law, net force equals the mass multiplied by acceleration.

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

Substituting the values, we solve for 'H', yielding:

Hence, the ball ascends to a height of 2.87 m above the top of the tube.