Compressed air is used to fire a 60 g ball vertically upward from a 0.70-m-tall tube. The air exerts an upward force of 3.0 N on the ball as long as it is in the tube. Part A How high does the ball go above the top of the tube

2.87 m

Explanation:

Given:

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

Height of the tube (h) = 0.70 m

Upward force by compressed air on the ball (F) = 3.0 N

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

Final velocity of the ball at the end of tube (v)  = ?

Acceleration of the ball (a) = ?

Weight of the ball is equal to the product of mass and gravity. So,

Weight (W) =

Therefore, the net force acting on the ball is equal to the difference of upward force and downward force. So,

Net force = Force by air - Weight

Now, from Newton's second law, the net force is equal to the product of object's mass and acceleration. So,

Now, acceleration (a) = 40.2 m/s²

Using equation of motion, we have:

Now, let the maximum height reached be 'H'.

So, applying energy conservation from the top of pipe to the maximum height.

Decrease in kinetic energy = Increase in potential energy.

Plug in the given values and solve for 'H'. This gives,

Therefore, the ball reaches a height of 2.87 m above the top of the tube.

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2