All categories

3 years ago

In a game of pool why will the balls eventually stop after a collision

1 answer:

7 0

Everything we see happening on Earth has friction along with it.
For a pool ball, it's the work the ball does bending all the little cloth
fibers out of the way as it rolls along the table, and also the work it
does pushing air out of the way. These things rob a little bit of each
ball's kinetic energy, so that it eventually stops rolling.

Note: Gravity is not the answer.

You might be interested in

How many fire extinguishers must be carried onboard power boats, with installed fuel tanks, between 26 and 40 feet in length?

irga5000 [103]

It is either Two B I or one B II

6 0

3 years ago

Describe how cool and warm air move in the atmosphere.

ruslelena [56]

Answer:

i think cool air moves slow and hot air moves fast

8 0

3 years ago

Aluminum, carbon, and calcium are examples of what type of matter?

Vanyuwa [196]

Element would be it man

4 0

3 years ago

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the vertical spee

MissTica

Answer:

Explanation:

The nice thing about parabolic motion is that the object launched from a certain height will have the same velocity coming down when it reaches that height again, just in the opposite direction. For us, that means if the velocity of the ball right off the ground is 30 m/s, then right before it hits the ground again it will be -30 m/s (the negative just means that the direction is the opposite). Your choice is D.

4 0

3 years ago

Col. John Stapp led the U.S. Air Force Aero Medical Laboratory's research into the effects of higher accelerations. On Stapp's f

Lerok [7]

Answer:

A. $a=203.14\ \frac{m}{s^2}$

B.s=397.6 m

Explanation:

Given that

speed u= 284.4 m/s

time t = 1.4 s

here he want to reduce the velocity from 284.4 m/s to 0 m/s.

So the final speed v= 0 m/s

We know that

v= u + at

So now by putting the values

0 = 284.4 -a x 1.4 (here we take negative sign because this is the case of de acceleration)

$a=203.14\ \frac{m}{s^2}$

So the acceleration while stopping will be $a=203.14\ \frac{m}{s^2}$ .

Lets take distance travel before come top rest is s

We know that

$v^2=u^2-2as$

$0=284.4^2-2\times 203.14\times s$

s=397.6 m

So the distance travel while stopping is 397.6 m.

8 0

4 years ago

Read 2 more answers