If you add air to a flat tire through a single small entry hole, why does the air spread out to fill the tire

2 answers:

7 0

The tire fills up just like any thing else that holds air when u pump a ball or tire up it fills all the way up cause it is a small confined space and after filling it with air the atoms of the air fill the tire up

mr Goodwill [35]2 years ago

5 0

Answer:

Explanation:

Air is made up of small particles which occupies all the place if filled in a close body.

A flat tire has a small nozzle to fill in air. As air pumped into it, it tends to spread itself into the complete tube.

This air gives shape to the rube making it perfectly circular. The particles of air have a greater inter molecular spacing and can acquire any shape.

This air given to the tire spread out in every part.

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Block A can slide relative to block B which, in turn, can slide on a perfectly smooth horizontal plane. If the initial velocity

Anna11 [10]

Answer:

the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

the distance that A slides relative to B is $S = \frac{v_o^2M}{2 \mu g (M+m)}$

Explanation:

From the diagram below;

acceleration of A relative to B is : $a = - ( \mu g + \frac{ \mu mg}{M})$

where

v = u + at

$0 = v_o + ( - \mu g - \frac{\mu m g }{M})t$

Making t the subject of the formula; we have:

$t = \frac{v_o M}{(\mu g )(M+m)}$

$v^2 = u^2 +2 as\\\\0^2 = v_o^2 - 2 (\mu g ) (\frac{M+m}{M})S\\\\$

$S = \frac{v_o^2M}{2 \mu g (M+m)}$ which implies the distance that A slides relative to B.

The final velocities of the two blocks can be determined as follows:

v = u + at

$v = v_o - \mu g \frac{v_oM}{\mu g (M+m)}\\\\v = \frac{\mu g mv_o}{m+M}\\\\$

$v = \frac{mv_o}{m+M}$

Thus, the final velocity of the two blocks is $v = \frac{mv_o}{m+M}$

4 0

2 years ago

The eagle drops the trout a height of 6.1 m the fish travels 7.9 m horizontaly before hitting the water what is the velocity of

Semenov [28]

The velocity of the eagle is 7.0 m/s

Explanation:

The motion of the fish is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction, where the horizontal velocity of the fish is equal to the horizontal velocity of the eagle

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion, using the following suvat equation:

$s=ut+\frac{1}{2}at^2$