In order to construct our own environment in the glass jar known as bell jar system, which can be used to explore and consider our larger environment on Earths, for an instance. Here a glass jar that hinges on an airtight rubber basis i.e seals appropriately. At the top of the jar, a bung is connected to it which passed via a metal tube. It has an adjacent flexible tube that goes to a hand vacuum pump and the best hand-powered pump was made with a wine preserver.

When the pump extracts the air from the bell jar, the pressure inside the balloon naturally decreases. The balloon usually has a air pressure around it, which restricts its size, but when this air is extracted and the pressure around it decreases the gas in the balloon will expand and the balloon seems to be inflating. When you release the air back into the bell jar, it will once again compress back to its actual size.

8 0

3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

2 years ago