Consider a squeeze bulb. When we force some air out of the bulb and seal the mouth with hand, the bulb remains compressed. How do we explain it in terms of pressure?
Consider a Cartesian driver. We put the eyedropper inside the bottle fully filled with water and squeeze the bottle, then the water gets inside the eyedropper. Is it because the air gets more compressed when we squeeze the bottle, so it becomes dense enough to push water inside the eyedropper? But it doesn't quite make sense to me.
How does a siphone tube like the following work? What force really pushes the water up in A? Does the exit speed depend on the height of tube entrance at A?
Air approximately follows the ideal-gas-law: $$pV=nRT$$
- In your case when squeezing, temperature $T$ is constant. Pressure $p$ doesn't change since you let it equalize with the surroundings. But you reduce volume $V$ and also the amount of air molecules $n$ inside. The gas constant $R$ is also a constant.
- When letting go, the volume $V$ wants to return to original size due to elastic forces in the material. But you are preventing molecules from moving back in, so you are forcing $n$ to stay constant. Since both $n$, $R$ and $T$ stay constant, we have to reduce $p$ when increasing $V$ according to the ideal-gas-law. Quickly the $p$ decreases (a "suck" from the inside) just enough to counteract and balance the elastic forces trying to expand the volume. The motion stops and the bulb stays compressed.
The Cartesian Driver experiment is explained here:
Squeezing the bottle causes the diver to sink because the increased pressure forces water up into the diver, compressing the air at the top of the eyedropper. This increases the mass, and density, of the diver causing it to sink. Releasing the squeeze decreases the pressure on the air at the top of the eyedropper, and the water is forced back out of the diver.
Yes, air inside the eyedropper does get compressed, but that's not the important point. The important point is that more water enters the eyedropper. This means that there is more mass $m$ inside the eyedropper but still the same total volume $V$. Therefore the density $\rho$ rises: $$\rho=\frac mV$$ Something less dense will float in something more dense. The density of the outside water doesn't change significantly (there is no air to compress, and water compresses very, very little - which is why it enters the eyedropper in the first place, since it has nowhere else to go). So if the density of eyedropper increases enough to become larger than that of the outside water, it sinks.
- The siphone tube works by using pressures to "suck up" more liquid.
When liquid moves ahead inside the tube, the low pressure behind it sucks up more liquid.
But it doesn't work like you have described it: Liquid won't go upwards from the lower B to the higher A cup - it will only go downwards from the higher to the lower. The trick is not moving liquid up hill, but it does move liquid over obstacles on the way towards a lower point.
Have a look at this film.