Liquid Water does not change it’s density much with pressure, but yes.
If you take a water balloon filled normally at the surface of the ocean and take it down, like 100 meters would it deflate due to the higher water pressure? What if you popped it?
Submitted 3 weeks ago by Crackhappy@lemmy.world to [deleted]
Comments
Zwuzelmaus@feddit.org 3 weeks ago
JakenVeina@midwest.social 3 weeks ago
Water doesn’t compress (for all intents and purposes), so no. I.E. the water pressure of the ocean would indeed push inward against the balloon much more strongly, but the water inside the balloon would push back by the same amount.
Nothing special would happen if you pop the balloon, the rubber would just rapidly contract, basically the same as it would above the surface.
plant@lemmy.dbzer0.com 3 weeks ago
Water is incompressible.
M137@lemmy.today 3 weeks ago
Not it’s not, common myth/misunderstanding.
From Wikipedia (couldn’t find a way to link directly to that part of the page): ImageBassTurd@lemmy.world 3 weeks ago
Ok, so for all intents and purposes, water is incompressible.
Scrogu@lemmy.zip 3 weeks ago
Virtually incompressable
FuglyDuck@lemmy.world 3 weeks ago
Depends on how much air is in there with the water.
Water itself is incompressible so the balloon won’t shrink that much, but there’s some air in there so it would shrink a bit.
The balloon’s popping action comes from being filled and expanded. It’s an elastic, so if something breaks it, it snaps back.
The only way that wouldn’t happen is if there was mostly air in there, and you went to the appropriate depth where that air was compressed down to its original volume.
someguy3@lemmy.world 3 weeks ago
At 100 m the compression would essentially be nil.
If you take it deep enough for there to be measurable compression and then pop it, nothing will happen because it’s already compressed.
nooneescapesthelaw@mander.xyz 3 weeks ago
Let’s first state our assumptions:
Very thin membrane, the balloon skin cannot be compressed more than it already is
Zero trapped air
Isothermal conditions, water temperature at the bottom of the ocean is the same as the top, that way we don’t need to deal with thermal expansion
Balloon is a perfect sphere
Calculations:
Pressure on the balloon is densitygravitydepth ~= 100010100 = 1 Mpa or 10 atmospheres
The bulk modulus of water is 2.2 Gpa so (change in volume)/(initial volume) = (change in pressure)/(bulk modulus) = 0.0046% decrease in volume
0.0046% decrease in volume is like a 0.015% decrease in radius
magnetichuman@fedia.io 3 weeks ago
is there a /m/theydidthemath ?
_NetNomad@fedia.io 3 weeks ago
/m/theydidthemonstermath