/sci/ - Science & Math

nah, pressure wont affect gravity. the same pressure would be above and below the object, so there is no extra force being exerted by the pressure, just the force of gravity, which is only dependent on the objects mass.

no

OP here.
thats what I was thinking but not being a science major I wanted to double check. thx

unless you suggest to compress air into liquid I think shit will fall at the same speed trough it

>>3442557
>not sure if CP...

>>3442557
Virgin here
whese should I put my penis?
I know the rough coordinates, but I don't know where EXACTLY it fits

Weigh more, no. Fall slower, yes. High pressure means denser air means greater air resistance.

Objects would be compressed, and air would be denser. So things would fall more slowly.

>>3442572
lol, if virgin fag start with the lower hole.

>>3442572

in the hole

>>3442569
>not sure if retard
No wait, I'm pretty sure.

>>3442580
I ask WHERE in that hole

>>3442586
All the way in, mofugga.

>>3442586
dude wat? you never even watched pr0nz?

>>3442589
>>3442588

I'm not asking wheter I should showe it into her asshole

I ask where in that slit is the RIGHT hole

>>3442594
>the RIGHT hole

Well, that all depends on whether it's right to stick it in her urethra.

The vag is at the base of the slit though, toward the anus.

>>3442606
oh thank you this is what I asked for

>nah, pressure wont affect gravity. the same pressure would be above and below the object, so there is no extra force being exerted by the pressure, just the force of gravity, which is only dependent on the objects mass.

then how does a blimp fly?

check n mate

Higher pressure = higher density = more buoyant force
This means in terms of weight on a scale, yes, it would weigh less. But not much unless you had some really dense atmosphere.

Also higher density = more drag, which would further reduce speed when falling, as others have said.

>>3442572
that's where you want to aim. if she's on her back, it's right at the bottom of that slit. if you want to stroke her clit, it's almost at the very top of that slit, just a little bit down. (pic related)

[i apologize for all the latex errors that will undoubtedly follow, but i'll try to correct most of them...]

>>3442544
yes, the pressure will affect the falling object somewhat.

usually when we write <span class="math">y(t) = x_0 + v_0 t + \frac {1}{2} a t^2[/spoiler], we are ignoring things like air resistance or air pressure.
while higher pressures won't affect gravity (like this anon said: >>3442552), there WILL be more force exerted on the object (unlike this anon said: >>3442552).
here's why:
let's say that the object is a cube, a meter long on each side. if the object were falling in air with a lower pressure <span class="math">P_{low}[/spoiler] than in air with a higher pressure <span class="math">P_{high}[/spoiler], the pressure difference at it's top and bottom will be different.
this is because the pressure increases linearly as the density of the fluid (air, in this case). specifically:
<span class="math">P(\rho, h) = \rho g h + P_0[/spoiler],
where <span class="math">h[/spoiler] is the distance below a set level where the pressure is <span class="math">P_0[/spoiler], and <span class="math">\rho[/spoiler] is the density of the fluid.

so let's find the pressure difference on top and on bottom of our object in the <span class="math">P_{low}[/spoiler] and <span class="math">P_{high}[/spoiler] situations. remember that the air has a higher density in the <span class="math">P_{high}[/spoiler] situation.

<span class="math">\underline{ \mathrm{for } P_{low} }[/spoiler]:
the pressure on the top of the object is something (we don't know, nor need to know), but let's call it:
<span class="math">P_{low,top}[/spoiler]

(...>>3443140 continued) (forgot pic last time. never mind, pic wont work)
for the pressure on the bottom of the object, it is the pressure on the top, plus however much the pressure has increased a meter down from there (since the object is a meter on each side):
<span class="math">P_{low,bot} = P_{low,top} + \rho_{low} g (1m)[/spoiler]

so then the pressure difference (net pressure) is:
<span class="math">P_{low, NET} = P_{low,bot} - P_{low,top} = P_{low,top} + \rho_{low} g (1m) - P_{low,top} = \rho_{low} g (1m)[/spoiler]

<span class="math">\underline{ \mathrm{for } P_{high} }[/spoiler]:
i won't walk through this again, but you should see that it's the exact same treatment, except that <span class="math">\pho_{low}[/spoiler] is now <span class="math">\rho_{high}[/spoiler].
and then you should see that:
<span class="math">P_{high, NET} = \rho_{high} g (1m) > \rho_{low} g (1m) = P_{low, NET}[/spoiler]

so, the pressure difference IS greater in the high pressure system. by how much?
<span class="math">\Delta P_{NET} = P_{high, NET} - P_{low, NET} = g (1m) \left ( \rho_{high} - \rho_{low} \right ) = g (1m) \Delta \rho[/spoiler]

if you want to see how much greater the force is, simply multiply by the cross-sectional area of the object:
<span class="math">\Delta F_{NET} = \Delta P_{NET} (1m^2) = \Delta \rho g (1m^3) [/spoiler]
so the net force on the object is greater in the high pressure system than in the low pressure system by <span class="math">\Delta \rho g (1m^3) [/spoiler]

>>3443140

it could high pressure but the pressure would still be constant P=P_0

(... continued >>3443152)

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you can see this pretty intuitively if you think about it in the extreme case.
think of an object falling in air; it falls relatively quickly. but now think about the same object falling in water; it falls much slower. why?
air is a fluid just like water, water just has a much higher density. so the pressure difference at the top and bottom (or net force) is much higher in water than in air.

actually, it is nice to note that this force difference <span class="math">\Delta F_{NET} = \Delta P_{NET} A_C[/spoiler] (where <span class="math">A_C[/spoiler] is cross-sectional area) is actually an expression for buoyancy! by that i mean that it IS the buoyant force. the typical definition of the buoyant force is that it is equal to "the weight of the fluid that has been displaced by the object".
mathematically, that's:
<span class="math">B = W_{fluid} = m_{fluid} g = \rho_{fluid} V_{fluid} g[/spoiler], where <span class="math">\rho_{fluid}[/spoiler] is the density of the fluid. and since the object displaces as much fluid as it's own volume:
<span class="math">B = \rho_{fluid} g V_{object}[/spoiler] and now, if the object has some regular shape (like a cube):
<span class="math">B = \rho_{fluid} g A_C a[/spoiler], where <span class="math">a[/spoiler] is the height of the object. and if it actually is a cube, we can make that into: <span class="math">B = \rho_{fluid} g a^3[/spoiler].

i do all this simplifying just to show that the buoyant force is simply the differences in pressures (well, forces really - or pressures times area). and you can easily see that is the same equation as <span class="math">P_{NET} A_C[/spoiler] using one of the <span class="math">P_{NET}[/spoiler]s from above.

(... continued >>3443159)

so it is no surprise that the pressure difference (and thus the net force) should be different for these two situations, since it represents the same thing as the buoyant force.
(okay now i'm done)

okay well, one more thing. so not only would the buoyant force be greater in a higher density situation (which slows the object down), but the effects of air resistance are also greater (which slows the object down even more).
at high speeds, the force due to drag, <span class="math">F_D[/spoiler] is: <span class="math">F_D = \frac {1} {2} \rho C_d A v^2[/spoiler],
and at low speeds: <span class="math">F_d = 6 \pi \eta r v[/spoiler], where <span class="math">\eta[/spoiler] is the viscosity of the fluid, which varies with fluid density, and <span class="math">r[/spoiler] is the effective "Stokes" radius.
i do this just to show that drag also depends on the density of the fluid.

so you could say yes to your answer right there, but i also wanted to show that the pressure difference (times area - or buoyancy) is greater as well.

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thanks for (probably not) reading! i hope at least one of you found this interesting.

>>3443154
hm, could you be more specific?
i hope you're not contesting <span class="math">P = \rho g h[/spoiler]
pressure does vary with distance.

>>3443168

I mean the OP didnt really specify the scenario, by high pressure environment I consider constant, high, pressure, unchanging with height, if he means a place like the atmosphere then of course it changes with height

>>3443159
It's not really fair to compare air and water, viscosity and even surface tension can have huge effects. Small and light objects can be suspended by surface tension alone in water.

>>3443177
the pressure will always vary with height. the only way you could not make it vary is if you somehow manually relive pressure the farther down you go.

you could have pressure <span class="math">almost[/spoiler] not vary with height if you kept it as such high pressures and in a small volume such that a difference of 1 meter or so doesn't change the pressure very much. (like a difference of <span class="math">10^5[/spoiler] atm and <span class="math">10^5 + 1[/spoiler] atm.

as long as there is gravity and the fluid has mass, the pressure will change with height. i can derive it for you if you want.

>>3442812
I made some basic assumptions (150% increase in air density, density of a person near water, etc) and came up with a buoyant force change of like .11 lbs (obviously doesnt include pressure).

>>3443199
Let me rephrase that: unless the pressure change brings about a phase transition, you can't simply substitute water for high pressure air.

>>3443199
surface tension would be an issue if we were going from air to water. but since we never consider a surface, it doesn't matter.

both are fluids, and both have some viscosity. i actually treat that in part 4 when i do drag.
but the buoyancy calculations are independent of drag.

>>3443140
way to really draw out an easy concept.

>>3443210
um... yes you can.
you can compare them pretty fairly. both are fluids with some density. you don't really need anything else.

i suppose all you have to do is consider that the air is in-compressible like water. but we never consider a change in volume so that's not an issue either.

i mean, it's a pretty standard comparison in fluid dynamics...

So this is a thread about high pressure enviroment density and sexual adivce?

>>3443239
It's standard when the problem is scale invariant. In that case you'd usually keep the Reynolds number (or some other nondimensional number) constant. If you simply substitute water for air this can make the difference between turbulent and laminar flow, which is pretty significant.