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10021667 No.10021667 [Reply] [Original] [archived.moe]

Someone explain pic related to me

>> No.10021672

Portable Network Graphics is a raster graphics file format that supports lossless data compression. PNG was created as an improved, non-patented replacement for Graphics Interchange Format (GIF), and is the most widely used lossless image compression format on the Internet.

>> No.10021676


>> No.10021678
File: 5 KB, 250x174, q5OL30E.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.10021686

Pressure isn't energy. Since work is energy over distance and pressure alone doesn't only that the object is compressed.

E. G. If you put a balloon in the left case it will be crushed. In the right case it will only shrink a tiny bit since the 10ml only has little potential energy and as soon as the balloon shrunk 10ml it will have 0 potential energy.

>> No.10021699

>Dive 2m in a pool
>can swim easily
>Dive 2m in the ocean
>get crushed by trillions of kilograms of water
Huh why did I never notice this before?

>> No.10021890


>> No.10021896

moving goalposts
water would be removed/added on both to keep the height at 10m

>> No.10021906

Utterly based.

>> No.10022005

The wall of the tank will bend and thats why they have the same pressure. You massive faggot who cannot think yourself. You always need to steal somebody's brain power. Fuck you

>> No.10022061
File: 41 KB, 572x719, 1498017657092.jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.10022065

it's not the same pressure, its the same hydraulic head
the same gradient of flow

>> No.10022113

>Never learned PV=nRT in AP Chemistry

>> No.10022131

>it's not the same pressure, its the same hydraulic head
>the same gradient of flow
Wait what? Doesn't pascal's experiment show that both have same pressure? I'm really confused. Please enlighten me.
"Compressed" water isn't an ideal gas, silly.

>> No.10022144

Water is modeled as an object (ignoring cohesion and other memes) that is homogeneous, obtains the shape of the volume were it is, and it's incompressible. So what happens? If water is incompressible, it literally means that no matter how hard you push it, it will maintain it's volume so water has this weird property as it's obviously not solid, but a volume of it has internal forces that maintain it's shape, so because it's static, the average forces inside the liquid will be 0, but at the boundary, it's the container the one that cancels the overall forces, so depending on the geometry of the boundary, you will get your pressure, because at the end, it's the same force applied to the water, which is then applied to the surface. You can now consider that the rock, as it is solid, it will not diffuse in water, and so water (obviously) cannot occupy the same space as the rock, or in other words, a new boundary appeared, but it's the same force acting on it, so it's the same pressure. What is the difference from this and considering crushing the rock with more and more massive solid objects? Two solid objects are really only in contact at a single point, but more importantly, a uniform force will then make the solid exert a force in only the same direction without breaking the model, as the solid has also internal forces that maintain it's shape independent of forces acting on them.

>> No.10022153

Considering I failed that exam I'm not surprised I fucked it up

>> No.10022154
File: 141 KB, 1224x1445, 1469420749373.jpg [View same] [iqdb] [saucenao] [google] [report]

>ideal gas law
>liquid water

>> No.10022168
File: 49 KB, 406x364, 1523619103206.jpg [View same] [iqdb] [saucenao] [google] [report]

>ideal gas
>water of all things

>> No.10022173

To clarify, a solid doesn't need a medium to define it's shape. The idea of incrompresibility on water is that the volume will not change, but you can have different geometries with the same volume. So liquid can only be static if there are other forces maintaining it contained. A solid already has this condition, so it has no need to distribute applied forces isotropically for it to keep being a solid so it's own weight isnt going to distribute as a liquid. A liquid like water can change it's shape constantly, so to it makes no sense to look it as a single massive object, and we can only consider how water interacts with itself at sections of the fluid.

>> No.10022186

I know I'm a brainlet pls no rub in

>> No.10022190
File: 35 KB, 609x280, h2o.oil.is.replaced.by.water.as.the.major.fuel.png [View same] [iqdb] [saucenao] [google] [report]

The pressures are equal because water always seeks it's level, so that the initial resistance of traveling through various diameter or volume of cylinder does not change the net force applied to the widened bottom segment.

>> No.10022229

>water always seeks it's level
what did he mean by that

>> No.10022266

Two columns of water of equal vertical length but different horizontal width do not differ in pressure after filling their containers and thus reaching equilibrium.

>> No.10022475

then you would be adding energy to the system...

it's written badly, but the essence is correct.

>> No.10022494

retard here

Is it because pressure is in some way based on the surface area of the object being compressed?

>> No.10022582

same pressure on stone, different sum of pressure (forces) on the containers which makes up for the difference in mass for the two objects

>> No.10022635

That's some ancient greek magic shit, ask Archimedes

>> No.10023148

10m is 10m, otherwise you're solving a different problem that no one asked

>> No.10023207

thanks for explaining in 1 sentence what the brainlets above can't in 20 paragraphs. Based.

>> No.10023219

Or more succinctly, more force on left but more area that it is spread over, and since p = f/a same equivalent pressure.

>> No.10023247

I now understand <3

>> No.10023316


>> No.10023406


>> No.10023459

Pressure is the perpendicular component of a force over a surface. 1000L of water can exercise the same pressure over an object as 10 mL of water, depending on the extension of the surface in wich the force (gravity force of water) is applied.
assuming T= 25 C: water density is near 1 Kg/L
If the pressure is the same on the two object: P1=P2
A1/A2=1000Kg/0.01Kg=10^5 (keep in mind)
if h=10m: A1=V1/h=1m^3/10m=0.1 m^2 and A2=V2/h=10^(-5)m^3/10m=10^(-6) m^2
As you can see A1/A2 is 10^5 confirmed
in your figure P1 = P2 because A1 = 0.1 m^2 and A2 = 10^(-6) m^2
My english is bad, I know, but I wish this helped you

>> No.10023468

false analogy

>> No.10023733

no, you are adding a component that wasn't asked. you inbred PIECE OF SHIT AAAAAA RAAAAGGEEEEE

>> No.10023756
File: 17 KB, 552x648, I fucking hate op.png [View same] [iqdb] [saucenao] [google] [report]

Pressure is a vector that points in all directions, right?

Well, in the left side, the red arrows are the downward and sideward forces caused by the weight of the water above, whereas the green arrows are the forces that push the water up originating from the resistance of water to compression. The purple arrows are the forces exercised by the glass wall.

As you can see, on the left side it's the water columns that push down (weight) and the water that pushes up to resist compression.

On the right, it's the GLASS CEILING that pushes down onto the water. and the force is equal to the force with which the water pushes down in the narrow tube, else the water would just flow up the narrow tube.

Result? same pressure.

this is a backwards and stupidly simplified explanation. But I hope it might just be stupid enough that OP will understand.

>> No.10023795

The cross-sectional area of the object (a dimension of pressure) is the same as the cross sectional area of the column of water pressing down on the object. What's the technical explanation for why this works other than "the areas cancel"?

>> No.10023800

question is described in prompt
10m all the time

>> No.10023805

Pressure is force/area. On the left you have force 1N per big area (lets say 1m^2, on the other side you have force 0.001 N per small area (which will be 0.001m^2) which results in same N/m^2 pressure of 2Pascal (1Pa).
On the left (if we assume the object takes the entire circle and is not just a little piece inside for simplicity) you will have the same pressure because the circumference and area is the same, so you will get a pressure of 1Pa on the rock.
On the left side, there is a big difference, since the container gets wider after 1m to 1m^2 from 0.001m^2.
The thing is, the force still remains 0.001N so the pressure will drop off and you will get only aa pressure of 0.001 Pa on the rock on the right.

>> No.10023823

Everyone else gave brainlet answers

>> No.10023873

No. It is you who didnt read properly. And there can be tons of reason why the pressure is same. He just write down one reason and that is the only one that you can understand.

>> No.10023928

>t. brainlet

>> No.10023977

Good kind of autism

>> No.10023980

Why do people find levers easy and intuitive, but water pressure is somehow sorcery?
If it helps, think of the water column as a simple machine amplifying force.

>> No.10023990

Calculate the force/area and you'll see the pressures are the same at the bottom of the 10m line. This is a troll thread so I'm not wasting more time on it.

>> No.10024451

because the intuition is that the hydrostatic pressure of water comes from its weight. But this is obviously false as OP shows. Shits conterinuitive IMO, but i am a bit of a retard.

>> No.10024638

San somoeone quickly check what prestige level the green sub rogue artifact weapon requires? Is it still obtainable?

>> No.10024758 [DELETED] 

same pressure

pressure is perpendicular force/area
the force in this case will be the force of gravity (weight) from the volume of water above the reference point

p = density
v = volume
g = acceleration of gravity
W = weight
A = cross-sectional area
h = height

W = pVg
V = Ah (assuming cross-section is constant relative to height)

W = pAhg
P = W/A
P = pAhg/Ah
P = pgh

You would do the exact the same equation for both and get the same result. Even though the 1000L would obviously exert a larger force, it's doing so over a larger area so the pressure is the same.

The key here is that density is constant and we're assuming uniform cross-sectional area.

We make this assumption when dealing with large bodies of water.

If you were dealing with an irregular shape, you could use a multitude of integation techniques to sum all the different force vectors over differential areas, or use gaussian surfaces/fluxes to make it easier. Doing it by hand would be tedious so we numerical (linear algebra) methods.

Welcome to engineering.

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