Harder Physics Questions

[Mao]

Pressure at a certain point depends on the total downward force acting on it

Force exerted by atmosphere on water surface:

Force exerted by some mass of water occupying some volume:

Total force is hence


EDIT, it took some self-convincing that the first statement is true. I'm still a little unconvinced.

Gravity force at that point = 0 [mass is rho * A * dh ~ 0]
Normal reaction force is the same magnitude as the force downwards [it's in translational equilibrium], however, if you were a plate, you would only measure pressure on one side, hence calculating either of these forces would give the pressure

No idea how sideways pressure works,

[QuantumJG]

Correct QuantumJG! (at least I think it is)

There is another solution that doesn't involve calculus.

The pressure at the bottom is , and the pressure at the top is just

Since is a linear function of , the average pressure is .

Thus we have

That's neat!

[/0]

Nice derivation Mao, I'm not really sure how sideways pressure works either, I was hoping someone could tell me

[Cthulhu]

*facepalm*
When I used calculus I used the depth of the water as the radius of the pool instead of 10m.... thats what you get for doing calculus at 3am though -.-

[Cthulhu]

From Dresden Codak.
[img]http://dresdencodak.com/comics/2005-06-14-lil_werner.jpg[/img]

[/0]

LOL Cthulhu xD
Heisenberg is a legend!

[appianway]

Oh, and just in terms of sideways pressure, I think this is how you'd go about figuring it out - similar derivations are often used when looking at pressure differences and the speed of sound. This ignores gravity and the like, and it's probably only useful for considering ideal gases (not liquids!), seeing as I'm not accounting for how the particles with velocities on the z axis (defined as the "upwards" direction) undergo a change in momentum (this would require the surface of the water to exert a force downwards, and for the surface to be static).

Consider a particle moving at the average velocity of a substance v. When it hits a wall, a change in momentum occurs as it changes direction, and hence a force is exerted. The magnitude of this force = 2mv/t. However, as pressure is F(av)/t, you can take the time interval between hits, which should be 2x/v, where x is the length from one end of whatever it's in to the other. If you multiply this by the number of particles in the mixture, you should get the total force exerted, and if you then divide by the area, you'll find the pressure.

However, how can the sideways velocity be calculated? If the temperature is known, we can then find out the kinetic translational energy, as only the translational energy contributes to the temperature. Relationships between internal energy and degrees of freedom could then be used to find out the velocity of the water, and presuming the equipartition of velocity, the x direction of the velocity^2 = v(total)^2/3. Or something like that.

I don't think this even holds for liquids, because the intermolecular forces are too large and there's the whole problem of the downwards force at the top (it could be caused by surface tension, but I don't actually know anything at all about that, so I'm not going to make any assumptions). You could probably make a model taking these into account, but it seems like it'd end up being really fiddly...

Edit: I don't even know if gravity needs to be taken into account, because the whole pressure equation for downwards pressure takes F(net) = 0 (dp/dx = -rho * g or something...)

[Cthulhu]

I remember appianway mentioned thermodynamics earlier on and I found this question in a book(it's shortened a bit):

Calculate the temperature at which many hydrogen atoms will be in the first excited state (n=2).
( average thermal energy.)

[Cthulhu]

Bumpin it up.

[appianway]

Well, I presume you plug in the formula for the energies of the hydrogen atom with n=2 (I don't know it off the top of my head), and equate that to the internal energy of 3/2kT before solving for T. Although I could be wrong.

If anyone wants to have a look, IPhO 2009 has a nice, neat gravitational question. If I could do it, I'm sure you can too. I prefer the quantum question though

[Edmund]

This isn't that hard but I can't remember how to do it:

A piece of timber 2 metres long and of retangular cross section 30cm x 20 cm floats vertically in water of density 1000kg/m³ with 20cm protruding above the surface. Which of the following is closest to the average density of the timber?

800kg/m³, 850kg/m³, 900kg/m³, 950kg/m³

[appianway]

OK.

Fnet = 0, because it's not accelerating upwards. You can find out the volume of water displaced, and the buoyancy force is equal to the weight of this (density * volume * g). This force has to be of equal magnitude to the weight force of the log, and then the weight force is the density times the volume times g. You just solve for the density.

[Edmund]

So I did

Buoyancy force = 1000kg/m³ x (0.10*0.20*2) x 9.8

and

Weight force = density of log x (0.30*0.20*2.00) x 9.8

and make them equal to solve for density, is this correct?

[appianway]

Don't quite think so. The amount of water displaced is given by cross sectional area * (length - 0.2 m).

[Cthulhu]

Yes that is correct:


where h = the height of the block that is submerged (1.8m) and H is the total height of the block(2m) L and W are the length and width respectivley


I SURE HOPE THIS IS RIGHT.