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> would the water pressure at the bottom of the glass increase?

I think the water level in the rest of the glass is sinking, according to the amount of water you are removing/lifting.
The pressure at the bottom would decrease.

Of course you need energy to lift the straw with the water, but so far everything seems fine.

.
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paul Offline OP
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well yea but the height of the water in the glass would only drop by the amount of water that the straw displaced while it was submerged in the water in the glass.

which would not be much , when compared to the new height
of the water in the straw and glass.

the height has almost doubled , but the pressure has dropped at the bottom of the glass.

do you agree to this?


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I'm not sure smile

On the surface of the water the weight of the atmosphere is applying pressure. This pressure is also driving the water into the straw.

Inside the straw the elevated water has weight, and is trying to fall downwards.


The height of water in the straw marks a state of equilibrium: the force of the atmospheric pressure and the force of the weight of the elevated water + the pressure of the enclosed air at the top of the straw are equal.

The water level in the glass (apart from the straw) has dropped due to the now missing elevated water in the straw.
I still expect the pressure in the glass to drop.

I'm not entirely sure... smile

The pressure at the bottom of the glass will be: (weight of dropped water level + atmospheric pressure).
The pressure at the bottom of the glass underneath the straw will be: (weight of dropped water level + weight of the elevated water + reduced air pressure in the straw).

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Momos

the water in the lifted straw is not putting any pressure on the water in the glass.

it cant fall because of the vacuum at the top of the straw.

this same scenario occurs in larger pipes.

take a 2" pipe 8 ft long and put a cap on one end.

then submerge it in your swimming pool , then lift it straight up with the capped end pointing upwards.

the water will stay in the pipe until you completely pull the pipe out of the water.

ie...when the air can rise into the pipe.

if you hold your hand directly under the lifted pipe you will not feel any pressure from the water in the pipe.

untill air can rise up and allow the water to fall by removing the vacume in the pipe.

or use a 2 liter coke bottle , fill it with water then
turn it upside down in a coffee cup , as soon as the water in the cup fills up and touches the top of the coke bottle
the water will stop comming out.


just like a pet waterer , the pet drinks water , the water level falls , air moves up into the tank , water falls down again.

the water is held up by the vacume.



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The strange thing is: if there would be no pressure in the straw, nothing could float up inside it. Buoyance is nothing more then the resulting force of pressure differences.

AT the top of the water in the straw we most likely won't have vacuum, just low pressure. And exactly speaking the vacuum doesn't hold the water up (e.g. is not applying some upward force). It is the air pressure of the surrounding atmosphere pressing the water into the straw, because the low pressure in the straw can't compensate.

Thats why the water level in the straw rises, as long as the weight of the water column equals the pressure difference between the surrounding air pressure and the lower pressure at the top of the straw.

At least that's what I guess smile

This leads to the following conclusion:
If you have a perfect vacuum in the straw (no air pressure) there is a maximum height to wich we can lift the water along with the straw. This height is determined by the air pressure.

Since air pressure is 1 bar = 100,000 pa = 100.000 N/m.
A "straw" can hold at most a ~10m column of water, after that lifting the straw wont lift the water anymore.

Actually, for the same reason this is also the maximum lifting height of any liquid achievable by a suction pump.

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I think Paul has it spot on.

Yea things would still float in the straw. Because there's still a pressure gradient from 1atm where it enters the glass, to >0 at the top of the straw. The pressure gradient is the same as in the glass or under the sea, so bouyancy would be identical.


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I have no doubt things would still float in the straw.
So far I'm agreeing with both of you.

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paul Offline OP
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the pressure inside the straw is not the reason that something would float up inside the straw.

it is the density differential that would cause a object to float up inside the straw.

the water in the straw has no pressure in any direction.

at the top of the straw the plug is supporting all the weight of the air and water in the straw.

the weight of the water in the straw is causing a vacume in the air between the plug and the water.

the water in the straw is trying to fall , but cannot because water will not stretch , so its just hanging
there exerting no pressures at all.

picture a board that is nailed to the ceiling.

the board would fall if it were not supported by the nails.

the nails in the straw example is the vacuum in the straw.
the plug is the ceiling.
the water is the board.


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Vacuum = nothing. Nothing cant hold anything.
Try holding a board by NOT attaching it to the wall wink

And water is an incompressible fluid, there is virtually no difference in the density of the liquid.
Anyway: a minimal density difference in the water would be THE RESULT of pressure compressing the water.
If you were right the water in the straw would have no density gradient. Nothing could float there.

The water is hold by the air pressure, hence the maximum lifting height of suction pumps.

And buoyancy is the result of pressure differences between the top and the bottom side of the object.
This picture might give you an idea what I mean:
http://de.wikipedia.org/w/index.php?title=Datei:Auftrieb_Archimedes_1.svg&filetimestamp=20080324085043

red dot at the top: pressure of the liquid at top, force downward.
red dots at the sides: equal pressures, canceling each other out.
red dot at the bottom: pressure of all the liquid above it, this pressure is more then on the upside of the object.
Resulting in an upward force.

There has to be no density difference whatsoever.

Ah, found it: http://en.wikipedia.org/wiki/Suction
"Suction is the flow of a fluid into a partial vacuum, or region of low pressure. The pressure gradient between this region and the ambient pressure will propel matter toward the low pressure area. Suction is popularly thought of as an attractive effect, which is incorrect since vacuums do not innately attract matter. Dust being "sucked" into a vacuum cleaner is actually being pushed in by the higher pressure air on the outside of the cleaner.

The higher pressure of the surrounding fluid can push matter into a vacuum but a vacuum cannot attract matter."


This should be obvious. If vacuum had some magical attractive force, what would happen to the ISS? The realy BIG vacuum of outer space sucking in one direction, the tiny vacuum in direction of earth sucking with less force in the other direction....

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momos

Quote:
Vacuum = nothing. Nothing cant hold anything.


any pressure below 0 guage pressure is refered to as a vacuum.

http://en.wikipedia.org/wiki/Vacuum_gauge

there are low vacuums , medium vacuums , high vacuums
and ultra high vacuums and extremely high vacuums.

http://en.wikipedia.org/wiki/Vacuum

Quote:
And buoyancy is the result of pressure differences between the top and the bottom side of the object.


and the pressure is the result of density.
so density is the true reason for buoyancy.



http://en.wikipedia.org/wiki/Buoyancy

Quote:
This picture might give you an idea what I mean:
http://de.wikipedia.org/w/index.php?title=Datei:Auftrieb_Archimedes_1.svg&filetimestamp=20080324085043

red dot at the top: pressure of the liquid at top, force downward.
red dots at the sides: equal pressures, canceling each other out.
red dot at the bottom: pressure of all the liquid above it, this pressure is more then on the upside of the object.
Resulting in an upward force.


you are refering to a open system that is exposed to the atmospheres pressures , not a system that is exposed to a vacuum.

in a open system pressures are distributed in layers
and the only reason that water in a open system ever exerts pressure outward or upward is due to the weight of water
above it.

in a closed system pressure is distributed equally in all directions.

you continue with outer space which is also an open system
there is a difference between open and closed systems.

1) open systems are able to interact with atmospheric pressures

2) closed systems are not.

the straw system is a closed system and its bottom is the water pressure that is in a open system.

it should be clear to you that this straw in a glass system is a combination of both open and closed systems.

something you should try.

get a glass bottle fill it with water.
fill a pot with water , now turn the bottle upside down into the water in the pot.

now lift the bottle slowly out of the water observing the top of the bottle very closely.

the bottle can come completely above the surface of the
water in the pot , yet the water in the pot rises up to the bottle as you lift it.

still no water comes out of the bottle.

even though it is clear that there is not enough water pressure along the mouth of the bottle that could be supplied by the water pressures in the water in the pot
to support the weight of the water in the bottle.

so wheres the pressure that is holding the water in
the bottle?
remember in order to have water pressure you must also
have water at a elevation.
but theres no water higher than the mouth of the bottle.

so is it just atmospheric pressure that holds the water
in the bottle?

but if I continue to lift the bottle the water comes out.

so it cant be just atmospheric pressures holding the water in the bottle.

its the vacuum at the top of the bottle that holds the
water in the bottle , not the atmospheric pressure pressing down on the water in the pot.


I used a 1 gallon glass bottle when I tried this.

.


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Originally Posted By: paul

the bottle can come completely above the surface of the
water in the pot , yet the water in the pot rises up to the bottle as you lift it.
still no water comes out of the bottle.
even though it is clear that there is not enough water pressure along the mouth of the bottle that could be supplied by the water pressures in the water in the pot
to support the weight of the water in the bottle.
so wheres the pressure that is holding the water in
the bottle?


The pressure holding the water in the bottle is the athmospheric pressure. You can actually see it! If a high-pressure weather system coming along the water in the bottel will rise even more. You can use it as a barometer.

You can also calculate it: The weight of the elevated water in the bottle matches exactly the athmospheric pressure.


Originally Posted By: paul

so is it just atmospheric pressure that holds the water
in the bottle?


Yes.

Originally Posted By: paul

but if I continue to lift the bottle the water comes out.
so it cant be just atmospheric pressures holding the water in the bottle.


Sorry, but you are wrong.
The reason the water comes out when you rise the bottle above the surface is: water is a liquid. The atmosphere pressing against the water in your bottle will form bubbles. Each bubble displaces its own volume of water.

Try this: hold a cardboard over the opening of your bottle, just before rising above the surface.
Now the atmospheric pressure has some steady/inflexible board to press against. You will see, the atmosphere is pressing against the board with enough pressure to hold the water inside.

Originally Posted By: paul

its the vacuum at the top of the bottle that holds the
water in the bottle , not the atmospheric pressure pressing down on the water in the pot.


You are wrong. Please read again the section about the vacuum-cleaner and suction. Or ask any manufacturer of suction pumps. Or you could try it yourself:
You would need a tower of 11 m height, a hose of 11m length and a tub with water.
Even with a perfect vacuum at the top of the hose you can elevate water higher then the atmospheric pressure compensates (around 10.3 meters).




Lets try a thought experiment: Assuming you are correct and "nothing" (perfect vacuum) is somehow inducing a "suction force".

If you have 1 Liter of vacuum this should be some amount X of "suction force". 2 Liter of vacuum should induce twice as much suction force.

Lets consider a closed system: A box, completely sealed from surrounding pressures. Inside this box we have a vacuum. In this box is also a movable separator, dividing the box into two compartments. Like a piston?

If you were right about this "suction force" the piston would always tend to move into the middle. If the piston is somewhere else one of the two compartments will have more volume of vacuum this producing more suction.

Reality behaves differently: in reality the piston doesn't care on which side more vacuum exists.
Because vacuum (nothing) doesn't exert any force.

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Originally Posted By: paul

and the pressure is the result of density.
so density is the true reason for buoyancy.


Pressure is the result of a force applied. That's the very definition.

Force (in our case) is not the result of density.
You can have a stone (hight density) side by side to some piece of wood (low density) there won't by any force pressing the wood against the stone.

The force (in our case) is the result of mass and gravity.
In a Zero-G-Environment you won't have this force, so you won't have pressure and you won't have buoyancy.
That's independent of any density differences.

-----

http://theory.uwinnipeg.ca/physics/fluids/node10.html


But still there is a connection between density and buoyancy.
It just isn't "the true reason for buoyancy".


This is an applet simulation a submarine, showing the air-pressure, water-pressure and weight along with the resulting force.
http://www.didaktik.physik.uni-muenchen.de/materialien/inhalt_materialien/auftrieb/auftrieb.zip

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Momos

no , in our case it IS THE DENSITY.

it doesnt matter how you twist it around to fit your world.

the ONLY REASON THAT BUOYANCY OCCURS is the differences in



DENSITY

in our case were not in a zero g environment
so im not sure why you are refering to zero g

but since you mentioned zero g

if you pour out a gallon of water inside the ISS
a sphere of water will form.

if you push a ping pong ball into the sphere of water in the near zero g environment the pingpong ball will be pushed back out , because of the differences in the density of the water and the density of the ping pong ball.




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Kallog

you agree that it is the density differences that cause
buoyancy , so I suppose that we can continue our discussion and if momos wants to continue
thinking it is the pressures then that is his problem.

what do you think , do we have this covered already?





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I didnt know you had posted another reply momos so
I will respond to it.

Quote:
The weight of the elevated water in the bottle matches exactly the athmospheric pressure.


so all I would need to do to increase atmospheric pressure outside the bottle is to increase the water inside the bottle.

but what if I have 2000 lbs of water in a bottle?

are you saying that the atmospheric pressure outside the bottle would become 2000 lbs or are you saying that the 2000 lb of water looses its weight and becomes 14.7 lbs?

Quote:
Now the atmospheric pressure has some steady/inflexible board to press against. You will see, the atmosphere is pressing against the board with enough pressure to hold the water inside.


yes I tried this and then I poked a hole in the top of the bottle and the cardboard still held up all the weight of the water inside.

there is something magical about the atmospheric pressures.

then I woke up , but there was no water floating around because of the MASSIVE atmospheric pressures.

I went to get a cup of coffee , but stumbled over a block of atmospheric pressure along the way.


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Originally Posted By: paul

if you pour out a gallon of water inside the ISS
a sphere of water will form.
if you push a ping pong ball into the sphere of water in the near zero g environment the pingpong ball will be pushed back out , because of the differences in the density of the water and the density of the ping pong ball.


No it won't. No gravity = no weight = no pressure = no pressure gradient = no buoyancy.
This air bubble has less density then the surrounding water: still in zero g it stays submerged:
http://www.youtube.com/watch?v=ZyTwLAW-Z8c&feature=related

The density difference between the ping pong ball and the water doesn't "do" anything. Exactly like the density difference of a ping pong ball and a brick wall at its side doesn't exert any force.

Also there wouldn't be any mentionable density difference in the water (even with gravity). Water is basically incompressible. You can read this in any boom about physics.


You dont have to believe me, just take a take a look here (same source you got your "density" picture from:
http://en.wikipedia.org/wiki/Buoyancy#Forces_and_equilibrium
Or:
http://www.answers.com/topic/buoyancy
Or:
http://answers.yahoo.com/question/index?qid=20100401145652AAtLQew

Or if you don't like reading watch this:
http://video.answers.com/Q/learn_about_fluids_-_part_5_99162287 (actually these videos of kahn academy are quite good).


Originally Posted By: paul

so all I would need to do to increase atmospheric pressure outside the bottle is to increase the water inside the bottle.


Now you are playing stupid?
Of course not.
It means you can't elevate the water more then the air pressure allowes you to.
If you elevate the straw further then 10 meter the water would still stay at 10m.
You could lift up a straw of any length, kilometers if you like, the water column in the straw would still be at a height of 10 meters.


Originally Posted By: paul

are you saying that the atmospheric pressure outside the bottle would become 2000 lbs or are you saying that the 2000 lb of water looses its weight and becomes 14.7 lbs?


I'm saying: You can't support 2000 lbs of water in the bottle. At least not by using the "suction of a vacuum".
The pressure of the weight of the water in the straw will always match the surrounding air pressure.
If you lift the bottle/straw further, the water will stay where it is, you will just have a bigger empty place at the top of the straw.

Strictly speaking you can support 2000 lbs of water in the bottle: AS LONG as the height of the water column is less then 10.3m. You could just make your straw wider.
It's a question of pressure: Weight of the water / Area of the straw = downward pressure <= atmospheric pressure.

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momos

I was a little quick with my ping pong example.
I did not take into consideration surface tension in microgravity.

however even surface tension forms spheres in microgravity and even small spheres have gravity , pressures , and buoyancy.

you carried the discussion into outer space for some reason
and I made a speedy incorrect reply.

however if the water sphere were larger , or if the sphere were made of some other material such as tiny beads that would not have surface tension as water does , then the bead sphere would push the ping pong ball out if you tried to submerge it.

but only if the beads that surround the ping pong ball have a greater density than the ping pong ball.

even if you begin building a sphere starting with the ping pong ball then adding the tiny beads to the ping pong ball the mass that will form in the shape of a sphere in a microgravity
will eventually push the ping pong ball outward.

the problem with the water example is that the surface tension of the water attracts objects that come in contact with it ,
but the beads would only be attracted to each other by gravity

now in the bead and ping pong ball example there is gravity , pressure and buoyancy.

just like the earth has.

now you say that the reason that buoyancy occurs is only the pressure differential between the top and the bottom of an object.

and you say that Im wrong when I say the true reason for
buoyancy is only density.

the only reason that you have any pressure is because of gravity pulling a weight/density of a mass toward the earth.

without density all you have left is nothing.

no gravity no pressure no buoyancy.

I dont really think you believe that it is only pressures
that cause buoyancy , I think you are just refusing to admit
that the true reason that buoyancy occurs is density.

or maybe you truly do believe that.

if so , how would you propose that you could have any pressures without a object that has density acting against another object that has density?

as I said , I really dont think you believe that , I just think
you are holding out for some reason.

so why dont you show me an example where there is pressure without density and then I might believe you.


heres something to ponder while I wait.

take two identical sized spheres.

1 sphere has a weight of .0000001 lbs

the other sphere has a weight of 5000 lbs

they are both submerged and held at the same exact depth 10 feet below the waters surface level.

so the same pressure is being placed on the two spheres
on the top and on the bottom.

if you release the two spheres at the same exact time
what would happen?

I say that the 5000 lb sphere will sink because of its density.
even though there is a pressure differential between the top and bottom of the sphere.

and the .0000001 lb sphere will rise because of its density.
even though there is a pressure differential between the top and bottom of the sphere.

according to you they will both rise or they will both sink.
because they have the same exact pressure differential acting downwards and upwards on the two spheres.

and according to you density has nothing to do with buoyancy , it is ONLY the pressure differential between the top and bottom of a submerged object.

ie...post #34142
Originally Posted By: momos

Buoyance is nothing more then the resulting force of pressure differences.





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Haha wow you guys have really gone off on a tangent of miscommunication! I have a feeling you both know exactly how it works :P

Let me lay the smack down:

*** BUOYANCY IS CAUSED BY A PRESSURE GRADIENT APPLIED TO AN OBJECT ***

A pressure gradient is usually caused by density of the surrounding fluid and a gravitational field. On Earth, the pressure gradient is only a function of the fluid's density. So it's good enough to say the buoyancy force on an object is only a function of the surrounding fluid's density (and the volume of the object).

In the above I'm ignoring the self weight of the object, which of course needs to be added to convert buoyancy force into total, measurable force. Here we have a semantic issue. Paul, your two-spheres-underwater example assumes it's the total force, but I think it's more correct to ignore the object's weight (and density) when defining buoyancy force. Either way, it doesn't change reality when we apply whatever definitions we each use.



-----


In the case of a ball of water on the ISS, you raise an interesting point Paul. The water would have its own local gravity field, which would cause a pressure gradient inside it. That would cause boyancy of a ping-pong ball. Even tho this force would be amazingly tiny, what would stop it? The viscous friction of water approaches zero as speed goes to zero. So I'd think the ping-pong ball would actually move to the surface. But it could be extremely slow, and maybe other tiny forces like photons hitting it, or the microgravity of Earth would overwhelm buoyancy.

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Kallog

it doesnt matter how you word it.

when its time to turn the cards its density that wins.

you can say its the great purple rinoceros if you like.

you can call it what ever you choose , but density by any other name
or deffinition is still density.

you can call it weight if you like.

but in my world what causes buoyancy or the reason a object floats
or sinks in a fluid is the density dfferential.

and I dont care how far under water an object is , even if the waters pressure is
at 50,000 psi.

the surrounding water will have basically the same density per unit volume
as it has at the surface.

even if the water is inside a container that has a large vacuum thrown on it
where the water pressure at the bottom of the container is at -100 kpa

in this situation the pressure gradient you speak of is negligible or
non existant.

and the only two forces that apply in that type of situation are the force of gravity and the resultant force of buoyancy of an object due only to the density differential between the water and the object.

if I take a 10 ft tall pipe and attach a pressure guage and a vacuum guage at the bottom of the pipe and at the top of the pipe and seal it at the bottom and fill it with water and seal it at the top in 1 atm

then when I look at the vacuum guage at the bottom it reads zero
if I look at the pressure guage at the bottom it reads 4.33 psi.

if I add 1 lb of air pressure at the top

the vacuum guage at the bottom of the pipe still reads zero.
the pressure guage at the bottom of the pipe reads 5.33 psi

if I remove the 1 psi air pressure and throw a - 50 kPa vacuum on the air at the top

the vacuum guage at the bottom of the pipe reads -50 kPa
the pressure guage at the bottom of the pipe reads zero

but in each situation either in 1 atm , pressurized , or a vacuum
the only reason a object will have buoyancy is the differential between the
density of the water and the object.

and to try to continue our discussion if a HHO cell is at the bottom of the 10 ft pipe
more energy will be required to convert water into HHO at 4.33 psi guage pressure and even more
at 5.33 psi guage pressure

and less at -50 kPa guage pressure.

.


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I agree with kallog.
Maybe I expressed myself unclear: The measured buoyancy is the result of the (downward) force due to the weight of the object and the the resulting force of pressure differences.

And of course I don't deny the connection to the density of the liquid and the object. Since this density is the result of the volume (causing the pressure differences) and the mass in a g-field (causing the downward force).

But it's not the density difference as such causing the buoyancy! That's why I'm always referring to a zero-g environment.
Without gravity the density differences between liquid and object won't change. But there won't be any pressure difference in the liquid or downward force of the object anymore, resulting in a buoyancy force of zero. (I admit: I'm ignoring microgravity, thermal diffusion, surface tension, quantum mechanics and a lot of other aspects since they are not part of the principle in discussion).

Why am I always nagging about this?
Paul designed a device with an elevated water column, which is fine. I guess he will try to use some object (possibly hydrogen) to float along this column and sink again. That's ok.
When we are discussing the energy gained or lost during this process we will have to calculate the work necessary to achieve any change of density, movement and so on at different points.

That's why it is important for us tho have the same understanding which pressure and forces are applying to an object somewhere in the water column.

In my opinion the vacuum (or low-pressure air) at the top of the straw is not exerting any negative force (suction) at the water column below it. In case of a vacuum it just doesn't exert anything (how could it since "it" isn't there).
The water in the column has still its weight and would fall downwards. The air pressure on the open part of the system is not matched by the air pressure on top of the water column.
That's why this pressure is pressing the water into the straw, until the weight of the water in the straw balances the pressure difference between air pressure and the vacuum.

In the water column we have the usual pressure gradient (from 0 to 1 atmosphere) resulting in just the same buoyancy of any object as usual.

If Paul agrees to this explanation, I have to apologies since any misunderstanding could very well be on my part and a possible misuse of the word buoyancy smile


----

on a side note:
Since water is dielectric we could apply an electric field from the outside on the water column which exerts an upward force on any water molecule. Thus we could compensates the weight of the water (letting it really elevating it). The water column would have no internal pressure gradient, any object unaffected by the electric field would sink. Regardless of its density.

[Edit: found a video of levitating frogs and water in a magnetic field]

Last edited by Momos; 05/02/10 06:50 PM.
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