osram

also theories of technical analysis could be discussed there, like ewt and such things

We must secure our future here on Earth before we consider a future in the stars.

We have no future on earth. And if we have, we create it from space.

Sorcerer:

Make your simulator and we'll see who's making the profit. You have until 2007 to build and market it (Virgin is up in 2007, right?).

Welcome, another swede =).

"Engineering physics", you mean "teknisk fysik"? Isn't that more like "Technical physics", or am I wrong?

Hi obduro, I'm also 18, soon 19, and daydream all the time =).

I don't drink alcohol, don't smoke or take drugs. I even prefer not to take medication unless it's nesecary, usualy waiting untill headaches or other pain passes by itself (which remainds me that on average I have 1-3 headaches a year).

I drink alcohol, smoke cigars (occasionally) but don't take drugs.

I avoid medicines, exactly like you. As a result I haven't been ill for more than 5 years except two colds every year (fall and spring).

Earlier I had headaches very often, but now I don't. I think the best way to prohibit them is to sleep much, and avoid stress. However, my headaches mostly were migrain.

Hope to see you more =).

BTW, where do you live?

The water starts boiling if it's low enough pressure in the vacuum-machine, doesn't it?

Yeah, I'm not convinced that the equation is completely correct.

Which one? The one pulkit posted, or the one I made out of pulkits?

Toutatis' 4-year trek around the Sun ranges from just inside the Earth's path out to the main asteroid belt between Mars and Jupiter. The asteroid visits us every four years.

What do you mean with 600 years?

Or you mean it's really close every 600 years?

728 dynes/cm is 0.728 N/m - the two values are off by three orders of magnitude. That does change the answer.

Ok, thank you.

I actually thought the SI-unit for density was [MATH]kg/dm^{3}[/MATH] but I looked that up and found it isn't, it's [MATH]kg/m^{3}[/MATH]. That changed the answer to the ~10 m you are talking about.

But I get ~10 m with every radius I'm testing with >0,0005, that can't be correct.

Well, for starters, you need to put all of your values in the same unit convention. You have MKS, CGS and "other" all mixed together. (who uses dm for anything?) I get something close to 10m.

Well, I confused the units for surface tension cause I used multiple pages to look up the answer. According to the page however I think it should be 728 N/m? That doesn't change the answer. So what am I doing wrong?

You must have changed something to the better? =)

All you ever needed to know. It's a little more complex than "lamps" - they use cathode ray tubes (i.e. a beam of electrons).

That's why I used ", cause I didn't know exactly what it was =). Thank you for telling.

But a beam of electrons... can we see that? Don't you mean photons? =/

The electrons hit the screen, coated with phosphor which glows.

Red, blue and green "lamps" sends out color. The monitor mixes those colors inte the one you see. If you have a small drop of water on the screen you can see those "lamps".

Thanks for the answer, I'm trying to test what you say.

[MATH]\sigma = 728[/MATH] for [MATH]20^{o} C[/MATH] water

http://www.engineeringtoolbox.com/24_597.html

Not sure with the decimals. Not sure of anything actually, don't like my results.

[MATH]\rho = 1 kg/dm^{3}[/MATH]

[MATH]\sigma = 728 dynes/cm[/MATH]

[MATH]r = 0,001 m[/MATH]

[MATH]p_{o} = 101300 Pa[/MATH]

[MATH]g = 9,82 m/s^{2}[/MATH]

Changed your formula like this:

[MATH]h \rho g + \frac{2 \sigma}{r} =p_{o} [/MATH]->[MATH]h = \frac{p_{o}}{\rho g}-\frac{2 \sigma}{r \rho g}[/MATH]correct?

With the above values h resulted -274,949 m, doesn't seem realistic =/.

What am I doing wrong?

I live in Sweden and will soon be 19 years =).

Graduated from college this spring (school system in Sweden is different so college is not exactly the same as in the US, I've been to school for 12 years however).

I like most types of science, except biology. Think I like Classical physics the most...

Someone that knows how to calculate the surface tension?

exactly, maybe my description of the problem was bad...

It depends very much on how big the hole is, and as I said I'm not sure if the volume will affect it. But I guess it won't leak (think the hole must be round though).

However, if you make a hole in the bottom of the bottle too (the side that is up when the bottle is upside down =) ) it will leak, cause the pressure from above will push it through.

test for yourself, it won't

not if the hole is small enough.

think 5 mm is enough

As I guess everyone knows, when a full bottle with just a small opening is turned upside down it won't leak.

Not sure exactly why it's like that but I can guess.

Now to the question:

If the hole is to large, the water will leak out. What's the maximum diameter of the hole to keep the water in the bottle? Is that dependent of the volume of water or height of waterpillow?

Think you understand, just to make sure:

_____
|     |
|-----| <- height of water...
|     |
|     |
|     |
\    /
 \  /
 |-| <- not leaking

but wat?

But I guess I'm wrong, even if I don't understand why.

I don't get it... but...

A disc is bounded by a circle.

Yes, but if it's a circle only the circumference is affected by relativity. If it's a disk the whole plane within the circumference is affected.

That means it'll be misformed (if it's a disk).

But that's only for a disk and not a circle, or am I wrong?

Maybe it still applies?

The radius is just a distance between the center and the circumference. If the circumference is shortened, your measurement will show that the distance between the center and the circumference have shrinked (the radius).

Again: The radius is not an object and is thereby not affected by relativity.