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June 16

centripetal acceleration

http://www.feynmanlectures.info/

Just wondering, why would the force mv^2/R be directed away from the apex? Wouldn't there have to be a normal force N equaling mv^2/R directed towards the apex ie towards the centre of the circle? —Preceding unsigned comment added by 74.15.138.134 (talk) 00:43, 16 June 2009 (UTC)[reply]

You are right. The centripetal force is directed towards the center. I doesn't have to be a normal force. Any force towards the center will do. Dauto (talk) 01:14, 16 June 2009 (UTC)[reply]

Then why does the solution treat the force as pointing away from the center? —Preceding unsigned comment added by 74.15.138.134 (talk) 02:08, 16 June 2009 (UTC)[reply]

Note: I just realized that the link doesn't take people to where I want them to go. Please, go to exercises and go to "ball and cone". —Preceding unsigned comment added by 74.15.138.134 (talk) 02:10, 16 June 2009 (UTC)[reply]

I couldn't find any "ball and cone" problem in the exercise. It would be nice if you could give a direct link. Rkr1991 (talk) 04:44, 16 June 2009 (UTC)[reply]

I think (I haven't seen the question yet) the solution would have been in the reference frame of the revolving object, which causes a centrifugal force mv²/R away from the centre, balanced by the normal force, towards the center. A more complete explanation can be given after I see the question. Rkr1991 (talk) 04:48, 16 June 2009 (UTC)[reply]

woops, i meant particle in cone, it's the first pdf...and a direct link won't work for some reason. —Preceding unsigned comment added by 74.15.138.134 (talk) 04:47, 16 June 2009 (UTC)[reply]

Yes, I think I'm right. The other parts of the solution are self explanatoryRkr1991 (talk) 04:56, 16 June 2009 (UTC)[reply]

here's the link Rkr1991 (talk) 04:58, 16 June 2009 (UTC)[reply]

Also, don't forget to sign your posts, by typing four '~' marks at the end. Rkr1991 (talk) 05:07, 16 June 2009 (UTC)[reply]

Hmmm...well, if you treat it from the reference frame of someone else, it doesn't seem to work. How would you do this from an inertia reference frame? 74.15.138.134 (talk) 05:38, 16 June 2009 (UTC)[reply]

Well there isn't much of a difference. If you understand the above solution, that shouldn't be a problem. All you do is forget about that centrifugal force, all the other forces are just the same. The only difference is that in the normal direction, you will be writing sum of all the forces to be mv²/R instead of zero. That's all. So the equations essentially remain the same. And I see you have signed your post.:-) Rkr1991 (talk) 06:20, 16 June 2009 (UTC)[reply]

That I get, but where does the force which counters the mgcos(theta) (the force pulling the ball down) come from?74.15.138.134 (talk) 11:44, 16 June 2009 (UTC)[reply]

The published solution oddly omits the normal force - possibly because it has no component parallel to the cone's surface, but it would still be nice to see it mentioned. Working in an inertial frame of reference:

We have two forces acting on the ball; the normal force N and gravity mg.
The particle is not accelerating vertically, so N cos(θ) = mg.
The particle is not in equilibrium in a horizontal plane - we require a net centripetal force mv²/R. So N sin(θ) = mv²/R. (Note that centripetal force is horizontal, not normal to the cone's surface, because centre of ball's orbit is in plane of ball's motion, which is horizontal)
Eliminate N and you get tan(θ) = v²/gR as in the published solution. Gandalf61 (talk) 12:24, 16 June 2009 (UTC)[reply]

Well, what can I say - Gandalf has said it all. Just remember to keep gravity in mind. Rkr1991 (talk) 15:25, 16 June 2009 (UTC)[reply]

Damn I realized where I went wrong, I said that N=mgcos(theta)...thanks a lot for clearing things up :) —Preceding unsigned comment added by 64.254.236.120 (talk) 16:08, 16 June 2009 (UTC)[reply]

Actually wait. How can the normal force, which is perpendicular to the surface (and hence doesn't have a component parallel to the surface) cancel out the component of gravity parallel to the surface? 64.254.236.98 (talk) 17:08, 16 June 2009 (UTC)[reply]

It doesn't. Nothing "cancels" the component of gravity parallel to the surface. Nothing needs to because the ball is not in equilibrium; it is accelerating towards the axis of the cone. The only direction in which the net force on the ball is zero is vertically; this is because the ball's motion is in a horizontal plane, so it is not accelerating vertically. Gandalf61 (talk) 19:17, 16 June 2009 (UTC)[reply]

lol i got it now, thanks. 74.15.138.134 (talk) 20:01, 16 June 2009 (UTC)[reply]

Humans as Krikkets

So I'm reading the Hitchhiker's series...

There's a race called Krikkets who have spent their entire civilisation unaware of the 'universe'

I'm wondering when Humans first became aware that those twinkly things at night weren't like the clouds, that is, mere pretty decorations on this world (obviously clous are more significant than decoration, but my point is that they aren't off the planet)

When did we realize that whether or not "we're alone in the universe" there's a universe in which to be?192.136.22.4 (talk) 01:44, 16 June 2009 (UTC)[reply]

See History of astronomy. Tempshill (talk) 01:50, 16 June 2009 (UTC)[reply]

See Ancient and early modern ideas (about Extraterrestrial Life for thinking about "being alone or not". -- 07:36, 16 June 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talk • contribs)

One of my favorite alternate universes to contemplate is if the Earth was always overcast. This would mean no stars or planets would ever be visible and the Sun (and maybe Moon) wouldn't appear as a circle but only as a bright spot in the clouds. How would astronomy develop ? Would we even figure out that the Earth is round ? (We might still be able to see tall ships drop over the horizon as they sail away, but a continuously foggy condition could obscure even this observation.) StuRat (talk) 13:06, 16 June 2009 (UTC)[reply]

I think there'd be significant curiosity in what this difuse glowing object in the sky was. Also cartography would definitely conclude the Earth is round in not so much time. Factor in flight to all this and I don't think we'd be that impared on a cloudy Earth.Aaadddaaammm (talk) 16:50, 16 June 2009 (UTC)[reply]

The ability to safely navigate the oceans without astronomical navigation would be quite difficult, and a continuously foggy world would make for constant shipwrecks. As for planes, take-off and landing would be quite dangerous if it was always foggy, although perhaps a water landing could be done safely (but getting back to shore would be another issue). Navigation would also be quite a problem. Gyros and/or radio location beacons would need to be used. Cars would also be dangerous in "foggy world", so perhaps trains (and maybe barges pulled along canals) would be the only reliable form of transportation. StuRat (talk) 19:37, 16 June 2009 (UTC)[reply]

Takeoff doesn't require visibility, and from the early days of electronics blind landings have been possible --Polysylabic Pseudonym (talk) 07:10, 17 June 2009 (UTC)[reply]

It may be possible to take-off and land blind, but it certainly isn't safe, at least with current methods. For example, how do you ensure that the runway is clear, considering that even a small, flat scrap of metal like the one that crashed the Concord could be present ? Perhaps a runway sweeping device could be utilized between each use of the runway and radar used to detect if any people or animals have wandered out there since. StuRat (talk) 11:17, 17 June 2009 (UTC)[reply]

You cannot know that we can observe everything. For all you know, there may be a class of phenomena which we cannot observe that, if observed, would show us a much larger multiverse. Thus the beings that move from dimension to dimension pity us because we are restricted to a single dimensions and we are not even aware of it. Tautologically, this hypothesis is not scientific because it is not falsifiable. -Arch dude (talk) 13:28, 16 June 2009 (UTC)[reply]

If that were true - there would have to be an entirely new mechanism for viewing. We've explored the electromagnetic spectrum from end to end - and observed what there is to observe along the way (with varying degrees of precision of course). For there to be more "things" to see - the seeing would have to be with some unimagined energy/information transfer mechanism that's not related to gravity/electromagnetism/weak/strong forces. Also, the energy being transferred by this mysterious means would have to be accounted for in some new way because we're currently able to account for most of the energy involved in physical processes. However, for all of that, it's not impossible that what you say is true. This is what makes the whole 'dark energy/dark matter' business so interesting. It implies that there is indeed some entirely different kind of process going on that we only know about because of "accounting errors" in things we do know about. SteveBaker (talk) 13:59, 16 June 2009 (UTC)[reply]

The entirely new mechanism for viewing is here. Tempshill (talk) 15:57, 16 June 2009 (UTC)[reply]

That's not a new "mechanism for viewing", or you could say "sense". We can sense electromagnetic radiation (sight and touch/heat), chemical compounds (smell and taste), physical force (touch). This is a pretty small list, there are a whole lot of other things we could sense. Taste and smell don't even factor into the energy calculations so it could be anything. Aaadddaaammm (talk) 16:45, 16 June 2009 (UTC)[reply]

In those conditions, humanity still could observe the moon and the planets with powerful radar. And if the world is always foggy, radar would be incredibly useful for other purposes. 65.121.141.34 (talk) 13:07, 17 June 2009 (UTC)[reply]

If you mean active radar, where a radio signal is sent out and then reflected back, that certainly wouldn't work to view planets from the surface of the Earth. It might possibly work for the Moon, though. Passive radar, or simply radio telescopes, which detect radio signals emitted from or reflected from objects, would still work well in heavy fog, though, to view the planets and distant stars. So, astronomy would have had to wait until the invention of the radio telescope, in such a world. StuRat (talk) 13:55, 17 June 2009 (UTC)[reply]

The Arecibo Observatory does active radar observations of planets (out as far as Saturn, anyway). See Radar astronomy. I'm not sure how you would discover planets using active radar, though, it only really works if you know where to send the signal. Passive radio astronomy would be required to first discover the planets, radar could then be used to study them. --Tango (talk) 00:03, 18 June 2009 (UTC)[reply]

chemical potetial

Give me a derivation which show the variation of chemical potential with temperature.chemical potetial of a component of a system drecrease with increase in temperature - is it true ?Supriyochowdhury (talk) 06:54, 16 June 2009 (UTC)[reply]

Please do your own homework.

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. -- Captain Disdain (talk) 07:21, 16 June 2009 (UTC)[reply]

I'm not answering this specifically unless it turns out its not homework, but I've never had a problem helping people find the answers to their homework.
Have you tried looking in any of the basic physical chemistry textbooks such as Atkins' Elements of Physical Chemistry? Or perhaps your course textbook may have the information you are seeking. If not, try reading your lecture notes to find the answers in there. It's where I find them more often than not. If all these fail, then try the wikipedia article on chemical potential. However wikipedia articles on subjects such as these often assume chemistry knowledge beyond that assumed by an introductory textbook. If the problem really persists, email the lecturer or if you have a supervisor/mentor/tutor at your college or uni try them. They don't just have academic knowledge, they have teaching knowledge and it can always help to talk through the stages of a problem with someone.

This answer isn't to say that I don't know the answer to this, I just don't think you would gain much by me or anyone else just spouting a few formulae.

Alaphent (talk) 09:51, 16 June 2009 (UTC)[reply]

Insect identification?

Could anyone identify the insect in this photo, http://s696.photobucket.com/albums/vv323/bailey-tucker/?action=view&current=IMG_2146.jpg. The image was taken in New Hampshire on 2009-05-04. Thanks --Captain-tucker (talk) 13:21, 16 June 2009 (UTC)[reply]

Looks like some kind of mayfly, but IANAE. --Sean 14:04, 16 June 2009 (UTC)[reply]

Probably in the Baetidae family.Alaphent (talk) 14:09, 16 June 2009 (UTC)[reply]

Aches after exercise

Is myalgia the correct term for the aches unfit people may get the morning after exercising a little too strenuously? If not, what is? --Dweller (talk) 15:14, 16 June 2009 (UTC)[reply]

Delayed onset muscle soreness? --Sean 16:30, 16 June 2009 (UTC)[reply]

Myalgia usually refers to an underlying disorder or other long term situation (alcohol abuse, medication, RSI) that triggers it off (i.e. not 1 exercise session). Does microtrauma help, although that applies according to how hard you push yourself, not your level of fitness. --87.115.23.15 (talk) 18:02, 16 June 2009 (UTC)[reply]

Myalgia as a medical term is a synonym for muscle pain - it's that simple. Similarly, arthralgia is joint pain. --Scray (talk) 20:13, 16 June 2009 (UTC)[reply]

What is RVS Steel?

A Google search of RVS Steel reveals many products manufactured from this material (many of them Dutch) but none provides information on what type or grade of steel this is or gives any reference to a national or international material specification.

This site suggests it may be simply a translation of 'stainless steel' or at least a variant thereupon. - Jarry1250 ^{(t, c)} 17:04, 16 June 2009 (UTC)[reply]

Indeed, it seems. Stainless steel interwiki links to "Roestvast staal" with "RVS" mentioned. This would suggest the S stands for "staal" i.e. steel. - Jarry1250 ^{(t, c)} 17:09, 16 June 2009 (UTC)[reply]

I'm betting that Roestvast is roughly either "rust proof" or "rust free" ie. stainless steel. --Polysylabic Pseudonym (talk) 07:16, 17 June 2009 (UTC)[reply]

Can the Lotus position be bad for your knees?

I mean the full lotus position.--80.58.205.37 (talk) 15:48, 16 June 2009 (UTC)[reply]

We can't give medical advice, but I think it's safe to say you need to be quite flexible to be able to do this in the first place, and you shouldn't continue if it is painful. Alternate the legs occasionally to avoid discomfort.--Shantavira|^{feed me} 16:20, 16 June 2009 (UTC)[reply]

I don't think it is a medical advice request (and isn't "stop if it is painful" advice anyway?). I used to do a lot of meditating, and any position becomes painful after an hour, even after a lot of paractice. I found no lingering ill effect, but always wondered about the effect that long term sitting would have (over a life time). I usually sat kneeling, and i read an article once about Japanese elderly having more knee problems due to a life-time of kneeling. Does anyone know better sources for that?YobMod 10:02, 18 June 2009 (UTC)[reply]

Speaking for myself, the full lotus position is bad for my knees. And ankles. And hip joints. I think it depends on the individual. Some people have naturally loose and flexible joints, and other like me don't in spite of doing a lot of stretching exercises. ~Amatulić (talk) 17:17, 18 June 2009 (UTC)[reply]

Why are damselflies bright blue?

Many dragon flies have ultraviolet reflective patches on their wings in order to attract prey. Are damsel flies colored to attract prey? —Preceding unsigned comment added by Dranorter (talk • contribs) 16:17, 16 June 2009 (UTC)[reply]

Either prey or mates. There's only really three good reasons to be brightly colored: 1) To attract prey 2) To attract a mate 3) To look poisonous so other things won't eat you. Take your pick. --Jayron32.talk.contribs 17:35, 16 June 2009 (UTC)[reply]

I would think for advertising strength and virility since only the males are brightly coloured (usually). As an aside, the male dragonflies eyes are larger than those of the females. They actually touch at the top of the head! I would think that is so because he has too always watch for rival males. 67.193.179.241 (talk) 22:53, 16 June 2009 (UTC) Rana sylvatica[reply]

Word, botany

What is the word for plants that prefer shadow? "Umbrophilia"? 80.203.110.95 (talk) 20:50, 16 June 2009 (UTC)[reply]

Our article Shade tolerance discusses the subject, which seems surprisingly complex; but doesn't offer a Latinesque or Greekesque word to describe the plants that "prefer" shade. Tempshill (talk) 21:55, 16 June 2009 (UTC)[reply]

Photophobic. B00P (talk) 21:57, 16 June 2009 (UTC)[reply]

When is the Pleiades meteor shower this year?

Oddly, I'm having difficulty getting relevant search results. --Trovatore (talk) 22:06, 16 June 2009 (UTC)[reply]

I don't believe a such-named shower exists. Are you sure you aren't after something else in our list of meteor showers? Algebraist 22:13, 16 June 2009 (UTC)[reply]

Have you got the right name? Pleiades is a star cluster. Exxolon (talk) 22:36, 16 June 2009 (UTC)[reply]

Do you mean the Perseids? As always, they'll peak on the night of August 12-13, and the best time to see them in Toronto (where I live) is around 3 a.m. on August 13. --Bowlhover (talk) 22:55, 16 June 2009 (UTC)[reply]

I did mean the Perseids, thanks. Of course the peak is not exactly the same day each year, which is why I was asking. Apparently the morning of Aug 12 will be the best this year, though with interference from the Moon. --Trovatore (talk) 22:58, 16 June 2009 (UTC)[reply]

Often the exact time of the peak isn't as important as local factors in determining the best time to go out. It is generally best to view between midnight and dawn, local time, although I can't remember why (something to do with the directions the meteors will be coming in). You also want to view when the radiant is as high in the sky as possible and when the moon is out of the way. --Tango (talk) 23:36, 16 June 2009 (UTC)[reply]

Aug 12 - oops! I somehow got the impression that the peak is almost always somewhere around Aug. 13 morning, but I guess that's not true. --Bowlhover (talk) 00:15, 17 June 2009 (UTC)[reply]

Yes, unfortunately these are somewhat incompatible this year, as moonrise will be around midnight. But this link suggests there could be a decent show anyway. --Trovatore (talk) 23:39, 16 June 2009 (UTC)[reply]

For the Perseids, I find it's best to go out around 3 a.m. because the radiant is high in the sky at the moment, and because going out at 3 a.m. gives plenty of time to observe before dawn. Of course there are advantages to observing earlier: you get to see the slow and long-lasting earthgrazers, the ones that enter the atmosphere at a shallow angle, and this year the Moon won't interfere. --Bowlhover (talk) 00:15, 17 June 2009 (UTC)[reply]

Thanks much. I'm hoping to be backpacking in the Sierras that night, and if the weather cooperates we can just lie in our sleeping bags on top of a ground cloth and watch the sky. I did that once as a boy during a meteor shower, and it was unforgettable. --Trovatore (talk) 01:25, 17 June 2009 (UTC)[reply]