Wikipedia:Reference desk/Archives/Science/2017 September 16

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September 16

Lesch–Nyhan syndrome

Lesch–Nyhan syndrome causes a mess of problems: the patient cannot produce a specific enzyme, produces too much uric acid, experiences intellectual disability severe enough to prevent speech and ambulation, and is susceptible to a raftload of self-harming behaviors. I get the impression that the enzyme absence causes the overproduction of uric acid somehow, but what causes the rest? The article seems to suggest that the hyperuricemia is responsible for everything, yet that seems unlikely (why would too much of a toxic acid cause compulsive self-harming?) — but then I've looked over the "Pathophysiology" section without understanding much of anything after the first three paragraphs. Nyttend (talk) 03:35, 16 September 2017 (UTC)

From the article: "The etiology of the neurological abnormalities remains unknown." Ruslik_Zero 08:13, 16 September 2017 (UTC)

Nonetheless, it might be useful to research purine autism, which also involves hyperuricemia. Purine autism is interesting in that it is a kind of autism said to respond to allopurinol treatment, though I would have expected much more confirmation of that by now if it were reliable. [1] Note that autism can also cause self-harm activity and is mentioned in this regard in our Lesch-Nyhan article. Wnt (talk) 11:32, 16 September 2017 (UTC)

Hm, I didn't realise that "etiology" meant "origin" or "cause". Nyttend (talk) 12:33, 16 September 2017 (UTC)

Aquo complex versus hydrated compound

Is there a clear demarcation between an aquo complex and a hydrated compound; or does it simp.ly depend on context; or does it depend on the particular author? I am not talking about water of crystallization when I say 'hydrated compound'. Case in point: tetraaquocopper(2+) sulfate versus copper(2+) sulfate tetrahydrate. Does it have anything to do with the degree of lability of the aquo ligands, or perhaps whether the complex is homoleptic? Plasmic Physics (talk) 07:03, 16 September 2017 (UTC)

At the level of one of my freshman chemistry textbooks (General Chemistry by Ralph H. Petrucci 5th edition 1985, p.911) and this website [2] (Washington University at St. Louis), both would appear to be non-standard names for coordination complexes, although the first is close. Standard nomenclature would be tetraaquacopper(II) sulfate, indicating copper 2+ coordinated to 4 waters and not coordinated to the sulfate counter ion. "Copper(2+) sulfate tetrahydrate" would mean CuSO4.4H2O to me, which would be an example of water of crystallization. An example of where you have seen this would be helpful, but using CuSO4.4H2O terminology would be necessary in instructions for a copper sulfate solution preparation to get the right mass of starting reagent for the desired concentration of copper sulfate.--Wikimedes (talk) 17:45, 16 September 2017 (UTC)

Using the ionic charge instead of the oxidation state when constructing additive names seems to be the preffered alternative convention according to the latest IUPAC nomenclature conventions. From what I've read, when 'hydrate' appears in the name, it can refer to either an aquo complex or w.o.c. introducing vagueness. From your your concluding remark, I gather that which one to choose, depends more on the context? If that is the case, what defines the context in which either use would be appropriate? Plasmic Physics (talk) 23:28, 16 September 2017 (UTC)

Standard nomenclature seems unambiguous, and you probably can't go wrong by choosing the standard way of saying or writing things. (2+ vs. II or aquo vs. aqua wouldn't impede understanding and is just a tangential nitpick.) One of my fields is solid state chemistry; I'm not up on common variations of standard nomenclature in coordination chemistry and can't recall much anecdotally.--Wikimedes (talk) 02:29, 17 September 2017 (UTC)

(My appologies, it appears it is actually supposed to be 'aqua'). Solid-state chemistry? Interesting... How can you tell expermentally or theorhetically, if an atom is bound to a neighbouring atom in a bulk phase? Plasmic Physics (talk) 03:03, 17 September 2017 (UTC)

Experimentally, X-ray crystallography would be choice for crystalline compounds, and can give some information about amorphous solids as well. There are also the related neutron and electron diffraction. Solid-state NMR gives information on the proximities of NMR-active nuclei. According to table 3.1, p.48 of Solid State Chemistry and its Applications by Anthony R. West 1984, most types of spectroscopy give information on local structure, though you would have to do some research to find out which technique is best used for a particular situation. Theoretically, according to p.20 of the January 2017 edition of the MRS Bulletin, "... (molecular dynamics) simulations are routinely used to model the structure of materials with steadily increasing accuracy".--Wikimedes (talk) 18:58, 17 September 2017 (UTC)

Breaking boards

How much force does it take to perform a power break on an unpegged stack of 4 standard 1-inch-thick boards made from seasoned pine, dimensions 12x6 inches, grain along the 6-inch edge? How much more force is needed if the stack is increased to 6 boards with all the other parameters remaining the same? (This is NOT homework -- this is me trying to gauge my own strength!) 2601:646:8E01:7E0B:3DB7:8D6E:A762:14CC (talk) 08:02, 16 September 2017 (UTC)

You don't break them with force (at least, not in karate), you use impulse to do it. This is the product of force and time (strictly, their integral). There are two impulses under consideration here, the impulse given to the moving hand by the body, and the impulse the moving hand then transfers to the boards. Much of karate training is about increasing the impulse given to the hand, in the short time available. Considering human muscle generally, we're "strong but slow" - a mediocre performer in pure strength sports can produce more force than most karetaka can, but they can't deliver anything like the same impulse. The ultimate demonstration of this being of course the "inch punch" techniques.

This also indicates why an empty hand technique is limited (for pure striking effect) against anything with a weighted glove, or a kick (legs are heavier than arms). Human muscles are better at loading impulse into things slightly heavier than a single hand.

Then there's the matter of transferring the impulse into the board. It takes a certain amount of energy to fracture a brittle material like dry cross-grain timber, so six boards will need at least $6/4$ as much as four. But, one must also couple the impulse of the hand effectively into the boards. This is very difficult to model as a physical process, because it's fast and the materials (mostly hands) are flexible, so they turn the energy of the moving hand into compressed and displaced flesh, rather than bending (and breaking) the board. This is the second matter of karate training: which part to strike with and how to keep the hand rigid enough.

Then there's the physics of breaking boards. The easiest boards to break are a stack of moderately thin boards, already touching at the impact point. It is hard to break the same boards with spacers at their edges. If they're spaced apart for visibility, they sometimes have a 'coupler' in the middle too. It's very hard - soon impossible - to break a single board of the same thickness. Also don't demonstrate against a carpenter who gets to choose which boards to use, because there's visible and selectable variations in their strength. Boards are broken by exceeding their strain, not their stress. It's not applying a force greater than their ability to resist, it's bending the board past their linear ability to extend (why the materials broken are chosen for being brittle, i.e. having a poor resistance to strain, even if strong against a stressing force). If the boards are too thick, you can't bend them enough to achieve this strain - you may put a Herculean dent into the top, but they don't snap. If they're too thin, they become simply flexible and all you will do is bend them down and have them spring back afterwards - this is some of the theory behind composite materials. If the boards are in a spaced stack, then there is considerable bending in the upper boards (absorbing your impulse), but there isn't enough left for the lower boards to break them. Bending the upper boards downwards to touch the next board absorbs energy and this energy is lost to the strike when the board breaks (it turns into sound or an acceleration of the board halves starting to move independently). So a spaced stack of boards will require an impulse that increases more than the number of boards. Exactly how much more is tricky to work out, as it depends on many factors - such as the relative energy needed to bend a board down vs. the energy to break it. A stiff board in a spaced stack (such as tiles) is very hard to break.

Boards and tiles are broken for demonstrations because they demonstrate the effect of impulse over force. To show simple force strength (and not highlight the effects of speed), then punching a swinging bag and seeing how far it moved would show this better - and the karetaka would probably lose to Indian wrestlers. Andy Dingley (talk) 09:39, 16 September 2017 (UTC)

@Andy Dingley: I am suspicious you are thinking of something else besides impulse here. The impulse is force x time, and causes an acceleration of the target object. In other words, the swinging bag would be a perfect measurement of impulse. What you are describing seems more like simple speed, or perhaps some other quality, that allows the board to be pulled far off center before it has much of a chance to push back on the hand i.e. to apply its impulse to the hand. But I know nothing of how to break multiple boards with a hand, so I am likely wrong. Wnt (talk) 11:39, 16 September 2017 (UTC)

Would it be rate of transfer of impulse (= rate of change of momentum) that is important here? Dbfirs 12:05, 16 September 2017 (UTC)

And if so, that would be the force! (F*t)/t=F 2601:646:8E01:7E0B:3DB7:8D6E:A762:14CC (talk) 12:17, 16 September 2017 (UTC)

So it would! I've never tried to break boards (at least not seriously). Presumably, one needs both a large impulse and a fast rate of transfer (speed of strike; force) to achieve the effect. There is some discussion here but I'm not convinced by the approximations in the mathematics. Dbfirs 12:27, 16 September 2017 (UTC)

The link loops back to Wikipedia: Breaking (martial arts) for a source. Our article there has some data (also shows some signs of personal stress, like the uppercase link to SCIENCE). I am still not convinced it is very simple. For one thing, the displacement of the board depends on the force, but force is a tricky thing. Suppose you contrast my hand, modelled by a heavy sausage wrapped in confectionary marshmallow, versus the hand of a martial artist, modelled by the sausage wrapped in horn. Well, if I had a rocket booster handy to get my hand up to the same speed as the artist, the impulse would be the same. But the force wouldn't be the same because the whole mass of the marshmallow has to squish up (OUCH!) before all the momentum is transferred; but the horny hand of the artist presumably is more rigid and conducts that impulse rapidly upon the first few millimeters of contact (and impulse/time = force). Bent off center more rapidly by the increased force, perhaps the board breaks sooner, and therefore has much less time to transfer impulse back to the hand of the artist. So how much is speed, needed to build up the hand's impulse, and how much is rigidity, to deliver it? And what else have I forgotten? Wnt (talk) 18:07, 16 September 2017 (UTC)

So, anyone find an approximate number value for the force involved? 2601:646:8E01:7E0B:3DB7:8D6E:A762:14CC (talk) 07:44, 18 September 2017 (UTC)

Force is the wrong measure. Andy Dingley (talk) 10:13, 20 September 2017 (UTC)

Shortest Nobel Prize-winning paper

Today I learned that Watson & Crick's paper announcing their discovery (or at least, their part in the discovery) of the structure of DNA is a mere 834 words - exceptional brevity for such a significant finding. Is that a record? I appreciate Nobel Prizes aren't for papers but for research generally, but using published papers which described Nobel-winning research as a metric has anyone ever bettered 834 words? 51.9.138.245 (talk) 10:10, 16 September 2017 (UTC)

Watson & Crick's paper was so short largely because they were in such a hurry. Much of the background work was already known. Pauling had already published one postulated structure, which was largely held to be unworkable. They thought that others (i.e. Pauling) would reach their own conclusions for themselves if given much more time. Watson & Crick had two ideas, the double helix (rather than a triple structure) and also the "zipper" idea for replication. They wanted to be Wallace, not Darwin. Their paper wasn't publishing the results of long years of research and careful study (that was mostly out there already, or was filled out later) it was throwing two wild ideas out to claim clear priority on them, even if they turned out to be wrong later. The specific paper is a brilliant hypothesis, not careful research. Andy Dingley (talk) 10:36, 16 September 2017 (UTC)

• Since Watson and Crick were mentioned, Rosalind Franklin must be mentioned too. Fgf10 (talk) 11:39, 16 September 2017 (UTC)

One of the justifications for Franklin not being mentioned on this paper has been that the paper was based on the two "wild guesses" that were so urgent to make publicly visible and so to claim precedence for. Franklin's work had been the painstaking careful research that led up to this. Sadly and IMHO wrongly, the Nobel Prize was awarded for the narrow paper, and overlooked Franklin's contribution to making it possible to have those ideas. Andy Dingley (talk) 12:42, 16 September 2017 (UTC)

• Franklin is acknowledged in the paper (as well as Wilkins). --69.159.60.147 (talk) 22:14, 17 September 2017 (UTC)

I fully agree that Franklin should have been given more credit by the scientific establishment, but it's difficult to see how the Nobel Committee itself could have: Nobel Prizes are never awarded posthumously, and Franklin had died in 1958, 4 years before the award to Watson, Crick and Maurice Wilkins. Watson himself suggested that, had she lived, she might also have shared the award.

Perhaps those three should themselves have made more of her contribution, which Franklin may never have been fully aware of, as Wilkins had shown her critical 1952 photo to W & C in 1953 without her knowledge, and after she herself had moved on to a different college and different areas of research. {The poster formerly known as 87.81.230.195} 90.200.137.12 (talk) 16:06, 16 September 2017 (UTC)

Well her also sharing the award with Watson, Crick and Wilkins, if she had lived was surely about as likely as her getting the award after she died, i.e. really unlikely. Prize rules limiting a single prize to 3 people would have required either there were 2 awards in two separate categories or years, or there was something weird like the Randall X-ray diffraction lab. Mind you, I'm not sure anything but the Nobel Peace Price can be award to an organisation. Our article suggests so but the cited source doesn't really seem to say this and the closest I can find on the Nobel site is that in some places they mention the Nobel Peace Price has been awarded to organisations etc. Anyway, if they had wanted to award to an organisation in memory of her, I don't see that her death really stopped that. And if they were going to award 2 prizes, they could have simply said no living person deserved the second, although that may seem a little weird since more likely no living person and Wilkins should share the second prize. Mind you the no 3 people and posthumous rules [3] don't really seem to come from Nobel's will anyway [4]. Also although that clearly says the award is divided equally, among the recipients, [5] suggests it isn't always. I guess most likely this arises when 2 separate works receive an award, in which case the one work with 2 recipients gets 1/2 share which is divided among the 2 people, and the other gets 1/2 share which goes to that single awardeee. Nil Einne (talk) 17:48, 16 September 2017 (UTC)

My assumption is that Franklin would have been jointly awarded the Prize instead of Wilkins. Both of them had been working on DNA crystallography, and Franklin had been recruited by their mutual boss John Randall to either collaborate with Watkins (as he believed) or to take the work over and forward it (as she believed – Randall's poor management of them led to their misunderstanding). Franklin's improved data was crucial to Watson and Crick, but with her dead, it was not unreasonable to include Watkins as the "next biggest player" in the discovery. {The poster formerly known as 87.81.230.195} 90.200.137.12 (talk) 19:46, 17 September 2017 (UTC)

Here is the paper, and as you see, Franklin is acknowledged (as well as Wilkins).

Possibly. However your original comment said also without any mention of excluding any recipients so I was replying to that. Nil Einne (talk) 05:06, 18 September 2017 (UTC)

Acknowledgement isn't an authoring credit though. Andy Dingley (talk) 10:16, 20 September 2017 (UTC)

• The assumption I've heard is that if Franklin had survived and been able to promote her work (and that of her doctoral student Raymond Gosling), there would probably have been two prizes: Watson and Crick would have won the Nobel Prize in Physiology (for working out the reproduction mechanism of DNA) and Wilkins, Franklin (and maybe Gosling) would have won the Nobel Prize in Chemistry or maybe Physics (for pioneering x-ray crystallography methods that uncovered the structure of DNA). Smurrayinchester 10:18, 20 September 2017 (UTC)

Conway famously published a paper with the title "can n^2+1 unit equilateral triangles cover an equilateral triangle of side > n, say n + e" with the substantive body of "n^2 + 2 can" accompanied by two diagrams. — Preceding unsigned comment added by 2A01:E34:EF5E:4640:35DC:A78A:A81D:9CA4 (talk) 15:32, 16 September 2017 (UTC)

The question was about Nobel Prizes, though. --69.159.60.147 (talk) 22:14, 17 September 2017 (UTC)

Infinity

Jean-marie Adragna Vancouver B.C Canada

Everybody thing at the infinity at the space in expansion. For my the description of the Infinity is the ( The wave of the Big-Bang forever increase ) Sorry but I don't have any background in science ,i this possible the somebody told me if is possible For my ( fiery ). Sin cerement J-M.A — Preceding unsigned comment added by 70.79.181.150 (talk) 14:57, 16 September 2017 (UTC)

Yes, there are many infinities. There's infinitely large, infinitely small, infinitely forward in time, and infinitely backwards in time, for example. You might also be interested in the infinite worlds hypothesis. StuRat (talk) 15:06, 16 September 2017 (UTC)

Some people regard "infinitely small" as a contradiction in terms. The usual term is Infinitesimal. Mathematically, there is a hierarchy of infinities (see Stu's link above). For example, there are more real numbers than there are whole numbers, but the number of fractions is the same as the number of whole numbers. Dbfirs 16:40, 16 September 2017 (UTC)

But no matter how you slice it, "infinity" is not a quantity, it is not a number. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:33, 16 September 2017 (UTC)

It's actually a cardinality. Dbfirs 19:36, 16 September 2017 (UTC)

Yes, there are different "cardinalities" of infinity. Hence the terms "countably" infinite vs. "uncountably" infinite. But infinity is infinity. As Carl Sagan said, no matter how large a number you can imagine, you are no closer to infinity than is the number 1. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:58, 17 September 2017 (UTC)

Did you read the article on cardinality? Dbfirs 06:25, 17 September 2017 (UTC)cardinality

The term "infinitesimal", which is used to mean "infinitely small", is essentially a way of saying "infiniteth", as compared with "tenth" or "hundredth" or "thousandth".[6] ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:07, 17 September 2017 (UTC)

See also our Shape of the universe#Infinite or finite. -- ToE 17:46, 16 September 2017 (UTC)

See also here. Count Iblis (talk) 22:33, 16 September 2017 (UTC)

Nothing is infinite (with one exception). Everything is by design or definition infact finite as a concept of science. The only exeption is the literal Nothing in Sense of empty space which is the only "thing" that can be regarded as infinite without braking our laws of physics. --Kharon (talk) 03:21, 17 September 2017 (UTC)

Or breaking. But braking works too. :) And if space is "finite but unbounded" then your premise remains true. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:00, 17 September 2017 (UTC)

The metric expansion of space should be braking, but isn't, so the Americans are apparently the only ones to put the brakes on something metric. :-) StuRat (talk) 04:29, 17 September 2017 (UTC)

Ofcourse i meant Breaking. Im German but my englisch was always very, very good. No idea how that got so wrong in my head - or out of to be more precise. I promise I'll take more care for my writing. --Kharon (talk) 05:06, 18 September 2017 (UTC)

I'll give you a break and assume your misspelling and improper capitalization of "English" was intentional. :-) StuRat (talk) 16:49, 18 September 2017 (UTC)

What's the diff between a straight-A student, her teacher, and Ford Proving Grounds ? One breaks the curve on a test, the next curves the test on a break, and the last tests the brakes on a curve. StuRat (talk) 16:43, 18 September 2017 (UTC)

• Empty space is not literally nothing, neither ontologically nor physically. See virtual particle. μηδείς (talk) 20:50, 18 September 2017 (UTC)

What does D. Œ. A. V. stand for?

In working on the history of glaciology, I'm finding references to "D. Œ. A. V.", such as this: "Le succès des sondages de 1899 encourageant les plus grands espoirs pour la réussite du levé complet par la même voie, d'un profil transversal du glacier, le Comité central du D. Œ. A. V., qui avait subventionné les premiers travaux, consentit avec la plus louable munificence à faire les frais d'une nouvelle campagne de sondages dans ce but." which Google translates as "The success of the surveys of 1899, which encouraged the greatest hopes for the success of the complete survey by the same route, of a transverse profile of the glacier, the central committee of the D. Œ. A. V., which had subsidized the first works, consented with the most laudable munificence to pay the cost of a new survey campaign for this purpose." I also see this abbreviation in some old citations. Can anyone tell me what it stands for? The "Œ" quite probably stands for "Österreich" or some variation, since we're talking about the Alps; the A might derive from the Alps. Googling the abbreviation and searching Google Scholar for the cited papers hasn't gotten me anywhere. Any other way to find out? Mike Christie (talk - contribs - library) 16:24, 16 September 2017 (UTC)

Possibly Deutscher und Österreichischer Alpenverein. Cheers ✦ _hugarheimur 16:46, 16 September 2017 (UTC)

That's sure to be it. Thank you very much! Mike Christie (talk - contribs - library) 16:50, 16 September 2017 (UTC)

D. Œ. A. V. and D.Œ.A.V. now created as redirects. Nyttend (talk) 23:24, 17 September 2017 (UTC)

Atmospheric pressure

My Thermodynamics professor drew a free-body diagram for a lid being lifted by boiling water in the pot below. Applying downward force is gravity and atmospheric pressure, he said, while vapor pressure provides the lift. But doesn't the air inside the pot provide a lift as well due to an equal amount of atmospheric pressure as the air above? Thank you. Imagine Reason (talk) 16:49, 16 September 2017 (UTC)

It could be that your professor is assuming that water vapor has displaced all the air in the pot. Your professor is the best authority on the assumptions s\he has made in his/her model; this would be a good question to ask him/her.--Wikimedes (talk) 17:05, 16 September 2017 (UTC)

Until the lid is lifted, the gas underneath it should be at atmospheric pressure plus the water vapor pressure. So yes, the atmospheric pressure on both sides of the lid cancels out. StuRat (talk) 17:13, 16 September 2017 (UTC)

Since the sides of the pot are not involved in the scenario, should I assume that the lid has been lifted? In that case, will the vapor have completely displaced all the air? Thank you. Imagine Reason (talk) 23:13, 16 September 2017 (UTC)

After a little more thought, I see it this way: If the water is boiling, it means that the gas above the water in the pot is at (or less than) the vapor pressure of water for the temperature of the water. On the stove top, heat is put in until the water temperature rises to the point where this pressure starts to push off the lid. This upward pressure is equal to the downward pressure on the lid caused by gravity acting on the lid and the external gas pressure on the lid, which is atmospheric pressure.

Another way to look at it is that the gas inside the pot is isolated from the atmosphere outside the pot (and the weight of the ~100km column of air that causes atmospheric pressure), so there's no reason to expect atmospheric pressure to be a factor inside the pot.--Wikimedes (talk) 00:42, 17 September 2017 (UTC)

The Temperature vs. specific entropy diagram of Steam

The air is not displaced. Infact it will not be displaced but to the contrary remain and get saturated (with steam) in dependence of its temperature and pressure, in an multicomponent (Water and Air (Nitrogen, Oxygen Argon Carbon dioxide etc.)) Vapor–liquid equilibrium commonly named and known as "Steam" (See Steam T-s-diagram in added picture on the right). --Kharon (talk) 03:00, 17 September 2017 (UTC)

• Ask the professor, but a possible simplification would start with the lid directly atop the water with no air under it, and with no air dissolved in the water. As it happens, in real life before boiling starts there will be air under the lid and air dissolved in the water, but the total amount of air is small in proportion to the amount of steam that is generated, so its effect is mathematically negligible in a sufficiently large container.-Arch dude (talk) 04:03, 17 September 2017 (UTC)

I don't want to ask the professor because he introduced concepts like gage pressure without much explanation and refused to answer a student who asked about it. He didn't mention anything about cooking a vacuum. Also, if the atmospheric pressure from the air inside the pot is negligible, then so is the atm outside, no? Imagine Reason (talk) 04:46, 17 September 2017 (UTC)

You are at the start of thermodynamic lessons obviously, else you had not asked your question. It will be answered later on. You will learn to use the Gas constant, learn about Bernoulli's principle etc. Your professor probably tried to tease you to get you interested and wanting to learn. You still have to learn, to understand, which obviously seems much more boring then trying to bend your mind around it with what you already believe to understand. Have some patience, do your lessons and you answer this question and other new ones yourself. --Kharon (talk) 05:40, 17 September 2017 (UTC)

If the atmospheric pressure from the air inside the pot is negligible, then the atmospheric pressure from the outside should be neglected as well, no?

It's not really a matter of atmospheric pressure inside the pot being negligible (i.e. so small it can be ignored). Atmospheric pressure is caused by the weight of the atmosphere. After the lid is put on, the gas inside the pot is isolated from the weight of the atmosphere, so it doesn't make sense to use the weight of the atmosphere to model the pressure inside the pot (except perhaps when the lid is first put on and nothing has changed inside the pot yet). Outside the pot, however, the weight of the atmosphere is still pushing down on the lid (even after the pot heats up), so atmospheric pressure must be taken into account.--Wikimedes (talk) 19:43, 17 September 2017 (UTC)

The partial pressure contribution of the air molecules is negligible because they are a negligible percentage of the total gas in the pot after it commences boiling. Essentially all of the gas is steam, as the steam molecules increase while the number of air molecules remains fixed. ("air molecules": the usual mix of gases. "steam molecules": H₂O). -Arch dude (talk) 20:34, 17 September 2017 (UTC)

Forgive me for still not understanding. Right after you close the lid, the air inside the pot is lifting the cover with 1 atm of pressure. Then you apply heat to the contents of the pot. The air inside then should contribute more than 1 atm of pressure now, thus canceling the weight of the air on top of the cover. Imagine Reason (talk) 03:28, 18 September 2017 (UTC)

When the air inside the pot is contributing more than 1 atm of pressure, it no longer exactly cancels the external atmospheric pressure. At this point you have to consider them as separate forces acting on the lid, rather than ignoring them.--Wikimedes (talk) 05:09, 18 September 2017 (UTC)

See Dalton's law. If the gas under the lid is 50% air and 50% steam, then the equilibrium pressure of the mixture will be that of the steam, which will be given by the Vapour pressure of water. HTH, Robinh (talk)

Feynman Lectures. Exercises. Exercise 14-15 JPG

. .

...

14-15. A certain spring has a force constant k. If it is stretched to a new equilibrium length within its linear range, by a constant force F, show that it has the same force constant for displacements form the new equilibrium position.

—  R. B. Leighton , Feynman Lectures on Physics. Exercises

In Solutions they write: Let the spring is stretched by the force F₀. The displacement can be found from F₀ = k x₀. Let's stretch the spring more by x. Then new force is : k(x₀ + x) = k x₀ + kx = F₀ + kx. So extra force is the same as if the spring is stretched from undisturbed state.

But the Solution' author uses k=const from beginning of the proof. If k is some function of x, and F = k(x)•x only for very small x, then how to prove the exercise? In other words, we have the undisturbed spring length L₁ and we know the law for it: F_I = k₁ x for small x. And we have the stretched spring length L₂ and the law F_II = k₂ x for small x. From these two laws it is clear that F≠k₂(L₂ - L₁)≠k₁(L₂ - L₁). Username160611000000 (talk) 19:02, 16 September 2017 (UTC)

If a spring has a force constant, then k should be a constant.

It is ridiculous that we are asked to prove k = const, and this is given under the statement of the problem. No, I think we should not use k = const. Besides the exercise is to Lectures 13,14 "Work and Potential Energy".Username160611000000 (talk) 08:14, 17 September 2017 (UTC)

I think you're being asked to prove the constant is constant from the new position. Yes, it's pretty elementary, I mean, k(x+dx) - kx = kdx or something. They can't all be stumpers. Wnt (talk) 18:15, 17 September 2017 (UTC)

If k increases in the spring stretched under force, it seems apparent the new k will be the "force constant" for very small deviations, if the function derivative is continuous. i.e. FII = k2(x +- delta x) for small delta x. Wnt (talk) 19:35, 16 September 2017 (UTC)

It is not impossible, e.g. k = sin (x). Then for x = 0.5π, 1.5π etc. the derivative = 0 and so k(x) = k(x+dx).Username160611000000 (talk) 08:14, 17 September 2017 (UTC)

yeah, but ... if k = 0 at 1.5 m, how is the pendulum going to stay there when it's under some continuous force? Also, this isn't really an exception = for this FII = 0 (x +- delta x). For small deviations it's not a spring, but it's not an exception. Wnt (talk) 09:34, 17 September 2017 (UTC)

ancient glacial features

It's easy to point to valleys that were shaped by glaciers in the last million years. But what are the oldest known glacial features? Are there glacial valleys in continents that were near the poles in Mesozoic times? —Tamfang (talk) 20:35, 16 September 2017 (UTC)

Note that land didn't need to be near the poles to have glaciers. Glaciers covered much of Europe and North America in the most recent ice age, for example. StuRat (talk) 20:53, 16 September 2017 (UTC)

See here. Count Iblis (talk) 21:17, 16 September 2017 (UTC)

Global glaciations and atmospheric change at ca. 2.3 Ga. Count Iblis (talk) 22:14, 16 September 2017 (UTC)

As others have noted about, we have seen signs of ancient glaciations going back billions of years. That said, features like carved valleys will be eroded and changed over time. Given 10-20 million years or so, most features won't be easily recognizable as glacier derived except by detailed study from experts. Dragons flight (talk) 09:11, 17 September 2017 (UTC)