Wikipedia:Reference desk/Archives/Science/2011 April 30

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April 30

Sicily, the Gulf of Sidra and continental drift

I noticed that the Gulf of Sidra north of Libya where the fighting is going on looks like Sicily had been taken out of it. Is it possible that Sicily was once connected to this area?--X sprainpraxisL (talk) 00:14, 30 April 2011 (UTC)

You might want to read about the Apulian Plate. Geologists think that some crust that forms Italy and the surrounding region did break away from North Africa; but its a bit of a jump to conclude that Sicily "fell off" just based on coastline features alone. Nimur (talk) 01:04, 30 April 2011 (UTC)

I'm looking at paper sea floor maps because I can't find enough info on the net. But on these paper maps there is a fault line that runs from the Azores through the Gibraltar straight and all the way through the Mediterranian and to the red sea. It passes along the North African coast between Africa and Sicily. At that point it gets messy with a lot of break up spreading up through the Tyrrhenian sea on the Apulian plate. The Periadriatic seem is far north of that. This map does indicate the Apulian plate is moving northish away from Africa, and Sicily is on the Apulian plate. So your idea seems at least plausible.190.148.136.203 (talk) 02:07, 30 April 2011 (UTC)

The tectonic history of the Mediterranean is pretty complex, but all the plate reconstructions that I've seen for the central Mediterranean suggest that Sicily (at least the deeper parts and ignoring Etna) have remained pretty much in the same place with respect to Africa since the Late Triassic, about 230 million years ago, so any shape match is probably coincidence. Mikenorton (talk) 08:07, 30 April 2011 (UTC)

Look at New Zealand and Australia, they have complementary coastlines, which suggests that they split off from each other. Geological records show that while this is true, far less than 10% of the current landmass was present at the time of the split. The remainder was generated after the split; the current complementary shape has very little to do with the split and is purely coincidental. Plasmic Physics (talk) 08:13, 30 April 2011 (UTC)

Also if you're appealing to continental drift to match fragments of continental crust to a neighbouring continent, the intervening area (the Gulf of Sidra in this case) must be made of oceanic crust, which it isn't. Mikenorton (talk) 09:49, 30 April 2011 (UTC)

...and the modern Mediterranean Sea and its shorelines are a quite recent feature on geologic timescales - see Zanclean flood. Gandalf61 (talk) 10:11, 30 April 2011 (UTC)

Yes the refilling of the Mediterranean seems to have been quite recent but it's not clear how that would have anything to do with tectonic movements other than perhaps a little lubrication for subduction.190.56.125.8 (talk) 15:25, 30 April 2011 (UTC)

Gandalf61 is I think referring to the shape of the coastline, which will vary a lot with water depth. As to lubrication, the Messinian salt layer has strongly affected the shape of the Mediterranean Ridge accretionary complex as subduction has continued - because salt is so weak under geological deformation rates. Mikenorton (talk) 15:32, 30 April 2011 (UTC)

Superconductors

Does superconductors obey ohm's law? Because when R is zero I will be infinity as I=V/R and the magnetic field will also tends to infinity and the material comes to normal state. —Preceding unsigned comment added by Balajir333 (talk • contribs) 06:06, 30 April 2011 (UTC)

V is not the total voltage applied but the voltage loss at R. So if no voltage is lost at the superconductor, V=0, and if you write V=IR to avoid mathematical troubles for dividing 0 by 0, everything is fine again. The actual I is then determined by the non-superconducting parts of the circuit. 95.112.236.165 (talk) 08:07, 30 April 2011 (UTC)

One way to state Ohm's Law is that the current that flows through a material is linearly proportional to the applied voltage. The constant of proportionality is the electrical resistance. Superconductors do not behave this way, so it can be safely said that Ohm's Law does not apply to them. Instead, the current that flows through a superconductor has no relationship to the applied voltage; the current that flows is subject only to conservation of charge flux (Kirchoff's current law). In some superconductors, especially if analyzed at very small scales, this is not even a very good description, because the electrons can not be individually discerned; in the case of some superconductors, the electrons must be described as an ensemble wave function. Nimur (talk) 15:50, 30 April 2011 (UTC)

Well, another way to state Ohm's Law is that the electrical potential difference (applied voltage) is linearly proportional to the current (V=IR). The constant of proportionality is the electrical resistance (denoted R). For the limit case R=0 this is, if taken literally, not equivalent to I=V/R. In case we are willing to take Ohm's Law to be V=IR, a superconductor with R=0 well satisfies this equation. Of course, superconductivity will break down if the current exceeds some maximum that depends on the material, but that is independent of the version of Ohm's Law- 95.112.236.165 (talk) 22:46, 30 April 2011 (UTC)

Kepler mission

is there perfect information about data collected with kepler mission ?--78.38.28.3 (talk) 08:37, 30 April 2011 (UTC)

Try the Kepler website. The most complete information should be under Science->For Scientists. There's also a data archive if you're interested. --Wrongfilter (talk) 09:34, 30 April 2011 (UTC)

What I'd like to know is what percentage of stars actually have planets that traverse their star's disc, as seen from our point of view. My guess is less than 1% of the stars that have planets, have planets who move in just the right plane. What's your guess? --InverseSubstance (talk) 02:25, 1 May 2011 (UTC)

Akbarmohammadzade:

It is modern technology helps man to find exoplanets , if found about 500 planet in one location it might be further than 1percent of stars which have planet . my question is about the goals and objective of that mission , if they try to find life in other place ,or want to find any way for good explaining the mechanism of formation of planets in solar system.I am interested second one , from finding same condition of stars equal with sun and planets same to earth,to comparing found planets with our system planets , finally I want to know if they find terrestrial planets or gas planets . it was in [[NASA - NASA'S Kepler Mission Discovers Its First Rocky Planet -by Ken Kremer (on January 10 2011]]that new rocky planet was found .the characteristics was not clear introductionlly it is wonderful ,properties shows it’s orbit is near to star less than Roche limit .other one found two times bigger than jupiter .only it was in one picture that was shown the orbital of planet in photo , never we can got picture fron eclips of any planet orbital .

--78.38.28.3 (talk) 03:14, 1 May 2011 (UTC)

The goals and objectives of the Kepler mission are described in our article Kepler (spacecraft) in its section Kepler (spacecraft)#Objectives and methods.

The Kepler mission is not designed or expected to detect extraterrestrial life (or life's effects, such as on a planet's atmospheric composition) directly, because we haven't yet thought of a good way to do that for most planets beyond our own Solar system. It is intended to detect planets which might be capable of bearing life "as we know it" by being what we think is the right size, composition and temperature to have liquid water on the surface: this is most likely if it orbits its star within what is called the "habitable zone", though other scenarios are possible.

By gathering more data about other planetary systems using Kepler, it is indeed hoped that we will be able to deduce more about how all such systems form, including our own. We have already been able to deduce much, as described in detail in the articles Nebular hypothesis (see particularly the section Nebular hypothesis#Formation of planets) and Formation and evolution of the Solar System, but there are aspects that are not yet fully understood. Note that all data gathered by instruments like Kepler are not instantly definitive and 100% "correct", but are subject to improvement and correction by further observations.

We had already discovered, prior to Kepler, around 500 Extrasolar planets, orbiting a wide variety of stars, and deduced from these discoveries that at least 10% of Sun-like stars have planets. These figures, however, are likely to be substantial underestimates because our current methods of detection must necessarily fail to detect many planets. Radial velocity observations are limited in accuracy by our current quality of instruments and their locations (Oh, for a Far side observatory!) and, as InverseSubstance alludes to above, the chances of a terrestrial planet transiting its star from our viewpoint, and thus being detectable by missions like Kepler (in whose article the problem is discussed), are only about 0.5%, so by this method we can only detect about 1 planet for every 200 that are actually out there.

All of our current detection methods are biassed towards detecting large planets close to their stars, which means hot Gas giants (unlike our own Solar system's more distant and cooler ones) rather than smaller rocky or Terrestrial planets like Earth; nevertheless our discoveries so far have included some more distant gas giants and some terrestrials, so allowing for the bias, the evidence suggests that terrestrials in earthlike orbits are not uncommon.

Any "pictures" of extraterrestrial planets are, of course, artistic renditions based on the data collected by Kepler (and other projects, not direct photos, but they are (or should be) consistent with our present scientific understanding.

I hope this covers most of what you were asking, as your English is not at all clear. {The poster formerly known as 87.81.230.195} 90.197.66.100 (talk) 06:54, 1 May 2011 (UTC)

English is not my mother tounghe ,thank you . --78.38.28.3 (talk) 09:25, 1 May 2011 (UTC)

Lead Cycle

Hello. Given:

1. Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2NaNO₃
2. PbSO₄ + 2KI → K₂SO₄ + PbI₂
3. PbI₂ + Na₂CO₃ → PbCO₃ + 2NaI

I wrote the net ionic equations. Iodide ions displace the sulfate ions in its lead(II) salt and carbonate ions displace the iodide ions. How does that suggest the ladder of solubility: PbCO₃ < PbI₂ < PbSO₄? Thanks in advance. --Mayfare (talk) 16:05, 30 April 2011 (UTC)

Because solubility is basically the relationship between the energy of attraction between a solute particle and water vs. a solute particle and another solute particle. To put it another way, consider these reactions (broadly speaking):

1. Pb²⁺ + H₂O --> Pb²⁺(H₂O)_n ΔH = X kJ
2. Pb²⁺ + SO₄^2- --> PbSO₄ ΔH = Y kJ

Reaction 1 is solvation and Reaction 2 is precipitation, and since both involve bond-formation events, both will be exothermic. Roughly speaking if Y is more negative than X (that is, reaction 2 is more exothermic than reaction 1), then the substance will be insoluble, because thermodynamically the precipitation reaction is more likely than the solvation reaction. If X is more negative than Y, the substance will be soluble Now, consider this third reaction:

• Pb²⁺ + I^1- --> PbI₂ ΔH = Z kJ

What did YOUR series of reactions say about the relationship between Z and Y; and by the commutative property, what does that say about the relationship between Z and X? (that is, how does Z-X compare to Y-X). What does THAT say about the solubility of PbI₂ versus PbSO₄? If you can answer those questions, you can answer your homework problem here. --Jayron32 19:24, 30 April 2011 (UTC)

Since the respective net ionic equations are:

1. Pb²⁺ + SO₄^2- → PbSO₄
2. PbSO₄ + 2I^- → SO₄^2- + PbI₂
3. PbI₂ + CO₃^2- → PbCO₃ + 2I^-

is it safe to say that the carbonate ion is more reactive than the iodide ion, which is more reactive than the sulfate ion? I'm not currently studying enthalpy. --Mayfare (talk) 15:33, 1 May 2011 (UTC)

Reactive is the wrong word. Think of it this way: Given a choice between water molecules and sulfate ions, the lead will bond preferentially to the sulfate ions, which is why lead sulfate is insoluble. Your second equation tells us that given a choice between sulfate ions and iodide ions, the lead would preferentially bond to iodide ions. So, knowing THAT a) Lead likes sulfate more than water and b) Lead like iodide more than sulfate, the question becomes how does the lead-iodide-water relationship play out compared to the lead-sulfate-water relationship. Lets think of it this way. Say that the lead-water bond is given an arbitrary number of "1" to represent the strength of that bond. Lets say that, because the lead bonds stronger to the sulfate ion, we give it a number of "2". Since the iodide bonds even stronger than the sulfate does, lets give the lead-iodide bond a value of "3". If negative values mean soluble, and positive values mean insoluble, which difference (lead-iodide minus lead-water OR lead-sulfate minus lead-water) gives a more positive value? That's the more insoluble compound. --Jayron32 19:38, 1 May 2011 (UTC)

Maximum possible temperature

I have calculated the root mean square speed equation by plugging in the following figures:

• vrms = 299,792,458
• Mm = 1.008 (the molar mass of monoatomic hydrogen)
• R = 8.3143 (the molar gas constant)
• T = unknown

Of course, the goal was to calculate the remaining unknown quantity, namely the temperature at which monoatomic hydrogen atoms would move at the speed of light. The result was: 3,661,133,160 K. At any temperature higher than this, monoatomic hydrogen would be travelling faster than light, which is impossible and thus means that monoatomic hydrogen cannot be heated further.

Heavier elements would be able to be heated up to higher temperatures, but they would still have an upper temperature limit since their atoms cannot travel at a speed higher than c. So now, my question is: why doesn't this appear in the article "Absolute hot"? Am I making a mistake in my calculations or my assumptions? Thanks.--78.126.167.239 (talk) 16:23, 30 April 2011 (UTC)

It's a neat trick, but its wrong. The RMSS equation is a classical equation, assuming the kinetic energy of a particle is $E_{k}={\frac {1}{2}}mv^{2}$. But if you use it at particle speeds close to the speed of light, you must take relativistic effects into account, and use the relativistic kinetic energy $E_{k}={\frac {mc^{2}}{\sqrt {1-(v/c)^{2}}}}-mc^{2}$. Since that goes towards infinity as $v$ goes towards $c$, you get any $T$ at some smaller $v$. --Stephan Schulz (talk) 16:50, 30 April 2011 (UTC)

Either that, or Einstein's relativity theory is wrong. Isn't this an instance of the logical fallacy which says "your theory must be wrong, because it doesn't work under my theory"?--78.126.167.239 (talk) 16:58, 30 April 2011 (UTC)

(ec) It would just be "my theory vs. your theory" if we were mere philosophers, but in actual fact, we are scientists, and we rely on experimental observations to verify the equation that Stephan Schulz posted. We don't have to resort to "guessing." For example, this collection of experiments validate that the lorentz-corrected energy equation correctly and accurately models the kinetic energy of a moving object. My favorite test is the high-atmospheric muon production, explained in our Muon article. Relativistic theory is valid; its equations predict natural phenomena correctly; and more than a century of scientific measurements have verified such equations to many decimal places. Nimur (talk) 17:15, 30 April 2011 (UTC)

Well, you use the speed of light limit from Einstein, so you should probably use all of that theory. You approach is not consistent with Newtonian physics (because in Newtonian physics 299,792,458 m/s is just another arbitrary speed) or in relativistic physics (because you use the wrong kinetic energy). Moreover, the fact that the relativistic mass is is better described by the relativistic than by the Newtonian formula is easily observed and has been verified over and over and over (and over) again. So, sorry, no bananas. Historically, "or Einstein's relativity theory is wrong" has usually been resolved in favor of the other explanation ;-)" --Stephan Schulz (talk) 17:11, 30 April 2011 (UTC)

Actually there is a limit in the sense that neutrino cooling hastens core collapse in the Type II supernova. The hotter the star "tries" to get, the faster it cools down by shedding neutrinos. The result is that the collapse accelerates rather than slowing down due to heat pressure. Hcobb (talk) 16:57, 30 April 2011 (UTC)

But that is just a limit for the temperature at the core of the hottest stars, not for temperatures in general. --Stephan Schulz (talk) 17:13, 30 April 2011 (UTC)

We can rephrase this in plain English without using any equations: to date, we have not observed any natural behavior which leads us to think that there is an upper bound to the kinetic energy of a particle. We can continue to add more energy to a particle, ad infinitum. The particle's speed continues to increase, but we never see any indication that the particle will ever reach the speed of light (even though it is eking its way towards that speed). We have observed this behavior experimentally in numerous particle accelerator experiments; in fact, we even see this behavior in ordinary (old-fashioned) television tubes. Because most definitions of temperature are essentially average measurements of particle energy, we can conclude that there is no upper bound to temperature, either. Nimur (talk) 17:25, 30 April 2011 (UTC)

Well, there at least aren't any limits to temperature according to special relativity. But getting above the Planck temperature probably doesn't work; see absolute hot. Red Act (talk) 17:59, 30 April 2011 (UTC)

I thought that according to our whole Supersymmetry/Higg's boson discussion, there is an absolute temperature at which electrons, protons, neutrons, W and Z particles all become massless particles moving at the speed of light. Wnt (talk) 20:32, 30 April 2011 (UTC)

There is such a temperature, yes, but it isn't a maximum temperature. If it were, the idea of the universe cooling down and symmetry breaking when it crossed that temperature wouldn't make sense. --Tango (talk) 22:54, 30 April 2011 (UTC)

There is a difference between the kinetic energy of an isolated particle and the temperature of a system of interacting particles. For example the temperature and kinetic energy of every isolated electron is zero, in it's own frame of reference. Any sufficiently hot group of particles that is large enough to reach Thermodynamic equilibrium will be limited by neutrino cooling. Hcobb (talk) 21:44, 30 April 2011 (UTC)

Why? Loss of energy by neutrinos requires something the produce those neutrinos. In a collapsing star, that is nuclear fusion. Just because a group of particles is very hot doesn't mean it will undergo nuclear fusion. For one thing, it would need to be the right kind of particles. --Tango (talk) 22:54, 30 April 2011 (UTC)

Nah... Hcobb is right. Neutrino cooling will happen. What he is neglecting though is that neutrino cooling happens because the neutrino bath's temperature is significantly lower then the star's temperature. There is no physical principle that says this must always be the case, and if it isn't, neutrino cooling won't happen. Dauto (talk) 14:01, 1 May 2011 (UTC)

What's a neutrino bath? And what determines its temperature? --Tango (talk) 16:50, 1 May 2011 (UTC)

The neutrino bath is the collection of neutrinos present in the environment from other sources. In our universe most of the neutrinos come from the big-bang and have a temperature of about 1.8 Kelvins, which is cool enough to allow for neutrino cooling :P. Now, imagine a star inside of a box made of neutrino-reflective walls. Once the star starts neutrino cooling, it will fill the box with a high temperature neutrino bath which will reach a thermodynamic equilibrium with the star's core. Once that equilibrium is reached, neutrino cooling stops. Dauto (talk) 18:34, 1 May 2011 (UTC)

Preparing DNA sample for crystallography?

I'm looking for lay descriptions of the above process - don't need heaps of detail - I've found recipes for deriving DNA from biological material, but how do you then get that goop ready for crystallography? I'm esp. interested in the early days of DNA crystallography.

Thanks Adambrowne666 (talk) 22:05, 30 April 2011 (UTC)

This paper from 1969 describes the crystallization of DNA, and while that's certainly not the earliest it may meet your requirements. The steps are basically:

1) Extract the DNA, dissolve into water, and sonicate (to break apart) any clumps

2) Add sodium cacodylate to high concentration (up to 0.5M) and then add ethanol dropwise until the DNA precipitates out as a complex

3) Filter out the precipitate (DNA), heat it rapidly (to 75C) to melt it

4) Slowly cool the DNA and it crystallizes out!

Hopefully that is what you wanted Micah J. Manary (talk) 03:05, 1 May 2011 (UTC)

Reference:

CRYSTALLIZATION OF DNA FROM DILUTE SOLUTION

G. Giannoni,† F. J. Padden, Jr., and H. D. Keith

BELL TELEPHONE LABORATORIES, INC., MURRAY HILL, NEW JERSEY

Thanks very much, Micah; it's ideal - meets my requirements exactly. Adambrowne666 (talk) 05:43, 1 May 2011 (UTC)