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Is banana not good for daibatic people? — Preceding unsigned comment added by 219.73.12.77 (talk) 03:31, 22 November 2011 (UTC)[reply]

Ripe bananas are relatively high in sugars compared to other fruit: [1]. But, it really depends on what you do with them. Just eating a whole bunch as a meal might be bad, but using one to sweeten your oatmeal or cereal is probably fine, since you are then using the banana in place of another form of sugar, rather than in addition to it. The banana has lots of nutrients in it which many other sweeteners lack, so in that sense it's a good choice. StuRat (talk) 03:37, 22 November 2011 (UTC)[reply]

Given the danger inherent in mismanaging your diabetes, if you have concerns about ANY food and its effect on your diabetes, please ask a doctor or other qualified medical provider, do NOT trust the advice of random people on the internet with your health. --Jayron32 03:50, 22 November 2011 (UTC)[reply]

They never said they have diabetes. StuRat (talk) 04:22, 22 November 2011 (UTC)[reply]

No, but on the off chance they are, we shouldn't be telling them that anything is OK without first consulting with a medical professional. If they die because your advice that something is "probably fine", well done! I'd rather err on the side of not killing people. --Jayron32 04:27, 22 November 2011 (UTC)[reply]

See http://www.askdrbernstein.net/.

—Wavelength (talk) 04:29, 22 November 2011 (UTC)[reply]

So what would happen if the spin of a large object reached relativistic speeds? I will ask about the following two circumstances:

• A non-black hole object (e.g., a pulsar) which started to spin this fast. Would there necessarily be a lot of drag on the star because the inner parts could spin faster than the outer parts? What would be the result of the drag - a quick slow down?
• A non-charged black hole object, whose outer edges reached relativistic speeds due to the influx of matter? I'm inspired by this article [2] which describes the observation of a black hole whose spin at its event horizon is 800 rps (!). What would happen with the shape of the event horizon? Could this cause a naked singularity? Would the black hole also slow down due to its drag, or is this not possible due to the other relativistic effects?

71.58.69.237 (talk) 04:25, 22 November 2011 (UTC)[reply]

I'm no expert, but frame-dragging would be one result. Clarityfiend (talk) 10:32, 22 November 2011 (UTC)[reply]

See Kerr metric for a description of what happens in the black hole case. One notable feature of a rotating black hole is the presence of an ergosphere, within which spacetime is dragged around at a speed greater than the speed of light. There are overextreme Kerr solutions in which a naked singularity appears, but my guess is that such situations don't actually wind up existing in reality. Red Act (talk) 14:02, 22 November 2011 (UTC)[reply]

Wouldn't a "non-black hole" object rip itself apart at such speeds ? StuRat (talk) 15:09, 22 November 2011 (UTC)[reply]

I'd go more fundamentally, would this be even possible for a non-black hole object? What energies would be required for the spin-up? --Ouro (blah blah) 17:28, 22 November 2011 (UTC)[reply]

Gravitationally, a neutron star is capable of reaching a small fraction of a relativistic spin (e.g. 1-30% of c at the equator, depending on mass) without the rotation being so strong that it simply flies apart. The energy required is astronomical, but not entirely crazy, e.g. 1% of the energy emitted during a supernova. The fastest known pulsar, PSR J1748-2446ad, is actually believed to be moving at 24% of the speed of light at its equator. Dragons flight (talk) 18:03, 22 November 2011 (UTC)[reply]

The record for a pulsar is 716 rps. read millisecond pulsar. Dauto (talk) 02:02, 23 November 2011 (UTC)[reply]

As far as the original question goes, why would anybody expect the inside to spin faster then the outer layers? Dauto (talk) 02:05, 23 November 2011 (UTC)[reply]

I think the OP's reasoning was that, since the inner parts have a smaller radius, they can have a higher angular velocity without their linear velocity exceeding the speed of light. I doubt that would actually cause the inner layers to spin faster, though - aren't neutron stars extremely rigid as a result of their high density? (They're described as being made of a neutron-degenerate gas, but I don't think our usual understanding of what a gas is is applicable.) --Tango (talk) 12:10, 23 November 2011 (UTC)[reply]

On the contrary, there is a very real possibility that the interior of the neutron star, being made of degenerate matter, will be some kind of superfluid. Dauto (talk) 14:51, 23 November 2011 (UTC)[reply]

Why does the result of mixture of some metrials make a mixture with new properties? Exx8 (talk) 19:17, 22 November 2011 (UTC)[reply]

I'm not sure I understand your question, but you may find some useful answers in the following articles:

• Suspension (chemistry)
• Colloid
• Solution
• Mixture

If you can restate your question or give more context, maybe someone can give you a more specific answer. --Jayron32 20:06, 22 November 2011 (UTC)[reply]

I suspect that you are using the word "mixture" in the general sense, meaning "the result of mixing any two or more things together", and not in the chemistry sense, where it specifically means a mixture where no chemical reactions have occurred between the constituents. Am I correct ? StuRat (talk) 20:29, 22 November 2011 (UTC)[reply]

Given that certain isotopes of curium are fissile, would it be possible to create a nuclear weapon using curium as its fissile material? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 19:52, 22 November 2011 (UTC)[reply]

The article and section Fissile_material#Nuclear_fuel covers the additional requirements needed for a fissile material to be useful as a nuclear fuel. That would imply that fissility is not, of itself, a sufficient property for such use. Since there would therefore be some fissile materials which are not useful for nuclear fuel, those isotopes of curium may or may not be useful in that regard. I'm not familiar enough with the specific properties of those isotopes of curium, but if you are you can compare it to that information and work it out for yourself. --Jayron32 20:03, 22 November 2011 (UTC)[reply]

You might be better off asking if it would be useful in a 'practical' nuclear weapon. It would need a active cooling system and much sheilding. --Aspro (talk) 20:11, 22 November 2011 (UTC)[reply]

Why is that? It has a half-life of 8500 years - much longer than that of some other radioactive isotopes used in nuclear weapons. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 20:21, 22 November 2011 (UTC)[reply]

? Plutonium-239 has a half-life of 24,100 years. The shorter lived isotopes are considered a contaminant. A party balloon's worth of tritium gas is used in fusion devises, but that has to be replenished every 6 years or approximately, half its half life. It beta decays (and in these quantities) without much heat. The rest of the short half-life isotopes are generated after initiation – from lithium.

Presumably you would produce Cu-235 or Cu-248, which, so far as I can tell, are not processes that would necessarily create lots of the shorter-lived isotopes of Curium like 232-234. I'm not sure you'd need active cooling or heavy shielding, but it would depend on which isotopes we are talking about and assumed impurities. --Mr.98 (talk) 16:49, 23 November 2011 (UTC)[reply]

I don't see any reason to think you couldn't do it in theory if you could isolate relative pure amounts of certain isotopes of curium at the exclusion of some of the other isotopes. Many of the isotopes of curium have what seem to be superficially positive properties: low critical mass requirements, high neutron absorption cross sections, long half-lives. There are likely practical difficulties. It is not easy to produce in large quantities. The amount of spontaneous fission in some of the isotopes would make pre-detonation an issue (in the same way Pu-240 does for plutonium). The main practical difficulty is that curium is difficult to produce, and the processes that produce it also produce plutonium in great quantities, so why go with curium if you already had plutonium? This ignores the possibility of acquiring 10 kg of curium through some other source, though. --Mr.98 (talk) 20:39, 22 November 2011 (UTC)[reply]

(EC) Doubtless it would be possible, but it would likely also be pointlessly difficult and expensive. As the article on the element mentions, most if not all isotopes of curium can only be obtained by treating other (already-radioactive) elements inside a nuclear reactor, so amongst those elements and the reactor elements we already have material more easily and cheaply available for such a purpose. Also, the yields of the various curium isotopes so produced have (up to now) been mostly far too little to make bombs. It would be like making matches by carving each pine tree into a single matchstick (as per the cartoon we've all probably seen sometime). {The poster formerly known as 87.81.230.195} 90.197.66.65 (talk) 20:41, 22 November 2011 (UTC)[reply]

It is worth noting that as far as I can tell, curium is not a safeguarded material, unlike obviously uranium and plutonium, but also neptunium and americium, which are more obscure fissile materials. This indicates probably that nobody is too worried about it, probably for good reason — all of the ways of producing it are very cumbersome and create only very tiny amounts relative to other fissile materials produced, and there do not appear to be weapons-level quantities available. The ways of producing it from plutonium seem like they only produce isotopes of very short half lives. --Mr.98 (talk) 00:41, 23 November 2011 (UTC)[reply]

If one wants to look towards the impossibly exotic, then nickel-56 promises the best firework display. Get a good pair of Ray-Ban's first though!--Aspro (talk) 22:35, 22 November 2011 (UTC)[reply]

In my introductory physical class, which I like to call "Physics - The Science of Absolute Truth" we were asked solve this problem:

• Two separated solid spheres comprise the "plates" of a capacitor.
• They are made of the same solid, conductive material (pick one, aluminum, i.e.)
• The spheres have the same density and temperature.

• Their electrical inputs are center-fed at the inside the spheres.
• You can ignore the influence of the entry points for the feeds.

• The first sphere has radius s
• The second has radius t
• s is constrained to be >= 2t
• The centers of the "plates" (spheres) are separated by a distance > s + t but < 2s
• Call that separation u (between the sphere centers).

Calculate the capacitance.

I have no clue. Can you point me in the direction of the calculation? I got as far as the volume of each sphere, the surface area of each sphere and the charge on the sphere under various conditions. I also tried to relate the properties of a single sphere to a point charge, using the center of the sphere as the "point". But that seems to get me nowhere re: the capacitance.

My professor, Dr. Jack, refuses to help. Says it would "give it away". The net seems to provide a lot of reference material on two spheres, one inside the other, and a few on variant shapes of flat plates. Other than that, I don't seem to find anything.

I'm pretty sure this problem can be solved by a method akin to the Method of image charges. Dauto (talk) 14:44, 23 November 2011 (UTC)[reply]

As "talk" said, some of the information may be superfluous and intended to confuse. Jack is a crafty dodger.

Got this in reply to my question at the capacitance talk page:

• "Centre-fed isn't entirely clear; the capacitor can only be charged by transferring charge from one sphere to the other - along some actual path external to the spheres. While u > s + t is a necessity, s >= 2t and u < 2s seem irrelevant - likewise the density.
• "Have you seen the article bipolar coordinates? What you actually want is bispherical coordinates, but the first-mentioned article has a much clearer picture. The spherical surfaces are your equipotentials (and two of them correspond to your capacitor plates - note that these are not centred on the poles of the coordinate system). The flux paths are along the toroidal surface (which meet the spherical surfaces at right-angles). Find V(r), thence E and Vc, thence D and Q --catslash (talk) 02:22, 22 November 2011 (UTC)

To which I replied thusly:

• " ""s >= 2t and u < 2s seem irrelevant - likewise the density."" Relevant or not, they were part of the problem statement given to me.

• "The """external""" charge, so called, was imparted to the sphere at its center. For example, if it was coax, it would run inside the sphere and terminate in the center. Center-fed I guess is the best phrase I could come up with."

• "So you gave the formula for the solution, or not?" — Preceding unsigned comment added by 72.95.47.115 (talk) 10:27, 22 November 2011 (UTC)

And talk replied:

• "Not. On closer inspection, it seems that these coordinate systems do not give you an exact solution in 3D (it works in 2D - cylinders rather than spheres). Perhaps s >= 2t and u < 2s are relevant because an approximate solution is expected. Anyway take your question to the science reference desk." --catslash (talk) 15:23, 22 November 2011 (UTC)

And I:

• "Thanks! I appreciate the effort and your honesty." — Preceding unsigned comment added by 134.223.116.201 (talk) 17:28, 22 November 2011 (UTC)

I have made some minor edits and additions to my comments above, although I have not modified theirs.

Any thoughts? — Preceding unsigned comment added by 72.95.47.115 (talk) 23:19, 22 November 2011 (UTC)[reply]

From a physical understanding: Wouldn't the charges tend to concentrate on the points as far from the other sphere as possible, in coordination with minimizing the repulsion between like charges on one sphere (i.e. not all the charge would be located at the one point: there would be a gradient with lowest charge at the point nearest the other sphere and highest charge at the point farthest from the other sphere). How would a sphere then still be an equipotential surface, as would be a conductive sphere in the absence of an external field? Many capacitor problems have classic trick solutions, but this one does not ring a bell. There are lots of problems on the web for parallel plate caps, for concentric sphere caps, and for spheres isolated in space, but not so many for separated spheres of unequal radius. Are you (or anyone) able to access a ref from our Capacitance article, "Note on the Capacitance of Two Closely Separated Spheres?" It sounds like it might be right on point, if it can be extended to spheres of unequal radius. Edison (talk) 23:33, 22 November 2011 (UTC)[reply]

Rereading my post above, I was describing the situation as like-charged spheres, whereas the situation for a charged cap would be opposite-charged spheres. Edison (talk) 15:30, 23 November 2011 (UTC)[reply]

I guess I don't understand what the lines of force would look like on the back (distal) hemispheres of the two. The lines on the proximal side seem straightforward enough. But then you start looking at the interaction between the two hemispheres on the same sphere, and that logic seems to break down. Maybe the answer is to solve the 4 two-hemisphere problem first ( the four problems being for hemispheres of sphere 1 and sphere 2 respectively, 1-distal/2-distal, 1-proximal/2-proximal, 1-distal/2-proximal, 1-proximal/2-distal ).

Attempting a solution of the polar hemispheres does not seem helpful to this problem. It might be interesting in its own right, however.

Based on the abstract, I'd say that article is exactly what I am looking for. I'll have to find it at the library. $32 for the PDF is a little steep for a lowly student. Thanks! — Preceding unsigned comment added by 72.95.47.115 (talk) 02:21, 23 November 2011 (UTC)[reply]

You might want to give a little back to the educational system by teaching your professor the difference between "i.e." and "e.g." Deor (talk) 09:32, 23 November 2011 (UTC)[reply]

• Very good! I did not know that. For those who share my ignorance (which I freely admit):

e.g. stands for the Latin phrase “exempli gratia,” which means “for the sake of example.” You should use it when presenting examples or more possibilities for the term in question. An easy way to remember this is to associate e.g. with “example given.”

i.e. Stands for the Latin phrase “id est,” which means “that is.” You should use it when explaining or rephrasing a sentence. Usually it has the same meaning as “in other words.”

• I thought "i.e." stood for "in example" which was erroneous. — Preceding unsigned comment added by 134.223.116.200 (talk) 12:43, 23 November 2011 (UTC)[reply]

I'm pretty sure that this problem can be solved by a method akin to the Method of image charges. Dauto (talk) 15:22, 23 November 2011 (UTC)[reply]

The "images" approach is the trick I was alluding to above. The book "Applied electromagnetics" (1978) by Martin A. Plonus has a good section on solving various capacitor problems, but it is buried somewhere in the attic. Edison (talk) 15:30, 23 November 2011 (UTC)[reply]

I think the images approach will work, even though most of the examples I see are a point charge and a sphere interacting. I guess I'll give up on my quaternion approach. It wasn't bearing much fruit anyway. Thanks again. — Preceding unsigned comment added by 72.95.47.115 (talk) 18:43, 24 November 2011 (UTC)[reply]

If the OP solves it or is given the solution later by the prof, it would be gratifying to have it posted here. Edison (talk) 19:51, 24 November 2011 (UTC)[reply]

DO aliens Really EXIST!? — Preceding unsigned comment added by Akshar218 (talk • contribs) 23:48, 22 November 2011 (UTC)[reply]

Probably. 109.149.81.210 (talk) 00:21, 23 November 2011 (UTC)[reply]

Yes they do. And we are they. The real problem is finding the natives. — Preceding unsigned comment added by 72.95.47.115 (talk) 02:23, 23 November 2011 (UTC)[reply]

The odds that there is intelligent life elsewhere in the universe at the moment are probably equal to 1 (100%). The odds that we will be able to communicate with them in a meaningful way are probably a lot lower (though estimating that requires estimating a lot of things we don't know much about). The odds that we will be able to visit with them and vice versa, based on current physics, before our species (or theirs) goes extinct, is probably a lot lower than that. (That's my interpretation of the Drake equation, anyway.) --Mr.98 (talk) 02:38, 23 November 2011 (UTC)[reply]

Yes they are among us as we speak, cunningly. disguised as Australian musicians (who would have suspected that possible?) Cuddlyable3 (talk) 04:20, 23 November 2011 (UTC)[reply]

Nope, they are working in the California cottonfields.  ;-) Now why ain't there no article about that Merle Haggard hit song (I mean "California Cottonfields")? 67.169.177.176 (talk) 04:57, 23 November 2011 (UTC)[reply]

If you've ever tried to communicate on some discussion page on Wikipedia where there's a dispute you would not rate our chances of communicating with intelligent aliens very highly. ;-) Dmcq (talk) 10:41, 23 November 2011 (UTC)[reply]

do dinosaurs and neandertals exist ? not anymore- which highlights my point that alien life may well have been or will be, but the chance that they are existing in the same moment as you and I in this incomprehensable time span, very unlikely.