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April 30

Likely that Antarctica has large amounts of oil and gas reserves?

Is it likely or unlikely that antarctica has large amounts of oil and gas reserves? Are most major energy resources already found? Pass a Method talk 00:20, 30 April 2013 (UTC)[reply]

There's definitely coal to be found there, as the land mass once had a much milder climate (before Australia broke away from it). Where there's coal, there's often oil as well. More research needed, though. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:59, 30 April 2013 (UTC)[reply]

Actually, the article you linked describes Antarctica's natural resources. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:04, 30 April 2013 (UTC)[reply]

According to [1] it is likely that it has some reserves, but they would be very difficult, both technically and politically, to exploit. 202.155.85.18 (talk) 06:23, 30 April 2013 (UTC)[reply]

Agreed. The problem with drilling through glaciers is that they move quickly enough to destroy the well. So, you'd need to drill where there are no glaciers, like a rock outcrop, or where the glacier isn't moving, like in a crater. As for the environmental risk, any oil spilled won't be cleaned up by natural forces any time soon, like it would in a more temperate climate where bacteria can get at it. So, you might have an oil slick on the glacier for thousands of years, until it eventually drains to the sea. StuRat (talk) 10:10, 1 May 2013 (UTC)[reply]

I know, let's warm up the atmosphere to melt all those fangled glaciers, by releasing a whole lot of carbon dioxide. That should solve both problems. Plasmic Physics (talk) 08:14, 2 May 2013 (UTC)[reply]

Brilliant, let's suck up the last splosh of hydrocarbons to be really sure that the planet fries and color the ice beaches black while at it ;-) Electron9 (talk) 00:18, 5 May 2013 (UTC)[reply]

Abwehr

In the article "Operation Mincemeat", it says that "The Germans had the means to read the letter without opening the envelope". How did they do it? 24.23.196.85 (talk) 06:09, 30 April 2013 (UTC)[reply]

I wouldn't know, but I can think of a way: I have made etched copper electronic circuit boards from masters printed in electronics magazines. Such masters have been printed in ordinary black ink, with the text on the other side of the page printed in green or blue ink. Making the etch circuit involves coating the copper with a photo-sensitive varnish, placing the magazine page on top, and then exposing it to the sun for 30 minutes or so. The sun's ultraviolet light is blocked to a usefull degree by the black ink but passes through the paper and green ink, and hardens the photo-sensitive varnish. You then wash it with an alkaline solution and only the varnish not exposed to unltraviolet remain. You then dunk it in acid to etch away the copper not protected by varnish. The result is not all that good as the paper causes some blurring, and the contrast is not really sufficient, but it works.

Therefore the Germans could have read letters by exposing photographic film to ultraviolet light (or the suns' rays) passing through the envelope. Making the paper wet would help, and using an ultraviolet filter to block visible light & infrared would have been a good idea.

Ratbone 124.178.140.70 (talk) 06:45, 30 April 2013 (UTC)[reply]

"Darned" if I know for sure, but one method floating around is to slide two knitting needles under the flap, one on each side of the letter and twirl them so that the paper becomes rolled up tightly enough to remove without disturbing the seal too much. I saw this on some TV show or movie. Another claim is that liquid freon can be used to make the envelope transparent for a short while without leaving a trace afterwards. Clarityfiend (talk) 07:01, 30 April 2013 (UTC)[reply]

I've actually used that once. I was down to my last envelope, I needed to post a check, stuck it into the envelope and sealed it...then realised that I'd forgotten to sign it...and it was my last envelope, so I couldn't tear it open. It was really easy to extract the check from the envelope, using a pencil with the sticky strip from the top of a post-it note wrapped around it sticky-side-out. You put the pencil in through the small gap at the top corner of the envelope where there is no adhesive...roll the paper around it into a tight cylinder - then remove the pencil. Putting the the (now signed) check back into the envelope was just as easy. I don't think the "tampering" would be evident...although maybe the check would still have some curvature to it when it arrived at the destination that might alert a suspicious victim...maybe not though. SteveBaker (talk) 20:25, 30 April 2013 (UTC)[reply]

Letter-unsealing without leaving obvious traces was a very well-developed skill by 1940. The British opened and read a large portion of mail going from the US to neutral countries in Europe, and used the info the found to uncover German espionage rings using mail drops in Portugul and elsewhere. (Original research) I tried "steaming" an envelope open using a teakettle, but the paper is quite soggy before the adhesive lets go. A sharp knife, even heated also did not open a letter without tearing the paper. Apparently the spies who received the mail did not detect that it had been opened, so WW2 counterespionage folks must have had really good teakettle skills, or other special chemicals.E-How results on Google just suggest the teakettle routine. It would seem like a good idea to make the letter "tamper evident" by using ink which would smear or blot if the envelope were steamed. Suggestions that the envelope can be opened when it has been in a deepfreeeze run counter to real life experience that letters which have been in a mailbox at -20 F are still securely sealed. Edison (talk) 19:08, 30 April 2013 (UTC)[reply]

It's possible that the material used for envelope adhesive has been improved since the 1940's. I don't see any indication of that in our articles though. SteveBaker (talk) 20:25, 30 April 2013 (UTC)[reply]

The article goes on to describe how the Spanish read the documents - "The Spanish removed the still-damp paper by tightly winding it around a probe into a cylindrical shape, and then pulling it out between the envelope flap, still closed by a wax seal, and the envelope body." and later returned them to the envelopes - "The documents were re-inserted into their original envelopes, reversing the process by which they were removed,". This sounds similar to the knitting needle technique mentioned by Clarityfiend. Note that the British were aware that the documents had been tampered with. Mikenorton (talk) 21:03, 30 April 2013 (UTC)[reply]

What I don't understand is how they could get the paper flattened out afterward so it wasn't apparent it had been rolled up/tampered with. Stick it under some heavy books perhaps? Clarityfiend (talk) 02:03, 1 May 2013 (UTC)[reply]

What I don't understand is how anyone is supposed to get even the faintest hint of what this question is about from the heading "Abwehr". I think I'll run some courses in creating useful and meaningful headings. -- Jack of Oz ^[Talk] 19:22, 1 May 2013 (UTC)[reply]

If only there were some sort of readily-available comprehensive online reference work that might help you to resolve such difficult situations....

Abwehr as a section title makes perfect sense, as the question is about techniques used by the World War II German intelligence service of that name. If there were any confusion about the context or meaning, the OP helpfully linked to the article, Operation Mincemeat, that prompted his question.

Could the OP have used a longer or more detailed section header? Sure—but it still names a relevant subject area, it has the virtue of brevity, and it's a damn sight better than the ever-popular generic "Question". Is knowing the name of a German intelligence service a bit of specialist knowledge? Sure—but it's the sort of semi-specialist knowledge one might expect a reader who could answer the question to have. Someone who doesn't recognize the term Abwehr almost certainly isn't going to be able to write knowledgeably about WWII-era German spycraft. TenOfAllTrades(talk) 20:41, 1 May 2013 (UTC)[reply]

You’re parsing the issue completely backwards. You’re looking at the question and understanding why the OP chose the word Abwehr for the header. We can pretty much always work out how people come up with their strange headers, e.g. why the previous question was headed "Likely"; or why a question that's about the numbers of people killed and injured in a mining disaster in Alabama in 1917 would be headed "America". That's not rocket science, but does that help anyone to know what the question is actually about? Well, not at all. The header is supposed to tell us what the question is about, at least broadly. Searching for this question in 5 years time, who is going to remember that it's headed "Abwehr", given that that word never appears in the question? You might say the header meets the "broadly" criterion. But it doesn't. The question was framed in the context of Operation Mincemeat (a historical event), but it was placed on the Science desk because it’s primarily about the technical aspects of espionage, and only marginally about the history of the Abwehr. Some respondents have shared their own relevant experience of the science involved, and I very much doubt any of them got that experience from the time they spent as members of the Abwehr. -- Jack of Oz ^[Talk] 21:29, 1 May 2013 (UTC)[reply]

(un-indent) Thanks for the input, everyone! So I gather that the knitting-needle trick was the method actually used -- but if I ever use this in my novel writing, I think I'll go with the UV projection technique because it has more of a sci-fi flavor to it. 24.23.196.85 (talk) 00:00, 3 May 2013 (UTC)[reply]

Current = Conductance * Potential difference

V for potential difference, R for resistance, I for current, and G for conductance.
We know, I ∝ V. After removing proportional sign, it is I = G*V. If we write, V ∝ I, then after removing proportional sign, it is V = R*I. Please, check whether my above statement is right or wrong. Scientist456 (talk) 07:55, 30 April 2013 (UTC)[reply]

Yes, you'll see all that at Electrical resistance and conductance. Dmcq (talk) 08:29, 30 April 2013 (UTC)[reply]

Energy in Prokaryotic Cells without mitochondria

We know that a eukaryotic cell gets its energy requirements from a cell organelle called mitochondria. But what's about a Prokaryotic Cell, from where it get its energy requirement?Scientist456 (talk) 09:10, 30 April 2013 (UTC)[reply]

All prokaryotic cells are capable of some form of cellular respiration (many if not most are capable of aerobic respiration, just like eukaryotic cells). They simply don't have specialized organelles in which to carry out the reactions. This is similar to how prokaryotic cells can carry out transcription and DNA replication even though there is no nucleus. If I removed all the interior walls from your house, you would still have a bed to sleep in, even if you could no longer call it a bedroom. Someguy1221 (talk) 09:22, 30 April 2013 (UTC)[reply]

Are you aware of the Endosymbiotic theory? Under that theory mitochondria were originally separate prokaryotic organsisms that formed a symbiosis with another organism. I know this isn't directly relevant to your question, but as you are thinking about related things I thought it might interest you. Equisetum (talk | contributions) 09:44, 30 April 2013 (UTC)[reply]

Binomial plant names contain special character in url (×) and give a 404 if you put a regular 'x' instead

This question has to do with all plants on wikipedia that are hybrids, and therefore can contain a special symbol called a cross (×) as part of their binomial name. While it's correct to have the special character '×' in the binomial name of a hybrid, it's neither consistent across all plants on Wikipedia, nor is it terribly useful for users trying to find a certain plant when Wikipedia gives a 404 if I put an 'x' instead of an '×'.

For example, I can reach the plant Rubus × loganobaccus with this special character (×) (http://en.wikipedia.org/wiki/Rubus_%C3%97_loganobaccus), but not with an actual 'x' (http://en.wikipedia.org/wiki/Rubus_x_loganobaccus).

Could you tell me how I can make links to plants that contain this special character where a person could put a normal 'x' instead?

Thanks Mskogstadstubbs (talk) 12:25, 30 April 2013 (UTC)[reply]

You can do this using a redirect. I've done it for the requested page, to do so on other pages (after reviewing the linked guideline) just create a page with #REDIRECT [[Page name]] Jebus989 ^✰ 12:43, 30 April 2013 (UTC)[reply]

I've taken the liberty of raising this issue at the Village pump (technical) [2]. Wnt (talk) 14:32, 30 April 2013 (UTC)[reply]

...and I've now filed a bug request for someone to update the search. Andrew Gray (talk) 15:09, 30 April 2013 (UTC)[reply]

Unsolved numericals related to electricity

Today, I solved hundreds of numericals related to electricity, but I got stuck at these three -
1. If electrons are caused to fall through a potential difference of 10⁵ volts, determine their final speed if they were initially at rest. (ANS: 23 * 10⁷ m/s)
2. An electric bulb connected to a 220 volt supply line draws a current of 0.05 A. Calculate the amount of coulombs per second flowing through the bulb. (ANS: 0.15 C/s)
3. Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watts of power. What should be the power dissipated if the resistors are connected in parallel across the same e.m.f? (ANS: 90 watts)
This is not my homework. I tried my best to solve these, but I couldn't. Someone give me hints so that I can solve these. Thank you very much! Scientist456 (talk) 16:47, 30 April 2013 (UTC)[reply]

Please show us your work, and how you arrived at a different answer from the stated one, or how far you got before you were stumped, rather than just giving the stated problem and the stated answer. Then folks here will be delighted to help you. We would even point out if a stated answer is incorrect! Edison (talk) 18:47, 30 April 2013 (UTC)[reply]

I'll take number 3. E across each resistor (in series) = E_total/3. Current through each resistor R is the same (in series) so call that I; where (E_total/3)/R = I. so the power for each resistor = (E_total/3)*I And so we have original total power = (power for each resistor)*3 = (E_total/3*I)*3=10, says the problem.

Now, if you connect the resistors in parallel rather than series, each resistor sees E_total, therefore the current through each resistor is I*3 (because the new voltage, E_total, is 3 times the original, E_total/3, therefore the current will be 3 times as much) = ((E_total/3)/R)*3. so the power for each resistor will be E_total*(I*3), and the total for all three will be (E_total*(I*3))*3 which is nine times the original power ((E_total/3*I)*3) which was defined as 10 in the problem, so the new power equals 90.

or you could do it by looking at the total. in the original you have E_total across a total resistance of 3R (where R is the resistance of 1 resistor), giving you a current of I = E_total/3R, and power=E_total*I = E_total*(E_total/3R)=10 watts. then you've switched to E_total with a total resistance of R/3 (because they're in parallel) so you've got a current of I = E_total/(R/3), and power=E_total*I = E_total*(E_total/(R/3))=90 watts, again substituting from the original power.

or, once you get an intuition of where it derives from and how it works, you don't need to break it up into each resistor/current any more, you just use power=(E_total squared)/R_total; E_total remains the same, R_total goes from 3R to R/3, so new R_total = old R_total/9 so the power goes up 9 times. (That's a general rule, by the way; if you have N identical resistors in series and you switch to parallel, the total power goes up by N squared. Or down by N squared, if you switch from parallel to to series; if you understand the above you'll see how that works, the voltage and current on each resistor each go up N times) Gzuckier (talk) 18:49, 30 April 2013 (UTC)[reply]

Falling electrons: volts are joules per coulomb. You know how many volts you have, how many coulombs per a single electron, therefore you can calculate the kinetic energy the electron must obtain. You know its mass so you can solve for velocity. I suspect there's an easier way, but that will do it.

An ampere is a coulomb per second, so the answer you parenthesize there mystifies me entirely.

For resistors in series, the resistance adds up. For resistors in parallel, the conductance adds up. Just think of them intuitively as narrow channels with water flowing through them. Wnt (talk) 19:53, 30 April 2013 (UTC)[reply]

The book answer for 1 has the electron accelerating to 3/4 the speed of light. Is it still ok to ignore relativistic effects at that speed? I agree that the book answer to 2 is too high, and 3 is pretty straightforward given Ohm's law and the formula for power and for series versus parallel resistances. Edison (talk) 22:05, 30 April 2013 (UTC)[reply]
- I think that the answer to 2 may depend on the lamp being AC, but I can't see how to get a value of 0.15 C/s either. The _net_ number of coulombs per second is zero (assuming there's no DC component), ~~the _total_ number of coulombs per second is 0.283 (peak current = rms current * sqr(2), total C/s = peak current * 4)~~. Tevildo (talk) 23:20, 30 April 2013 (UTC)[reply]
  - Sorry, not thinking properly. 0.283 is the number of coulombs per _cycle_, so the number per second will be 14.14 at 50 Hz and 16.97 at 60 Hz. Tevildo (talk) 23:29, 30 April 2013 (UTC)[reply]

Our article on kinetic energy gives both Newtonian and relativistic formulae, so the OP can work them both and compare the results if desired. (because this is a homework question I've been holding off on doing the calculation) But our article on electron says it has a rest mass of 511000 electron volts, and we've just handed it 100000 eV in kinetic energy - if half its mass is kinetic energy it should be pretty relativistic. Wnt (talk) 23:41, 30 April 2013 (UTC)[reply]

I have written here the same questions and answers written in the book. I also think the second Q is wrong as the Wnt pointed. For the first Q, I used the formula: charge (q)* potential difference (V) = Half mass (m)* velocity (v) squared. But after using this, I found that the answer differs a little bit. Then, what should I do? Scientist456 (talk) 00:24, 1 May 2013 (UTC)[reply]

According to [3] (the formula under "thus the work expended" and/or "for higher speeds") 1/sqrt(1-v^2/c^2) = (total mass/rest mass). The rest mass is 511 keV, the kinetic mass is 100 keV. So (1-v^2/c^2) = (511 keV/611 keV)^2 = 0.837; v/c=sqrt(1-0.837) = 0.405 * 3*10^8 m/s = 1.21*10^8 m/s. Hmmm, but they got 2.3*10^8? I'll leave it to you to figure out where I was off - since it's a homework question and all, yeah, that's why. :) Wnt (talk) 01:10, 1 May 2013 (UTC) (some numbers corrected per response below)[reply]

Electron's mass is 511 keV, not 51.1 keV, making that electron safely non-relativistic. Dauto (talk) 14:03, 1 May 2013 (UTC)[reply]

Wait a second, the problem states 100,000 eV, not 50,000 eV which makes the electron borderline relativistic. The classical answer will be off by about 10 to 20%. Dauto (talk) 14:55, 1 May 2013 (UTC)[reply]

Dang it, that should teach me to try to do math in a rush when I'm half-hearted about whether I ought to! Worst part is, I corrected my math above, and now it's way off from the book answer at the beginning. Wnt (talk) 15:58, 1 May 2013 (UTC)[reply]

DNA

In a DNA double helix how do you tell the Template strand from the Information strand? Are they different in some way? — Preceding unsigned comment added by GurkhaGherkin (talk • contribs) 17:02, 30 April 2013 (UTC)[reply]

From our DNA article: A DNA sequence is called "sense" if its sequence is the same as that of a messenger RNA copy that is translated into protein. The sequence on the opposite strand is called the "antisense" sequence. Both sense and antisense sequences can exist on different parts of the same strand of DNA (i.e. both strands contain both sense and antisense sequences). Both strands can contain 'template' sections. AndyTheGrump (talk) 17:31, 30 April 2013 (UTC)[reply]

You might also consider during replication, when you can tell the newly synthesized strand from the old strand by labelling it with 13C/15N (the Meselson-Stahl experiment; the technique is actually still used more than 50 years later [4]) There are also some differences even in naturally grown cells in DNA methylation and histone posttranslational modification when the strand is first synthesized, and occasionally afterward. Wnt (talk) 19:46, 30 April 2013 (UTC)[reply]

The short answer is that the location of the transcription start site tells you which is which. The TSS and associated sequences (like the TATA box) are sequence specific, and not identical to their reverse complement, so on only one strand will it be of the correct sequence to start transcription. As Andy mentions, either strand can (and does) contain TSSs for various genes, but for any particular gene the initiation sequences will be on one strand or the other. -- 205.175.124.30 (talk) 01:07, 1 May 2013 (UTC)[reply]

Thanks guys. I appreciate it.GurkhaGherkin (talk) 14:08, 1 May 2013 (UTC)[reply]