Science desk
< November 2	<< Oct \| November \| Dec >>	November 4 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 3

motion

In all standard books motion is defined change in position with time and displacment is also defined as change in position and in many closed path the displacement is consider to be zero , so according to the defination of motion change in displacement is zero hence their should be no motion but we say that body has shown motion .. but how plz explain but do not explain with the help of graph? — Preceding unsigned comment added by Bhaskarandpm (talk • contribs) 05:26, 3 November 2011 (UTC)[reply]

I have a hard time understanding your question, is this what you meant ?

"If an object moves and then returns to it's original position, wouldn't it's motion be zero since that is displacement over time and no net displacement has occurred ?"

In one sense, the answer is yes, the average speed is zero. That is, if it moves north at 100 kph for an hour, then moves south at 100 kph for another hour, returning to the origin, the average speed is zero ((100+(-100))/2 ). Here it might be more useful to determine the average velocity, which, unlike speed, doesn't have a sign (velocity uses a vector instead to show direction). In this case the average velocity is 100 kph ((100+100)/2).

Similarly, while there is no net displacement, the total displacement would be 200 km. In this case the sum of displacements (ignoring signs) over each leg of the trip gives you the total displacement. For more complex paths (regardless of whether they return to the origin or not), you must use the total arc length of the path, not the distance between the starting and ending points, for total displacement and average velocity calculations.

Note that both the average speed and average velocity have their uses. Let's say you are making a multiple day car trip, and want to know how fast you were going. Assuming that the roads aren't all straight and in the same direction, the two values should differ. One tells you how fast your car was moving, which is useful in avoiding tickets and accidents, and the other tells you how quickly you will be able to reach your destination. StuRat (talk) 06:01, 3 November 2011 (UTC)[reply]

Note that StuRat has crossed up "speed" and "velocity"; speed (a scalar) is the signless component (and thus average speed in his example is 100 kph) while velocity (a vector) retains the notion of direction and thus sign. — Lomn 13:08, 3 November 2011 (UTC)[reply]

please explain why a graph won't help. And in answer to your question, if I start off in London on Tuesday morning, catch the train to Birmingham, spent the night there, then go back to London to arrive at home in time for supper, I have moved - my displacement over time has changed. The fact that I'm back where I started is merely a side-effect of my motions (relative to London) cancelling each other out. The displacement isn't zero, except in as much as London (which isn't a fixed point in spacetime) is arriving where I am at the same time as I do. AndyTheGrump (talk) 06:07, 3 November 2011 (UTC)[reply]

Am I misreading these replies, or is there some confusion between distance and displacement? Average speed is total distance travelled divided by time (no negatives allowed), and can never be zero if you have moved at all. Average velocity is a vector quantity and is net displacement divided by time, and can be zero if you return to the start. Dbfirs 08:41, 3 November 2011 (UTC)[reply]

organic chemistry

explain in detail about nucleophiles — Preceding unsigned comment added by Bhaskarandpm (talk • contribs) 06:27, 3 November 2011 (UTC)[reply]

Please read Nucleophile. Dualus (talk) 06:31, 3 November 2011 (UTC)[reply]

A nucleophile is a particle (usually an ion or molecule, but certain elements like chlorine are nucleophiles in itself) that is attracted by the positive charge of an atomic core (nucleus - usually a carbon core as nucleophiles are primarily of interest in organic chemistry). The attraction results from the fact that the nucleophile always has a free electron pair, which acts as a localized "negative charge" and is thus being attracted by localized positive charges (like that of an atomic core). This free electron pair can form a covalent bond between the nucleophile and the carbon ~~core~~ atom. How much more would you like to know? Phebus333 (talk) 22:34, 3 November 2011 (UTC)[reply]

No, that's fairly inaccurate. Nucleophiles are not attracted to the "atomic core". They are attracted to empty or partially empty valence-level orbitals. A nucleophile is basically a lewis base and it is attracted to a lewis acid. The electrons of the HOMO (highest occupied molecular orbital) of the nucleophile will fill into that of the LUMO (lowest unoccupied molecular orbital) of the target carbon atom. The carbon atom can be primed to accept a nucleophile by itself being bonded to an electronegative atom (which will tend to pull electrons away from the carbon, and thus partially "empty out" the LUMO, making it a better electron acceptor). That's what occurs in SN2-type reactions. In SN1-type reactions, there is a carbocation which forms, giving a completely empty LUMO to form a new bond with. --Jayron32 00:29, 4 November 2011 (UTC)[reply]

I cannot identify this bird, please help me

I saw this bird in Genoa, northern Italy. It was hiding behind a street vase, and was no bigger than a blackbird. Please, let me know what kind of bird is this. Thank you. — Preceding unsigned comment added by 85.18.173.4 (talk) 13:17, 3 November 2011 (UTC)[reply]

Could be a water rail, Rallus aquaticus. DuncanHill (talk) 13:39, 3 November 2011 (UTC)[reply]

I would agree with what DH says and only add that it might be Rallus aquaticus aquaticus which is a European nominate subspecies. But this is well over the edge of being picky :-). Richard Avery (talk) 14:27, 3 November 2011 (UTC)[reply]

DuncanHill and Richard Avery: thank you so much, you are wonderful people — Preceding unsigned comment added by 85.18.173.4 (talk) 14:49, 3 November 2011 (UTC)[reply]

Genetic studies of modern Steller's sea lion populations suggest...

"Genetic studies of modern Steller's sea lion populations suggest that this sea mammal likely hauled out on the rocks along Beringia's island-studded south shore. So the migrants may have had their pick not only of terrestrial mammals but also of seafaring ones."

Article is available here, if anyone can access it. A little bit of it can easily be read here. Link to Steller sea lion.

Question: By what means can "Genetic studies of modern Steller's sea lion populations suggest that this sea mammal likely hauled out on the rocks along Beringia's island-studded south shore"? Bus stop (talk) 14:34, 3 November 2011 (UTC)[reply]

Having no access to the source, you might want to check the relevant references listed on it instead. However, as with human genetic studies it is possible to use genetic "markers" to trace population divergence among animals. Modern sea lion populations are generally divided into two populations that diverged after the disappearance of the Beringia land bridge, the western population is found in and around the Bering Sea and Asia while the eastern population is found in the western coast of North America. As Beringia does not exist anymore today, you would not be able to find direct evidence of the existence of Steller's sea lions in Beringia, which I guess explains the wording used in that article. However, since Steller's sea lions exist in both Asia and North America with substantial genetic differences, scientists can use genetic evidence to link the two populations together to an ancestral population in the former landmass now beneath the sea. As the linked study says, it may not be the first time the populations have been divided by the rise of sea levels. The two populations separating today is the result of the end of the last ice age and the loss and subsequent population movement to find other rookeries that Beringia once provided.-- Obsidi♠n Soul 15:19, 3 November 2011 (UTC)[reply]

I checked the Scientific American article but it doesn't really discuss it and doesn't have any references. I searched google scholar for "steller sea lion "beringia"" though and found this and this which are probably what the SA claim is based on. I'll leave it to a population geneticist to explain it though! SmartSE (talk) 17:46, 3 November 2011 (UTC)[reply]

I'm afraid I asked a question the answer to which is over my head. I do find this interesting nevertheless. Bus stop (talk) 20:00, 3 November 2011 (UTC)[reply]

atom

why octet in most of the cases are one the stable configuration of atom — Preceding unsigned comment added by Bhaskarandpm (talk • contribs) 18:02, 3 November 2011 (UTC)[reply]

Definitely answerable by reading our article about the octet rule. DMacks (talk) 18:28, 3 November 2011 (UTC)[reply]

I've just read that article, and it doesn't answer the question. It says the 3rd shell can contain up to 18 electrons, so why is having only 8 in it a stable configuration? --Tango (talk) 20:30, 3 November 2011 (UTC)[reply]

The key to understanding this lies in understanding how atomic orbitals work. I suggest you read up on that if u really want to understand why, otherwise its much easier just to accept that it is, as this is pretty complicated... Phebus333 (talk) 22:44, 3 November 2011 (UTC)[reply]

The 3^rd shell is subdivided into two lots, the lower valence, and the upper valence. The lower valence contains 10 electrons, the upper valence contain 8 electrons. Plasmic Physics (talk) 23:05, 3 November 2011 (UTC)[reply]

Now that does not really answer the question for someone who is not familiar with the energetic principles of orbitals anyways, does it, Plasmic?

Can you explain that subdivision? My understanding is that shells are defined by where there are big changes in the energy level of electrons. Is there a larger than normal change in energy level in the middle of the 3rd shell that isn't quite large enough for people to consider it a new shell? --Tango (talk) 00:05, 4 November 2011 (UTC)[reply]

This is a pretty complicated subject, as I said. If you study chemistry (as I do) it takes you about 2 months to get the basics you need to understand this. Phebus333 (talk) 23:12, 3 November 2011 (UTC)[reply]

No, it takes about a paragraph or two to explain it so someone will understand it; if done properly, so long as the reader has had some exposure to things like electron configuration and atomic orbitals. Here goes:

In most atoms, the lowest energy atomic orbitals are the "s" and "p" orbitals. While the valence level may also contain " d" and "f" orbitals as well, those orbitals are significantly higher in energy, and thus valence level "d" and "f" orbitals do not carry electrons in the ground state for most atoms. Thus, since most atoms only need to fill their "s" and "p" orbitals to reach a stable configuration, and since there are one s and three p orbitals on any one level, AND since each orbital can take two (spin opposed) electrons; that gives us two electrons in each of four orbitals, which is 8 electrons, thus the octet rule. This rule holds very well for elements which are small (first 3 periods on the periodic table) and in the main group of elements (the "A" groups, or the "tall" columns). The octet rule tends to be easily violated for the transition elements (those in the middle section of the periodic table) and for larger atoms, like those towards the bottom of the main group; especially large nonmetals.

Now, if someone has never heard of atomic orbitals or electron configurations or anything like that, Phebus333 is correct. But insofar as the OP mentions a familiarity with the octet rule, hopefully that will make some sense. --Jayron32 00:20, 4 November 2011 (UTC)[reply]

"so long as the reader has had some exposure to things like electron configuration and atomic orbitals" - this is an assumption that I would not have made here ;-)

"those orbitals are significantly higher in energy" that`s someting I would expect the OP not to accept as "god-given" so he would like to have that explained as well, and understanding why that (from a quantum mechanical POV) is would require some serious time explaining as a (German) B.Sc. chemistry student in the 5. semester (me) is definitely not able to answer that question. Phebus333 (talk) 00:58, 4 November 2011 (UTC)[reply]

Not really, conceptually you can simply think of it as an electrostatic problem. Electrons lose energy when they are closer to the nucleus (positive and negative charges attract) and Electrons gain energy when they are closer to each other (negative and negative charges repel). The "d" and "f" orbitals pack a LOT of electron density into a relatively small volume; which explains why they tend to be of significantly higher energy. For example, the 4f orbitals are higher in energy than even the 6s orbitals, due to the fact that cramming 14 electrons into the 4the 4th energy level is a difficult proposition. Certainly, this way of modeling the electron distribution glosses over the QM implications, but it is a fairly good heuristic way to think about it, in that it leads the introductory student to the right conclusions regarding the relative energies of the various orbitals and their filling order. The Wikipedia article on this topic, the Aufbau principle, discusses this in some more detail, and one so inclined could follow links from there to more detailed coverage. But just thinking of it in terms of electrostatic interactions between electrons and the nucleus, and among the electrons themselves, gets the broad picture pretty well. --Jayron32 02:48, 4 November 2011 (UTC)[reply]

To put it simply (perhaps too simply), s/p/d/f are determined by the angular momentum of the electron around the nucleus, which is closely related to kinetic energy, while the energy levels as I understand it represent, essentially, potential energy. So both numbers represent increases to the total energy of the electron. Wnt (talk) 09:55, 4 November 2011 (UTC)[reply]

Instead of electrical current could there ever be a situation where protons or positrons are made to flow as a current?

I know there can be proton and positron beams but is there any positive charge carrier that doesnt travel through a vacuum and flows as current Also do beams of particles meet no resistance in the vacuum so in theory is this a better way to transmit electrical current — Preceding unsigned comment added by 82.38.102.199 (talk) 21:07, 3 November 2011 (UTC)[reply]

Naked electrons travel in a vacuum in a thermionic valve. -- Finlay McWalterჷTalk 21:11, 3 November 2011 (UTC)[reply]

In a dense, charged plasma, positive ions may flow. If there is a net motion of ions, there may be a DC current. The situation is complicated by the fact that free electrons may counter the motion. Full solutions to plasma require a simultaneous solution of the Maxwell equations and the continuity equations for all species present. Nimur (talk) 21:31, 3 November 2011 (UTC)[reply]

The problem with charged beams flowing through vacuum is that all electrons (or positrons) will repell from each other. So you only get very divergence beams. But of course you could try something like the wehnelt cylinder. But then a cable is easier to transport charge.--Svebert (talk) 21:37, 3 November 2011 (UTC)[reply]

In a hypothetical "perfect vacuum" there would be no matter at all, so a traveling particle would meet no resistance at all (aside from gravitational effects through stellar bodies and stuff like that). However, such a "perfect vacuum" has never been observed. Even in space there is some amount of matter, its just so few that its density is almost negligible. A traveling particle would at some point collide with some of that matter, causing resistance. But given our current technological abilities, I would think that this resistance would be far less than that of any practical (electric) conductor. Phebus333 (talk) 22:53, 3 November 2011 (UTC)[reply]

Regarding the first question; that of positively charged particles flowing to act as a transmitter of electric current. You don't need an exotic high temperature plasma for that. Such things happen just fine in electrolytes; like, for instance, in the salt bridge of a galvanic cell. Cations will carry charge in such situations. --Jayron32 00:08, 4 November 2011 (UTC)[reply]

This is of course true, but I believe the OP had the concept of long distance conduction in mind, and salt bridges are pretty impractical in that context. Phebus333 (talk) 00:36, 4 November 2011 (UTC)[reply]

Positive ions (as well as electrons) flow between the electrodes in a thyratron. (Of course, few people have even heard this word in this solid-state-electronic day and age.) 67.169.177.176 (talk) 06:13, 4 November 2011 (UTC)[reply]

No one seems to have mentioned yet that one could presumably set up a current whose charge carriers are positrons, in a wire made of anti-copper. --Trovatore (talk) 00:44, 4 November 2011 (UTC)[reply]

The protons in a neutron star are believed to flow without resistance, providing the highest temperature superconductor known to natural philosophy. [1]

Proton conductor. --Heron (talk) 17:51, 4 November 2011 (UTC)[reply]

Thanks for your answers guys so could somebody explain more about this thyratron? Do positive ions flow in high temperature plasma? Is there a situation where there is negative and positive charge carrying current? — Preceding unsigned comment added by 82.38.102.199 (talk) 22:13, 4 November 2011 (UTC)[reply]

A thyratron is an ancient type of electronic device that was used in times immemorial to convert high-voltage, low-frequency AC current to DC, as well as to switch such currents on and off. It was a big, sealed glass or metal tube with three electrodes (cathode, anode and control electrode), kind of like the vacuum tubes found in ancient radios, but usually a lot bigger and filled with an inert gas instead of a vacuum. So when you applied a voltage across the tube and then energized the control electrode, this created both free electrons and positive ions in the gas, which then carried the current across the gap. (Which answers your other questions: yes, in a plasma, both positive and negative particles carry the current.) This was very convenient for high currents, because having positive charge carry some of the current really cut down on the resistance losses; the drawback, however, was that this device couldn't be used with high-frequency current, because it took too long for the positive ions to travel from the anode to the cathode. The thyratron is not used much anymore -- it was replaced by the thyristor and the triac, which don't have this limitation (not to mention that they're so much smaller and lighter, and don't burn out); the only place you might still see one is at your local electric substation. 67.169.177.176 (talk) 01:32, 5 November 2011 (UTC)[reply]

Somehow I find "ancient" rather grotesque when applied to things that were common in my childhood. --ColinFine (talk) 17:03, 7 November 2011 (UTC)[reply]

Thank you for all your answers it is interesting that letting positive ions do some of the work resistance losses are cut