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January 20

freckles

Question removed: See talk page. SteveBaker (talk) 04:55, 20 January 2010 (UTC)[reply]

can't see it on the talk page? --TammyMoet (talk) 10:58, 20 January 2010 (UTC)[reply]

Sorry: look at 71.100 is back and 71.100 is back (again). SteveBaker (talk) 13:24, 20 January 2010 (UTC)[reply]

Ah! Thank you. --TammyMoet (talk) 15:35, 20 January 2010 (UTC)[reply]

Ohm's Law

1) Ohm's law says that J is proportional to E...but is this E the E field created by the battery, or something else? Because when this expression is manipulated to get V=IR, this V is said to be the voltage drop caused by the resistor, which would seem to suggest that the E is not the electric field due to the battery. But then what is it?

2) When a current passes through a resistor, the electrons lose energy. Wouldn't this mean that, due to the reduced kinetic energy of the electrons, they would travel less quickly and thus the current intensity would drop? Why does the current stay the same? —Preceding unsigned comment added by 173.179.59.66 (talk) 02:59, 20 January 2010 (UTC)[reply]

The only difference between these forms of Ohm's Law is that one divides both sides of the equation by unit-length. As far as whether the voltage drop is due to the resistor or due to the battery... well, don't get stuck on what causes the voltage drop. The voltage drop merely exists (because there is a potential energy difference between the battery terminals, caused by a chemical reaction). And the voltage drop must exist somewhere - that somewhere happens to be across the resistance. Whether the voltage drop is caused by the resistor or by the battery is semantic - the voltage is a potential difference between the battery terminals, and it exists across the terminals of the resistor. As far as the kinetic energy detail, the assumption of a linear ohmic material (like a resistor) is that the electric potential energy of the electrons is not related to the kinetic energy of their drift velocity. Specifically, drift velocity is proportional to the applied electric field. This necessarily means that the kinetic energy of the electron drift velocity is not equivalent to the electric potential drop. (Rather, the kinetic energy due to electron drift motion is a tiny fraction of the total energy that the electron gains from the electric potential field). The electric potential energy manifests in other ways, including thermal motions and interactions with the conductor's atomic lattice. Mostly, though, it is potential energy, not kinetic energy. See electrical conductivity, and electron mobility, for more of the physics related to this principle. Nimur (talk) 03:35, 20 January 2010 (UTC)[reply]

Another and less eloquent perspective: Things that are equal are equal. The drop of potential across the resistor MUST equal the rise of potential across the battery, or the terminals (nodes) would have two different voltages at the same time. There is internal resistance in a practical battery, which led some scientists to claim that Ohm's Law failed, back in the day. Many materials do not follow Ohm's Law, including semiconductor devices and vacuum tubes, as well as light bulb filament, whose resistance changes with the temperature. Edison (talk) 05:14, 20 January 2010 (UTC)[reply]

1) E is the potential difference (= its voltage V) created by a battery. It can be communicated from place to place by conductors. Where there is no conductor E exists as a field but field is a concept that is useful here only in a homogenous environment without localised conductors, such as inside a material or in air or vacuum. Resistors do not cause any voltage by themselves; if they did then practically every object except a very few that are superconductors or perfect insulators (exist?) would produce a voltage. V=IR means a current I flowing in a resistance R causes a voltage V to appear across the resistor or a voltage V applied across a resistor causes a current I to flow. Which viewpoint you choose doesn't matter because they are just rearrangements of the same interdependance.

2) The E field gradient inside a simple resistor is linear so the drift velocity of electrons in the direction of the field is constant throughout the resistor material. Electrons lose their potential energy to the material, this causes Joule heating. Their velocity and the current intensity do not change until the current passes out of the resistor into a material of different resistance.

Cuddlyable3 (talk) 11:24, 20 January 2010 (UTC)[reply]

Ah okay...but if the V manifests in something abstract like a change in potential energy, then how is a voltmeter able to measure it? —Preceding unsigned comment added by 173.179.59.66 (talk) 12:24, 20 January 2010 (UTC)[reply]

Wait let me guess, does the E field create a magnetic field which is then measured? —Preceding unsigned comment added by 173.179.59.66 (talk) 12:28, 20 January 2010 (UTC)[reply]

Whenever using a voltimeter to measure the voltage you are actually placing a resistor that lets an small current through. (small enough that it won't disturb the behaviour of the circuit). By measuring this circuit the voltimeter can tell what's the voltage. Dauto (talk) 14:42, 20 January 2010 (UTC)[reply]

And what happens in the space between resistors? Let's say two resistors are connected in series. Then there is a V drop through the first resistor, and another V drop through the second resistor, but between the two resistors the voltage doesn't change (or negligibly so). I would guess that this implies that the E field is practically 0 in this area...so how does E field created by the battery dissapear here? Do the charges like rearrange to create a 0 E field? —Preceding unsigned comment added by 173.179.59.66 (talk) 12:37, 20 January 2010 (UTC)[reply]

Yes, the charges at the edge of the resistor rearrange themselves accordingly. Dauto (talk) 14:42, 20 January 2010 (UTC)[reply]

Space Shuttle Challenger disaster

Was a study ever done suggesting what might have happened had the leak in Challenger's SRB field joint been directed outward instead of toward the ET?

Chapter III: The Accident of the Rogers Commission Report details the timeline of events, with flame from the leak first visible at at +60s. During the next four or five seconds, before the LH2 tank was breached, the craft's control system started to react to the forces caused by the plume and the reduced right booster chamber pressure. The spacecraft broke apart at +72s, but both boosters continued to burn until they were detonated by the RSO at +110s, only 10 seconds before normal burnout. Was there any chance that the Challenger could have survived had the flame plume not been directed at the ET? 124.157.247.221 (talk) 06:25, 20 January 2010 (UTC)[reply]

Shuttle - Key Dates: January 24, 1985 The shuttle suffered from blow-by events of the O-ring several times during launches before the disaster. The engines compensated for the slight change in pressures. Zzubnik (talk) 10:11 January 20 2010 (UTC)

Yes - this kind of event had happened many times before - but it was only when the blowout hit something important that it caused the mission to fail. Sadly, the previous occasions were only spotted in hindsight after the Challenger disaster. SteveBaker (talk) 13:22, 20 January 2010 (UTC)[reply]

Your last sentence is untrue, according to our article STS-51-C and according to Zzubnik's link: "The shuttle flight 51-C of January 24, 1985, was launched during some of the coldest weather in Florida history. Upon examination of the booster joints, engineers at Thiokol noticed black soot and grease on the outside of the booster casing, caused by actual gas blow-by. This prompted Thiokol to study the effects of O-ring resiliency at low temperatures. They conducted laboratory tests of O-ring compression and resiliency between 50lF and 100lF. In July 1985, Morton Thiokol ordered new steel billets which would be used for a redesigned case field joint. At the time of the accident, these new billets were not ready for Thiokol, because they take many months to manufacture." Comet Tuttle (talk) 18:01, 20 January 2010 (UTC)[reply]

← While O-ring erosion and evidence of blow-by on previous missions certainly indicated the problem that led to the STS-51-L loss, the effects on flight dynamics cannot be compared. Whereas the worse previous case, STS-51-C, had resulted in "unprecedented penetration of the primary O-Ring and heavily charred effects on the secondary O-Ring" in the center field joint of the right SRB, the aft field joint of the right SRB of STS-51-L, which was recovered from the debris, showed a 27 in. x 15 in. hole burned through the SRB casing. (photo) While wondering if the vehicle could had survived had the flames not been directed onto the ET could be regarded as idle curiosity, such a question would seem to influence future probabilistic risk assessment. That is, would any such burn through necessarily result in the loss of vehicle? 124.157.247.221 (talk) 23:20, 20 January 2010 (UTC)[reply]

What mean east and west speaking of the Antartic ?

Hello you nice Pythias. Please excuse my poor English, I'm French.

I already asked this question in the French equivalent of your reference desk called the Oracle but I got poor answers.

Introduction : very often, speaking of the melting of the antartic cap (I know it's a continent, but there's ice on and around the continent), specialists speak of the western part of the Antartic.

Question : This western makes a sense for me for Europe, Africa and every where else in the world except Artic and Antartic.

What does it mean for the Antartic ?

Thak you very much for your explainations. Joël DESHAIES - Rheims in France - --90.7.206.69 (talk) 14:24, 20 January 2010 (UTC)[reply]

We have West Antarctica. The French ref desk is called the Oracle? Over here we are explicitly not a crystal ball. I think that excludes all forms of divination. 81.131.36.119 (talk) 14:31, 20 January 2010 (UTC)[reply]

More generically, an Oracle is one who gives wise answers to questions - possibly by consulting, not a crystal ball, but a relational database. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:27, 21 January 2010 (UTC)[reply]

When you are exactly at the pole (either North or South), it's true to say that the words "East" and "West" don't mean anything. Once you get away from the pole, they have their usual meaning. The problem with the artctic/antarctic is that any place you happen to be standing is both east of one place and west of another. However, the antarctic is conveniently split in two by the 'Transantarctic Mountains' - and it has been agreed that the side of the continent that's to the west of the Weddel Sea is called "West Antarctica" and the other half "East Antarctica". This is a bit confusing when you are on the side of the continent nearest to the Ross Sea because West Antarctica is to the east of you and East Antarctica is to the west. But that's the price you pay for living in such a geographically confusing place! However, that division makes some sort of sense because we talk about the Earth as having a "Western Hemisphere" and an "Eastern Hemisphere" - and West Antarctica is in the Western Hemisphere with East Antarctica in the Eastern Hemisphere. Of course the names of the two hemispheres is also a bit weird if you happen to live on the International Date Line and the Eastern hemisphere is to your west and the Western Hemisphere to your east...but that's what you get when you live on a sphere. Topology is a harsh mistress! SteveBaker (talk) 14:37, 20 January 2010 (UTC)[reply]

Thank you for for your good explainations, I think it's conclusive. Joël DESHAIES - Rheims in France ---90.7.206.69 (talk) 15:02, 20 January 2010 (UTC)[reply]

Obligatory xkcd link --Tango (talk) 02:36, 21 January 2010 (UTC)[reply]

That cartoon doesn't really work when you live in the UK...especially if you happen to be near Greenwich. SteveBaker (talk) 14:26, 21 January 2010 (UTC)[reply]

Speaking as someone who lives in Greenwich, I used to live in the west but I moved a mile and half away, so now live in the east. Mikenorton (talk) 17:01, 21 January 2010 (UTC) [reply]

Speaking of West Antarctica, it appears that Pine Island Bay, a major "tipping point" for the destabilization of the West Antarctic Ice Sheet has already passed its own tipping point for eventual collapse. ~A H 1^(TCU) 02:41, 23 January 2010 (UTC)[reply]

What kind of fish is this?

[1] [2]

Thanks. --✶♏‭ݣ 15:17, 20 January 2010 (UTC)[reply]

Chelmon rostratus, the Copperband butterflyfish. 124.157.247.221 (talk) 15:47, 20 January 2010 (UTC)[reply]

Thank you! --✶♏‭ݣ 15:55, 20 January 2010 (UTC)[reply]

Do Solids burn

Does a solid burn or does it change state to a gas first ? Purple pete2000 (talk) 19:37, 20 January 2010 (UTC)[reply]

Have a look at burning, especially the paragraph labelled 'solid fuel'. It says it does turn to gas. Richard Avery (talk) 19:42, 20 January 2010 (UTC)[reply]

If you are thinking that the heat from the flame vaporizes the solid and then the gas burns - then definitely no. For a chunk of solid carbon (like maybe a charcoal briquette) turn into a carbon gas would require a temperature of 3642 °C - but carbon burns easily at around 700 °C when there is nowhere near enough heat to cause the carbon to vaporize. However, maybe there is some other weird thing happening right at the point of ignition. Fire is weird stuff. SteveBaker (talk) 23:52, 20 January 2010 (UTC)[reply]

The answer is, of course, that it depends. Mostly, this depends on the solid fuel material, and the presence and quantity of oxidizer. Surprisingly, we have no article on flame dynamics, but here is Dynamics of Deflagrations and Reactive Systems: Flames (Progress in Astronautics and Aeronautics) ($104). Naturally I'm fascinated by this sort of thing. When designing my hybrid-fuel rocket [3], which used a liquid/gas-phase oxidizer and a solid-phase fuel, the ultimate question was to what extent can we control the thermodynamics of the phase transitions to optimize the specific impulse? I think the answer was, "50%," or something. The goal in our case was to keep the chamber temperature hot so that the fuel would gassify, and to keep the oxidizer flowing at the ideal rate to match the gassification rate of the fuel, and to do this without introducing oscillation. The trouble is, especially in the specifics of the particular chemicals we used, that the flow rate of the oxidizer is nonlinear (being choked at the inlet, and undergoing a phase change) - in fact, it spurts out in a combination of gas-phase and liquid vapor mist - each with different burn properties. Also, the fuel can burn in multiple modes - in solid form, in liquid form, in gaseous form (diffusion flame, roughly), and in "macrscopic chunks of fuel flying out the nozzle and igniting in flight." Anyway, the chamber can't start out hot - you have to heat it with flame! So the ignition must occur with a solid-phase fuel. To speed this along, we used a pyrotechnic igniter and blew hot fiery chunks of explosive which embedded into the walls of the fuel and ignited it in solid form right as we opened up the oxidizer flow. The process of switching flame modes from solid- to gassified- fuel is visible in some of the technical readouts and thrust curves, but it's hard to decouple from the effects of the oxidizer flow ramp-up. And if you're willing to supply a really powerful oxidizer, like fluorine gas, you can burn almost anything in solid form - even chicken. (Note that they don't even need an ignition source! Solid chicken will spontaneously ignite at room temperature in the presence of concentrated fluorine). So, the answer is, of course, it depends. Nimur (talk) 03:39, 21 January 2010 (UTC)[reply]

I like how they give a purity of the chicken starting-material. DMacks (talk) 04:17, 21 January 2010 (UTC)[reply]

In a technical sense, no, because as these fellows have stated, a flame is basically ignited gas, so if you have ignition you always had gas as a precursor. That said, yes, solids do burn, if you don't worry about the transitional phase between solid and flame. Phosphorous being the classic example. It spontaneously ignites in air. Vranak (talk) 13:13, 23 January 2010 (UTC)[reply]

Latitudes and longitudes with the most land or the most water

I don't think there is a single latitude or longitude consisting either completely of land or completely of water. But which latitudes and longitudes come the closest? JIP | Talk 19:51, 20 January 2010 (UTC)[reply]

Actually, while there are no longitudes consisting either completely of land or completely of water, there are such latitudes near the poles. Latitudes near the south pole consist entirely of land, and there might be some just to the north of them that consist entirely of water. Near the north pole it's mostly the same situation, depending whether you view the Arctic as land or not. Underneath all that ice is only water, no solid land, but because it's so cold there, the ice never melts, and thus no one actually gets to see that water underneath it. Other than the poles, which latitudes come the closest to being mostly land or mostly water? And what about the longitudes? JIP | Talk 19:54, 20 January 2010 (UTC)[reply]

To find lines of latitude with only water (or land) take a conventionally-oriented sinusoidal projection of the earth's surface (as in File:Sinusoid-projection.jpg) and lay a ruler horizontally across it. Features running horizontally are scaled correctly in this projection, so just find the widest point of the map that's all water (or all land). Eyeballing, the longest line of latitude without land is just south of the tip of South America, around 55° south. The longest line that's all land (and, indeed, the only place where this occurs) is somewhere in Antarctica.

For lines of longitude, you'd need to find a sinusoidal projection with the central meridian along the equator; I don't know where you'd get that. As JIP notes, all lines of longitude will pass through both the antarctic landmass and its surrounding oceans, so there are no completely wet or completely dry lines of longitude. TenOfAllTrades(talk) 20:19, 20 January 2010 (UTC)[reply]

What about 89 degrees south, that is all land, but covered with a lot of ice. And 89 degrees north is all sea, but covered with a lot of ice again. 92.29.57.199 (talk) 21:37, 20 January 2010 (UTC)[reply]

The International Date Line passes mainly through ocean, but it has to be diverted from longitude 180 degrees (W or E) to achieve this. Dbfirs 23:44, 20 January 2010 (UTC)[reply]

A look at the map shows that the longitude with least land is right around 170W -- there are only about 5 degrees of latitude of land there. The longitude with the most land is near 25E, with close to 120 degrees of latitude of land. Looie496 (talk) 00:25, 21 January 2010 (UTC)[reply]

Virginity

I'm curious, does anyone have statistics on the distribution of ages at which people experience their first sexual intercourse? For example, (pulling numbers out of the air) perhaps 10% of people lose their virginity at age 15, 15% at age 16, 15% at age 17, and so on... I sometimes see claims about the average age at which people give up their virginity, but such reports almost never present much real data. I'm wondering if people know of any detailed data sets. Dragons flight (talk) 21:28, 20 January 2010 (UTC)[reply]

Here you go. I found this by just googling "first sexual experience statistics" and hunting around a bit there. (Note that this is a U.S.-only dataset. I'm not sure global statistics are as available, but you can probably find similar things for all Western countries.) --Mr.98 (talk) 21:40, 20 January 2010 (UTC)[reply]

There was a similar question to this recently. 92.29.57.199 (talk) 23:32, 20 January 2010 (UTC)[reply]

There will also be issues gathering this data as it will be self reported data, and such data is often not fully accurate. Googlemeister (talk) 14:22, 21 January 2010 (UTC)[reply]

I'm puzzled by "Average age of first intercourse by gender" in the Kinsey report. It seems that at every age grouping a higher percentage of girls have had sex than boys yet, on average, boys lose their virginity earlier. Is that consistent?--Frumpo (talk) 22:09, 21 January 2010 (UTC)[reply]

I don't think so. However, there is one age group (20-21) at which the male percentage exceeds the female one, and data aren't given for ages over 24, so a strategically placed wiggle in the data at either or both of these levels could cause the effect. For example, perhaps most of the remaining 11% of males all lose their virginity before they're thirty, while more of the 8% of females do so at older ages. If this is true, though, that would be almost a more interesting fact about the distributions than the ones presented! It would be very nice to see a relative distribution (in the sense of Handcock and Morris). Neurotip (talk) 10:44, 22 January 2010 (UTC)[reply]