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November 20

School in Antarctica

I visited a school (for 18 students) held in a trailer on King George Island in Antarctica in 1992. How many children attend school today in all of Antarctica? - on which bases are the schools located? - how many students in each school? After researching I can only find one school on the Argentinian Esperanza Base, Antarctica with 21 students. Is it possible to contact the schools by e-mail or letter? If so, please provide contact info. Thank you very much from Mimi —Preceding unsigned comment added by Mimi Harrington (talk • contribs)

I'm seriously surprised the number isn't zero. My understanding has been that while many bases are continuously occupied, there are no people that are formally considered permanent residents of the continent, and given that, I am surprised to hear that there are any families living there with children that would need to go to school. Dragons flight 19:26, 19 November 2006 (UTC)[reply]

This page at last years' HMS Endurance Tracking Project gives the e-mail address of the "Julio Argentino Roca" school, as esperanzaantar@infovia.com.ar. Dr Mariano A. Memolli is the director of their Antarctic program/agency, Dirección Nacional del Antártico, at mmemolli@dna.gov.ar. Seejyb 22:39, 19 November 2006 (UTC)[reply]

I believe the reason that there's a school on Antarctica (and, further, I believe a woman has even given birth on Antarctica) is to bolster the Argentine land claim. ByeByeBaby 07:18, 20 November 2006 (UTC)[reply]

Physics Lab

Hi I have a question related to a lab I have to do tomorrow. We have to find the mass of our teachers car using tape measures, stopwatches, and bathroom scales. What myself and another class member were think was to put the scales at the back of the car, mark out a course in length, and apply a set force to the car, and calculate its acceleration. Then use F=ma to find the mass. I know it works in theory, but there's something that I don't like about it. Any help will be greatly appreciated. Thanks Deltacom1515 00:20, 20 November 2006 (UTC)[reply]

Do you actually have to DO it or just come up with a theoretical way of doing it? I don't see how you are going to apply a 'set force' to the car with only the instruments that you have. You aren't really using the tape measure or scales either. Is that the ONLY things you are allowed to use? If you could use some planks of wood you could make a balance type thing:

                                tttt
pppppppppppppppppppppppppppppppppppppppppppppppppp
  bb                                       bb
ssssss                                     bb

p = plank b = block s = scales t = tire (of the car)

Drive the front wheel of the car onto the plank from the block end until the scales approach the top of their register (200kg or whatever it is) then measure the distance from the scales to the tyre and from the block to the tire. Distance from the scales divided by the distance from the block will equal the weight on the scales divided by the weight on the block (x), the weight on the scales plus (x) will be the weight of the front wheel, multiply that by 2 and you have the weight of the front of the car, since weight should be roughly evenly distributed between the right and left, but NOT between the front and back so do the same with the opposite rear wheel and add them together for the whole weight of the car. But then you haven't used the stop watch so I don't know if that's right either... Vespine 00:58, 20 November 2006 (UTC)[reply]

Oh, yes we actually have to do this, and we'd apply the force to the scales at the back of the car and push the car a set distance and record the time it took to go that far getting the acceleration. And those are the only things we can use. Deltacom1515 01:05, 20 November 2006 (UTC)[reply]

Oh right, you're applying the force TO the scales, I see. If you apply a constant force you can assume the acceleration is constant and could infer it from the average velocity. But the method still doesn't take into consideration any friction, in the equation F= MA, F is the sum of all forces, including friction. I just had a brain wave!! What you do is after you have accelerated the car you LET IT GO and measure how long it takes for the car to come to a complete stop again! From that, you can roughly calculate the friction forces of the car. Add that to the F (or take it away, depending which way you look at it) in your F= MA equation and you will have a much more accurate result. 02:28, 20 November 2006 (UTC)

Is your teacher completely happy that his car might get wrecked? --Light current 02:35, 20 November 2006 (UTC)[reply]

How would her car get wrecked? Deltacom1515 04:07, 20 November 2006 (UTC) 03:14, 20 November 2006 (UTC)[reply]

There are many ways that the kids could wreck it! Use your imagination!

For example, disassemble the car into pieces small enough to be weighed on the scales, sum up the weights, account for local variation in the force of gravity to get the mass. Reassembly of the vehicle is left as an exercise for the instructor (something every physics student has always wanted to say). -- The Photon 03:52, 20 November 2006 (UTC)[reply]

HAHA, that's quite a good idea actually and our class would probably try to do that too :) Deltacom1515 04:07, 20 November 2006 (UTC)[reply]

Do you have at least four bathroom scales? The traditional method of finding the total mass and CG of a large object is to place each tire on one scale. The total mass won't necessarily be the sum of the four readings, but I can't remember how to calculate it - google should help. Also, this method won't work if each scale can't handle half a ton or so of weight. sthomson 16:10, 20 November 2006 (UTC)[reply]

With a level pavement, you could have someone push on the back of the car with a constant force, measured with the scale, with the car in neutral and the brakes off. (If my class had done this, I guarantee the driver would have kept the brake on at first as a joke!) With a constant force applied, as measured by the scale, s=1/2 *a *t squared. With the force and the acceleration known, and the mass can be determined. The mass of the driver and any passenger must be deducted to get the mass of the car. Weigh them on the scale and convert weight to mass. Friction is a confounding factor. To refine the experiment, someone in the passenger seat can drop a beanbag or other non-rolling object on the pavement at fixed time intervals to show the increasing distance in each interval, allowing a more accurate measurement, since friction may delay the initial motion of the car. If the force is too small the car may not move at all, and if it is large the pusher won't be able to go any faster after a short while. It will take practice to get the force application constant. If several scales are used by pushers (this is soundling like a drug cartel) the job will be easier and the forces can be added. Edison 16:20, 20 November 2006 (UTC)[reply]

Hot air balloon materials

For a physics class I'm taking, we're slowly building self-sustained hot air balloons on a miniature scale in small groups. We're in the very beginning stages currently, and groups of two are assigned to design experiments to test for the optimal balloon design. My group is supposed to test materials with which to construct the balloon's envelope. We can't use nylon like a real hot air balloon, so instead we're thinking of using the material from a space blanket (Mylar^®). We're also designing a balloon with tissue paper. We need two more types of materials. We could use trash bags or aluminum foil, but does anyone have some more interesting or creative ideas to test? It only has to fly, not fly well. :P Thanks in advance! --McMillin24 _contribs^talk 00:27, 20 November 2006 (UTC)[reply]

For some reason, latex comes to mind. I'm not sure how susceptable they are to heat though, but I can't imagine it being much worse than tissue paper or trash bags. --Wirbelwindヴィルヴェルヴィント (talk) 00:37, 20 November 2006 (UTC)[reply]

Give a look at a vacuum depositioning a thin film of some metal onto the balloon bladder. Most expandable (elastic) materials are notoriously porous and will be a source of leakage. --hydnjo talk 03:34, 20 November 2006 (UTC)[reply]

Paper mache like a pinata? Inflate a baloon and paper mache around it. --Tbeatty 04:53, 20 November 2006 (UTC)[reply]

How about Saran Wrap or the clear plastic bags that dry cleaning comes back in ? StuRat 05:49, 20 November 2006 (UTC)[reply]

By self-sustaining I imagine you mean that the machine travels with its heat source on board. Cooking (roasting) bags (I think one gets polyethylene and nylon-6 makes, I don't know the physical difference). Plus: Light, relatively impermeable, heat resistant (200-260°C). You would (of course) test materials for flammability as part of the project. Minus: Not easily stretched or shaped - may need a lot of cutting. Would probably require a superglue to join. Seejyb 13:11, 20 November 2006 (UTC)[reply]

Think low weight per unit volume. Aluminum foil or paper mache are WAY too heavy. Tissue paper is classic, A drycleaner bag is ideal for small weight and large volume. It does not neeed to be so strong and heavy as a trash bag or heavy mylar. There is a fire hazard if a source of ignition is inside; when camps have kids do this they use some sort of heat source like a garage heater: the bag gets a dose of hot air and lifts off without having a flame inside to cause it to land on someone's roof and start a fire. Edison 16:25, 20 November 2006 (UTC)[reply]

Thanks for all the suggestions! Some of these should be quite possible to use. --McMillin24 _contribs^talk 23:10, 20 November 2006 (UTC)[reply]

Black Holes

If a black hole swallows everything around it, and if most black holes are found in the center of galaxies, then do black holes eventually swallow everything in its galaxy? And, if so, what happens when it consumes everything? Where does it go? no one knows.one thing is that a black hole leads into aworm hole.

Yes, black holes will eventually swallow everything in the universe. But the distances between stars are huge, so it will take a very long time for all the stars to come close enough to a black hole to be swallowed. As for the second question, all matter that falls into a black hole will come out as Hawking radiation. It will take approximately a googol years (10 to the power of 100 years) for all the black holes to evaporate away. --Bowlhover 02:24, 20 November 2006 (UTC)[reply]

I thought that black holes were orbitable like any other massive thing with gravity. 84.66.218.66 04:10, 20 November 2006 (UTC)[reply]

Black holes are orbitable up to an interior radius which depends on its spin. The Schwarzschild metric shows that there is an ISCO (innermost stable circular orbit) which for a non-spinning black hole is 3x the radius of the event horizon.18.96.6.207 13:08, 21 November 2006 (UTC)[reply]

It definitely has not been proven either way if black holes will eventually 'swallow' the universe or not. See ultimate fate of the universe. Vespine 05:45, 20 November 2006 (UTC)[reply]

I believe Hawking radiation ('the hole's output' - sorry if that sounds yuck) is only a fraction of the hole's 'input' (even more yuck?). So it will grow, which means it attract more matter and grow faster, etc, causing exponential growth. And assuming that the first black holes started at roughly the same era in the universe's history, will there be a point in time when the growth is so quick that, say, from one human generation to the next we will change from ignorance to being swallowed up? Or is that a (gross) exaggeration? And if not, is there any indication of when this might happen? DirkvdM 11:33, 20 November 2006 (UTC)[reply]

The real question is: with all the BHs in the galaxy, why doesn't look it like Swiss cheese? Answer: because gravity isn't as strong. Your neighbor star can go BH and, aside from the explosion, you won't notice it because its mass simply hasn't changed - it's even less than before. The only difference is there is no longer radiation coming from it, and this is exactly the only thing that keeps a normal star from accreting the surrounding matter. -- 85.179.15.106 19:20, 20 November 2006 (UTC)[reply]

Bear problem...

Isn't there a word for the problem for when wild animals, such as bears, get conditioned to not be afraid of people and maybe search for food in trash instead in the way they normally would? I think it is different a little from tame because bears in this state can still kill people; I wouldn't call that tame. ike9898 02:25, 20 November 2006 (UTC)[reply]

Domestication perhaps? - Dammit 02:26, 20 November 2006 (UTC)[reply]

Habituation? -THB 02:37, 20 November 2006 (UTC)[reply]

Not a word I can think of but a phrase, they have "lost their fear of humans". StuRat 05:42, 20 November 2006 (UTC)[reply]

My dictionary says become tame. –mysid ☎ 10:17, 20 November 2006 (UTC)[reply]

question (fear)

why do we feel fear? Is fear a good feeling? in what part or the brain is fear located? how if works? —Preceding unsigned comment added by 65.148.116.148 (talk • contribs)

Please do your own research 1st, our article on fear already has most of the answers you are lookign for. And please sign your posts. :) Vespine 03:01, 20 November 2006 (UTC)[reply]

Fear exists to keep you from doing stupid things that are likely to get you killed or injured. StuRat 05:36, 20 November 2006 (UTC)[reply]

"We have nothing to fear, but fear itself." - FDR - StuRat 05:36, 20 November 2006 (UTC)[reply]

Yeah, you just keep on thinking that when you cross paths with one of the 'tame' bears in the above question. :) DirkvdM 08:49, 21 November 2006 (UTC)[reply]

Intercostal space

What arteries supply the first intercostal space? —The preceding unsigned comment was added by 71.226.139.16 (talk • contribs) .

The open up your textbook kind? Or you could try the links to the pages I added. Deltacom1515 04:10, 20 November 2006 (UTC)[reply]

IIRC that's the superior pwnd artery. Vespine 05:38, 20 November 2006 (UTC)[reply]

Pwnd ? StuRat 11:37, 20 November 2006 (UTC)[reply]

Yes, pwnd. –mysid ☎ 11:51, 20 November 2006 (UTC)[reply]

The superior "soundly defeat an opponent" artery ? StuRat 04:43, 22 November 2006 (UTC)[reply]

Which is right above the anterior ownd artery. X [Mac Davis] (DESK|How's my driving?) 05:09, 21 November 2006 (UTC)[reply]

Genetics

what is the name of the technique which shows you the size of DNA in a gene?

Chromosome banding ? StuRat 11:35, 20 November 2006 (UTC)[reply]

Your question is not very precise. If you want the exact size of the gene, you will need to to DNA sequencing. If you have restriction enzyme markers at both ends of the gene, you can determine the approximate size of the gene using agarose gel electrophoresis, a process which in this context is called Southern blotting. You might also want to have a look at the article on the polymerase chain reaction. --N·Blue^talk 18:09, 20 November 2006 (UTC)[reply]

electromagnetic wavelength

What is the wavelength of an electromagnetic frequency equal to the speed of light and is there anything special about it or its half-wavelength such as a correlation with the radius, diameter or circumference of a proton, neutron, electron or any atomic or subatomic particle? Adaptron 09:25, 20 November 2006 (UTC)[reply]

How exactly can a wavelength equal to speed? They are different quantities. –mysid ☎ 10:15, 20 November 2006 (UTC)[reply]

I don't understand your question. Have you tried reading the article on wavelength?--Shantavira 10:23, 20 November 2006 (UTC)[reply]

All the electromagnetic waves travel at the velocity of light in vacuum. The speed of light devided by the frequency gives the wavelength and vice versa -- WikiCheng | Talk 11:08, 20 November 2006 (UTC)[reply]

The wavelength is, of course, a length. There are units of length based on the speed of light, like the light-year or light-second. Did you have one of those in mind as the wavelength ? StuRat 11:33, 20 November 2006 (UTC)[reply]

At 60 mlles per hour a car goes 316800 feet (5280 * 60) or 88 feet per second (316800 / 60 /60). Its "wavelength" at this speed is therefore 88 feet per second. At this "wavelength" its "frequency" is (1 / 88) or 0.01136363... cycles per second. In other words it goes 1 foot in 0.01136363... of a second. Since the speed of light is 299,792,458 meters per second and the distance it travels in 1 / 299,792,458 seconds is 1 meter or in (decimal notation) 3.3356409519815204957557671447492e-9 cycles per second. Here unit of distance is used as a single cycle. Now divide this unit of distance by the speed of light to get the unit of distance I initially refered to as "wavelength" Now apply my question about correlation with any other measurement of distance or length such as the radius, diameter or circumference of a proton, etc. and what do you get? Adaptron 14:51, 20 November 2006 (UTC)[reply]

I doubt it has anything to do with any of those distances, as the second is a human invention, originally roughly derived from the rotation time of the Earth. –mysid ☎ 15:31, 20 November 2006 (UTC)[reply]

Adaptron, your value of 3.3356...x10^-9 is the time in seconds that it takes light to travel 1 meter, so its units are second per meter. You are trying to compare this with various lengths, measured in meters. If you compare values measured in different units, then the ratio between the values will vary as you change your units, and you can select your units to create any numerical coincidence that you like between the values. Therefore numerical relationships between values measured in different units cannot have any fundamental physical significance because they are unit-dependent. See pyramid inch for an example of a unit of length that was specifically chosen to make certain numerical relationships true and coincidences of units for other examples. Gandalf61 16:05, 20 November 2006 (UTC)[reply]

Adaptron, your definition of the wavelength of the car seems meaningless. But "from struggle comes comprehension." Edison 16:29, 20 November 2006 (UTC)[reply]

I guess I have really stated this wrong. The frequency of electromagnetic waves such as light seem to have a wide range of values and to be limitless. Cosmic rays seem to be at the top end of the spectrum with a frequency of 10^20 cycles pe second with a wavelength of 3*10^-12 meters. Let me ask it this way: which wavelengths of the electromagnetic spectrum correspond to atomic measurements of length (if any) such as radius, circumference, distances between shells, etc.? Adaptron 16:57, 20 November 2006 (UTC)[reply]

It depends on the substance. For example, the radius of a hydrogen atom is 25 picometers and the radius of a proton 0.8 femtometers. –mysid ☎ 17:23, 20 November 2006 (UTC)[reply]

Where is Planck? Seejyb 20:53, 20 November 2006 (UTC)[reply]

I don't understnd free space yet. How about this. Is there a scale (logorithmic or otherwise) that relates wavelength (light color, cosmic rays, etc) and atomic distances? Adaptron 21:03, 20 November 2006 (UTC)[reply]

I think he's asking: what frequency of light has approximately the same wavelength as the size of an atom, or neucleon. Ariel. 06:34, 21 November 2006 (UTC)[reply]

If so: EM radiation whose wavelength is about the diameter of a hydrogen atom would be classified as (soft) X-rays. If the wavelength is about the diameter of a proton, we are into (very energetic) gamma rays, which may be produced by colliding electrons with positrons flying against each other at high speeds, and are also occasionally seen in ultra-high-energy cosmic rays. --Lambiam ^Talk 21:22, 21 November 2006 (UTC)[reply]

drops

Are the number of drops per milliliter of liquids the same for all liquids or do they depend upon gravitation (which planet they are on), surface tension or the liquid or its viscosity? If so is there a formula that includes all of the factors that go into determining the number of drops per milliliter per the values for each of the relevant variables? 71.100.6.152 09:33, 20 November 2006 (UTC)[reply]

The maximum possible weight of a drop of radius a hanging from the end of a tube is m = (3πaλcosα)/g where λ is the surface tension of the liquid, α is the angle of contact with the tube, and g is the acceleration due to gravity (from Drop (liquid)). It may be worth noting that a drop is also a unit of volume (see Drop (unit)). –mysid ☎ 10:12, 20 November 2006 (UTC)[reply]

Thanks. This is exactly what I was trying to find out but since I don't think that drops will ever be dropped (excuse the pun for me as well as for StuRat below!) as a unit it is good to know that there is a way to determine how many drops per milliliter there will be for any given liquid given the values of the various variables so that instuctions oon medicine can be adjusted accordingly. Thanks. 71.100.6.152 15:01, 20 November 2006 (UTC)[reply]

It sounds like it should be dropped as a unit of measure. StuRat 11:29, 20 November 2006 (UTC)[reply]

One 'drop' of a certain liquid always has the same volume. It depends on its surface tension and density! Angle of contact I couldnt comment. Is angle of contact anything to do with viscosity or not? --Light current 00:09, 21 November 2006 (UTC)[reply]

It makes sense that angle of contact due acceleration due to gravity will have an effect on how large a drop can be formed (by weight) before it is dislodged from the tube and that if surface tension is different in different gasses another variable is needed. Since weight and volume for a given liquid are directly proportiona I suspect that a drop of the same liquid under differnt circumstances could have a different volume. Only if all the relevant variables are standard could one assert that a drop of the same liquid could have the same volume but that does not solve the problem unless drops per milliliter are stated as drops of a certain liguid under standard temperature and pressure and surface tension, etc. for that particular ligud which a standard formula should probably be used to derive. 71.100.6.152 06:51, 21 November 2006 (UTC)[reply]

properties of glass.

is glass an electrical conductor or not?—Preceding unsigned comment added by 202.88.229.83 (talk • contribs)

Not standard glass no. Not conductor.

In fact, glass is a common high-voltage electrical insulator. -- Plutor ^talk 14:33, 20 November 2006 (UTC)[reply]

I vividly remember an experiment (in the Nuffield A-level Combined Science course circa 1980) that I demonstrated to a class. A glass rod had bare mains wires wound round it, about 15cm apart. A mains light bulb was connected in series, and the whole circuit was plugged into the 240V mains. No light, because there was a 15cm break in the circuit. But heat up the glass rod with a Bunsen, and the bulb lights up! Usual caveat - don't try this at home...--G N Frykman 17:57, 20 November 2006 (UTC)[reply]

What a great fire alarm that would make. BTW was tha gap between the ends of the wire or between disconnected turns? Adaptron 21:49, 20 November 2006 (UTC)[reply]

What a rubbish fire alarm it would be, needing 240v mains, and temperatures of 1350 degrees celius to work. As opposed to the current ones that will detect smoke and run on batteries. Philc _T^E_C^I 23:12, 20 November 2006 (UTC)[reply]

Red hot glass is not usually used as an electrical insulator. But interesting result anyway! Any theories?--Light current 00:03, 21 November 2006 (UTC)[reply]

Or perhaps hypothesises or conjectures? X [Mac Davis] (DESK|How's my driving?) 05:08, 21 November 2006 (UTC)[reply]

I like it, I love it, I want more of it. I had not heard of this demonstration. Conductive red hot glass would be an interesting adjunct to iron becoming non-magnetic when it gets red hot and is above its Curie temperature. (Do not under any circumstances try this at home!) Edison 15:19, 21 November 2006 (UTC)[reply]

i would say once you return the glass to its liquid state, you give the constituent ions back their mobility, allowing them to carry a current. i wonder whether the same thing works with frozen salt water? would be quick to test. Xcomradex 11:03, 22 November 2006 (UTC)[reply]

I remember hearing about this somewhere too! Also I think you can switch off the burner once the current starts flowing as the ohmic heating in the glass is enough to self sustain a conductive state.--Deglr6328 12:17, 22 November 2006 (UTC)[reply]

I'll try to find a page reference for it in the text-book that I used all those years ago. I seem to remember that there was about a 15cm break in the coils of wire - enough for the bunsen to get in and heat the glass rod.--G N Frykman 19:29, 24 November 2006 (UTC)[reply]

Communication throught the earth

My grandad told me about a communication device that would send messages through the ground. It could be either through electricity or seismic vibrations. He remembered it being used during WWII. He would like to have more info. Could anyone enlighten us? Keria 13:05, 20 November 2006 (UTC)[reply]

At the start of WWII, Nikola Tesla was claiming he could split the earth in two, and communicate to anybody through the earth. If you research this, you can find many more claims. --Zeizmic 15:57, 20 November 2006 (UTC)[reply]

See Nathan Stubblefield and the material which was moved to the Talk page of that article. This inventor used voice frequency electric currents in the ground to broadcast wireless telephone messages in 1892, and his public demonstrations in Washington DC and Philadelphia in 1902 were among the first public demonstrations of wireless voice broadcasting. Normal telephone lines of the era used one conductor and the ground was used as the other conductor. (Today two metallic conductors are used, te reduce noise and interference). In trench warfare in WW1, sometimes the line was broken or cut, and telephones were connected to a pair of ground rods to talk through the ground for short distances. Stubblefield also used circular coils to send audio frequency voice messages through the air for distances of up to a half mile. Edison 16:39, 20 November 2006 (UTC)[reply]

More recently, extremely low frequency waves were used for Communication with submarines anywhere in the world. Rmhermen 00:36, 21 November 2006 (UTC)[reply]

VLF or ELF [1] communication?--Light current 00:43, 21 November 2006 (UTC)[reply]

That ELF article is enough to curl one's toes. StuRat 04:38, 22 November 2006 (UTC)[reply]

Making a big magnet smaller

Can you cut one long magnet into several pieces and then have several magnets from the one? In other words, does it ruin a magnet to cut it into several pieces?170.158.60.75 15:53, 20 November 2006 (UTC)Jessie[reply]

See Magnet. You get smaller (and less powerful) magnets if you cut a big magnet into little ones. It is ruined in the sense that you no longer have the big magnet, but if the small magnets are reassembled into the original configuration, they will come close to the power of the original big one. Each little one still has a north and south magnetic pole. Edison 16:41, 20 November 2006 (UTC)[reply]

Yes but sawing them can reduce the magnetisation due to the vibration--Light current 03:21, 22 November 2006 (UTC)[reply]

Fundamental forces

When I give kinetic energy to an object e.g. by pushing it, which of the four fundamental forces (gravity, electromagnetic, weak nuclear, strong nuclear) is it? Or am I misinterpreting the nature of a fundamental force? --217.43.240.251 16:07, 20 November 2006 (UTC)[reply]

Well, it's definitely not gravity, because you're pushing the object away from you. It's not the strong nuclear force either, because that only holds protons together in an atomic nucleus. I would say electromagnetism, because when you push an object away, the atoms in your hand (more specifically, the electrons) repel the atoms in the object. This is why your hand doesn't sink into everything you touch. --Bowlhover 16:52, 20 November 2006 (UTC)[reply]

Correct, the contact force is caused by the electrostatic repulsion of the electron clouds. Hmm, the article Touch could use a lot of work... —Keenan Pepper 17:04, 20 November 2006 (UTC)[reply]

Does this mean that all forms of energy taught in high school (chemical, sound, etc.) can be described in terms of those four? --217.43.240.251 17:21, 20 November 2006 (UTC)[reply]

Well, sort of. All forces are ultimately those four. To describe all forms of energy, I suppose you would have the potential energy arising from those four forces, plus kinetic energy. -- SCZenz 17:24, 20 November 2006 (UTC)[reply]

Yes, and except for gravity, anything you're likely to notice in day-to-day living is due to the E-M interaction. B00P 18:21, 20 November 2006 (UTC)[reply]

And its the electromagnetic force that's responsible for friction too. 24.39.182.101 20:00, 21 November 2006 (UTC)[reply]

Orbital speed of the moon / speed of shadow during total solar eclipse

I have been unable to find a statement of the maximum (theoretical) duration of an annular solar eclipse (and there are several different values quoted for a maximum total eclipse!) so I decided to attack the problem from first principles by calculating the size of the moon's shadow where it touched the earth and the speed the shadow moved across the Earth's surface.

The first bit was straightforward enough (using the theorem of similar triangles) and the answer I got was supported by a number of other Internet sources. The second part should have been simpler as during a maximum eclipse the shadow speed is just (to a good approximation) the moon's orbital speed minus the Earth's rotation speed at the equator. For a maximum total eclipse the moon must be at perigee (nearest to the Earth) and many references state that its orbital speed then is 1.082km/sec or about 3895km/hr. Subtracting the Earth's rotational speed (about 1670km/hr) gives a shadow speed of about 2225km/hr. Unfortunately, this speed just doesn't work when plugged into the formula: assuming a maximum shadow size of 273km the maximum eclipse time should be 7mins 22.5sec rather than the 7mins 31 or 40 or 45 (depending on the source!) sec given in references. It's actually shorter than the longest to have actually occured (7mins 29sec) so must be wrong. To get a value of 7mins 40sec the moon's orbital speed would have to be 1.0584km/sec. Furthermore, the 1.082 figure is actually too small for a maximum eclipse as it corresponds to the current lowest perigee of 363104km rather than the "all time" lowest figure of 356371. At this altitude the speed should be 1.104km/sec (I think) giving an even smaller totality time of 7mins 6sec. The same problem exists with annular eclipses by the way: the speed at apogee required for an adequately long eclipse is considerably smaller than the oft-quoted value.

So what's going on?? I really can't believe that the moon isn't going round the Earth at the speed everyone thinks it is so there must be some error in the theory I am using. It's not just me though - several sites I have consulted use exactly the same idea but either don't explain how they got their numbers (a listing of "longest eclipses" quotes a shadow speed which would correspond to a perigee orbital speed of 1.0365km/sec rather than the 1.1006 required at the quoted perigee of 357349) or fail to notice that their maths just doesn't work (one quotes a shadow width of 270km and a shadow speed of 1709km/sec, which would give a totality time of 9mins 36sec!!).

I have failed to come up with any explanation for the rather large differences detailed above and thus put my problem to all you experts out there. I won't mind if I am castigated for a simplistic theory or have a stupid error pointed out - I just want to know the answer!

Steve Holmes, UK 88.109.123.99 17:22, 20 November 2006 (UTC)[reply]

I don't really know what I'm talking about but still: do you need to account for Moon's inclination compared to Earth's equatorial plane? If you are standing on Earth's equator you move at an oblique angle compared to the Moon's orbit above you, so you can't subtract the whole 1670 km/h. I plugged in cos(18.29°) into your numbers and got 7 min 40 sec. But I really don't know if that's anywhere near right... Weregerbil 18:45, 20 November 2006 (UTC)[reply]

(edit conflict) First, a clarification: since you mention Earth's rotational speed, I'm guessing you're talking about the longest duration of totality at one location, rather than the longest period of time during which there is a total eclipse somewhere on Earth. Assuming that we're on the same page here, the likely culprit seems to me to be the motion of the Earth around the Sun; that will direct the Moon's shadow westward and lengthen the eclipse. In other words, we have to consider the Moon's velocity in a frame where the Earth and Sun are both stationary; call it its synodic velocity. Assuming for simplicity a circular orbit, this is

{27.321582\mathrm {d}  \over 29.530588\mathrm {d} }=0.92519600

times its actual velocity (relative to Earth), which I get to be 3604 km/h using your original figure. Then the shadow speed is 1934 km/h, which gives 8m28s as the eclipse time. (Using your 1.104 km/s number gives 2006 km/hr or 8m10s.) This is of course longer than your references! (It's also necessary to consider Earth's rotational velocity in the synodic frame, which is

0.997258(1674.4\,\mathrm {km/h} )=1669.8

, but the difference is obviously close to negligible.)

(after edit conflict, with Weregerbil's idea) First, we're close to a major standstill currently, so the Moon's inclination is actually 28.58° or thereabouts. Second, you make an excellent point, but we don't just want to take the cosine directly. Instead, we should consider the moon to be moving, say, southeast with its true orbital velocity, plus due west by an amount equal to the ideal defect of synodic velocity, while a point on the equator moves due east with its synodic velocity. The relative shadow velocity will be the vector difference of these two: we then have

\langle 3895\,\mathrm {km/h} \,\angle 28.58^{\circ }\rangle +\langle (0.92519600-1)3895\,\mathrm {km/h} \,\angle 0\rangle -\langle 1669.8\,\mathrm {km/h} \,\angle 0\rangle =\langle 1459.2,1863.3\rangle \,\mathrm {km/h} =\langle 2366.675205853\,\mathrm {km/h} \,\angle 51.93^{\circ }\rangle

; the Moon's shadow actually is moving faster south than east relative to the ground! With this relative shadow velocity, though, we get just 6m55s as the maximum totality. Perhaps the calculation at minor standstill is desired: then we get

\langle 2124.0\,\mathrm {km/h} \,\angle 35.1^{\circ }\rangle

, for a time of 7m43s. That seems to be the desired answer, and is also very close to what you (Weregerbil) said initially; my apologies if I only imagine you to have done something too simple! --Tardis 20:26, 20 November 2006 (UTC)[reply]

I didn't read all the details of the question, but I do remember that it requires a concord traveling faster then sound to keep up with a solar eclipse. Ariel.

Thanks to Weregerbil and Tardis for their improvements to my simple model. I had myself wondered whether the synodic period was the one to use but had entirely overlooked the "geometric" effect of the tilt of the moon's orbit. As mentioned by Tardis, the value at minor standstill is the one to use as then the effect of the Earth's rotation is as great as it can be. The cos factor certainly gets the value for a total eclipse much closer to expectations but unfortunately doesn't solve the annular eclipse situation where the model gives 11min 32sec as compared to a requirement of at least 12min 24sec. Clearly one could simply put in a much smaller minimum perihelion value but this isn't really the answer as for the eclipse in question (12min 24sec) one can find the true value from the JPL ephemeris calculator and it's nowhere near as small as that. Is there thus anything that might (due to their slightly different geometry perhaps) affect annulars and not totals??

Steve Holmes, UK88.109.123.99 10:53, 21 November 2006 (UTC)[reply]

All I can think of that's different about annular eclipses is that the shadow has, as it were, negative size on the ground. Presumably it can be bigger when negative than when positive, since you have a larger time. I might be able to say more if I knew what you had done to get it. --Tardis 17:06, 21 November 2006 (UTC)[reply]

My intention is to calculate the shadow width (simple geometry) and the shadow speed (not easy, it seems!) and thus get the duration. A maximum annular will always exceed a maximum total because the shadow cone in the former case is longer (as the moon is farther from the earth) and diverging: the umbral width is thus much greater. I think the problem rests with calculating the shadow speed: further consideration leads me to believe that a full vector subtraction is required, not just the "cos" factor suggested above, but unfortunately this makes things worse as the resultant speed is quite a bit larger than the "cos"-style value. Any further thoughts anyone??

Steve Holmes, UK88.110.180.219 23:49, 23 November 2006 (UTC)[reply]

Capacitor Power

Imagine a resistor is connected to a dc and ac source.The power is equal to product of voltage and current.(guess in the form of heat) Now replace resistor with capacitor. In dc, no power is output as no current flows, but in ac, some current flow due to induction.But assuming perfect capacitor, no heat is produced.SO how can we say power is dissipated?

Power need not be dissipated to exist. With a DC source, you will charge the capacitor, giving it some energy in a period of time (that is, at a certain power). But the energy can be retrieved from the capacitor, so it's not dissipated. With an AC source, you will alternately charge and discharge the capacitor, and energy will flow back and forth between it and the source at a changing power. Perhaps you're thinking of real power, which is 0 in the AC case with a perfect capacitor? --Tardis 17:57, 20 November 2006 (UTC)[reply]

This is an interesting question. What will happen with a capacitor connected to AC is you will have infinite current, and 0 voltage (as measured across the capacitor), because the voltage is 0 no power is consumed, even though the current is infinite. (Current is infinite if the power soure is infinite, and that the capacitor has 0 resistance, and the wires can handle it. Obviously that doesn't exist in the real world.)

On the other hand, if you hook up DC to the capactitor you will have infinite voltage (well, actually whatever the source voltage is), and 0 current, and again the power will be 0.

So if you hook a capacitor across the wires of a household outlet, you will trip the breaker, but your meter will not register any usage.

Ariel. 06:47, 21 November 2006 (UTC)[reply]

This isn't entirely true. Capacitors, even ideal ones, still have impedance though they lack resistance, so they will not draw infinite current from an AC source. With DC, the current is limited by whatever resistance a real capacitor has, and/or by the accumulation of voltage on the capacitor. Yet more fun would be something like a square wave driver... --Tardis 16:52, 21 November 2006 (UTC)[reply]

The current flowing in a pure capacitor leads the applied voltage by 90 deg. Since power is defined as VI cos (phi) where 'phi' is the phase angle between the voltage and current, power dissipated in any pure capacitor must be zero. And that, my friends, is why you can have voltage across, current through, but no power dissipated in a capacitor --Light current 07:03, 21 November 2006 (UTC)[reply]

so is it like heat is a form of work done by the resistor???

Energy is work. So heat energy is also work. Work is not done by the resistor, but energy is dissipated by it. The work could be said to be done by the source of voltage. IOW, electrical energy is transferred to the resistor from the source. This is then converted into heat energy.--Light current 08:49, 21 November 2006 (UTC)[reply]

Voltage (represented by E for electromotive force) leads current (abbreviated I for intensitè ) by 90 degrees in a purely inductive circuit. In a purely capacitative circuit, current leads voltage by 90 degrees. The memory trick to keep this straight is "ELI the ICE man." Edison 15:25, 21 November 2006 (UTC)[reply]

Electric Intensity

In fifth edition of APPLICATIONS OF ELECTRO-MAGNETICS by KRAUS and FLEISCH,(Page 50-60) a man stands under a transmission line and a tube light glows without power.So if Electric field is only a function of charge density,then will the same effect happen if we place a highly charged sphere which also has high electric field intensity???

If you suddenly bring a large quantity of charge near one end of a fluorescent tube, it will briefly light but then go dark. The continuous powering of the light arises because in the vicinity of an electric current there is a changing magnetic field, which gives rise to a changing electric field, which is capable of re-lighting the tube each time it changes direction, as if it were an AC source, albeit one that doesn't need to be attached to the tube's ends particularly. The fact that the tube can glow forever indicates that there is power present; in this case, it's the power of the electromagnetic radiation that comprises the changing fields. Normally almost all of the power is reabsorbed by the transmission line later in the same cycle, but the tube saps some of it merely by being present. --Tardis 18:27, 20 November 2006 (UTC)[reply]

A fluorescent tube would certainly glow if near a sphere energized with high voltage alternating current. Likewise cubes, and any other shapes. There is a severe danger of injury or death when working around high voltage electricity, so experimentation is strongly discouraged. Edison 21:22, 20 November 2006 (UTC)[reply]

IF the sun would be bought nearer to the eatrh, the earth would have to move faster(more gravitational potential). Threr would no be any need for sudden movement (oscillation) of the sun to cause this force.Since ELECTRIC Field intensity can be modeled the same manner,Then why do we need to Move(oscillate) the Heavily charged sphere????

teeth vs sugar substitutes

As i understand it, damage to a person's teeth is caused by sugar. Specifically, the dental caries article says this:

Tooth decay is caused by acid-producing bacteria which cause the most damage in the presence of fermentable carbohydrates, such as sugar. The subsequent acidic pH levels in the mouth affect teeth because of their high mineral content.

This makes me wonder about sugar substitutes (such as aspartame, saccharin, and sucralose). The back of any diet-soda can will tell me that these compounds contain about zero carbs, so i kind of doubt they could be considered 'fermentable carbohydrates'. Does that mean the bacteria in my mouth can't use them the same way they can use 'real' sugar (like sucrose or corn syrup)?

In other words, should i feel guilty if i drink a Fresca right after i brush my teeth? ~ lav-chan @ 18:13, 20 November 2006 (UTC)[reply]

First - flouride continues to protect your teeth after you brush. So, other than a weird taste, drinking sugary sodas is fine right after you brush your teeth, but it will wash away the flouride and then the sugar can fuel up the bacteria. Non-carb sweeteners cannot be used as a food source by bacteria. I won't be surprised when someone announces they've discovered a bacteria that can use it as a food source, but as of now the bacteria in your mouth cannot survive on it. --Kainaw ^(talk) 18:32, 20 November 2006 (UTC)[reply]

However, even if your fruit juice is sugar-free, it itself has acidic behaviour and will damage your teeth if left in the mouth. This can be prevented by taking a few mouthfuls of water after drinking juice. Recommended with cranberry juice, as it is particularly acidic. -- 85.179.15.106 19:03, 20 November 2006 (UTC)[reply]

Sugar substitutes are used to reduce fat, not so much for dental health.

There would be a problem if the body, tasting sugary things, would, behaving like Pavlov's creatures, try to keep much more fat from every food allowing it, eaten in the same hour. But this may never occur and never believe all that tell you the contrary :). -- DLL ^{.. T} 21:24, 20 November 2006 (UTC)[reply]

However, feeling guilty is not necessary, lav-chan. If in doubt about the effect of Fresca, just brush your teeth again. That takes less effort and time than asking a question here. JackofOz 23:39, 21 November 2006 (UTC)[reply]

fermion or boson?

A no-spin particle a boson? a spin of zero is still an integer isn't it?

Yes, bosons can have an integer or zero spin. The hypothetical Higgs boson has a zero spin. –mysid ☎ 19:05, 20 November 2006 (UTC)[reply]

A spin-0 particle is a boson, but be cautious about describing zero as an integer. Zero is a very strange (for want of a better word) number

compounds

is there a lewis structure that glycinate has that can act as a bidentate ligand?

With out going into details gyclinate can act as a bidentate ligand with the nitrogen and an oxygen both bonding to a (metal) atom. This forms a 5 membered ring. (See glycine glycinate is glycine with one proton removed.)

gases

Is propane gas heavier or lighter than air?

--Mmwanda 21:30, 20 November 2006 (UTC)[reply]

Try comparing a molecule of propane to a molecule of nitrogen first. Melchoir 21:37, 20 November 2006 (UTC)[reply]

The density of propane is 1.83 kg/m³ whereas the density of air is 1.2 kg/m³. Thus, propane is heavier than air. –mysid ☎ 21:37, 20 November 2006 (UTC)[reply]

EDIT conflict: From the article "Unlike natural gas, propane is heavier than air (1.5 times denser). In its raw state, propane sinks and pools at the floor." This agrees with my own experience. 152.3.73.203 21:41, 20 November 2006 (UTC)[reply]

Entropy

Let's say, I have a gas and I rapidly expand it, and then compress it back to it's original pressure, temperature, and volume. How does the entropy of the system change? It simply does not increase, but can we be more specific? --HappyCamper 01:38, 21 November 2006 (UTC)[reply]

The entropy within the contained gas would not have changed, but to do the work that would have allowed the expansion & compression would have created entropy elsewhere, right? It's been a while since college physics... -- Scientizzle 02:25, 21 November 2006 (UTC)[reply]

Correct; the entropy of the gas will remain unchanged. The entropy of the surroundings would increase. — Knowledge Seeker দ 02:52, 21 November 2006 (UTC)[reply]

Scientizzle and Knowledge Seeker are wrong. What HappyCamper is talking about is the basic engine cycle. Entropy is a representative measure of the work that the molecules of the system do on each other as they change position and arrangement over the course of the process. The fact that the intermolecular work energy of the forward process does not exactly cancel out the intermolecular work energy of the reverse process, a characteristic associated with the phenomenon of irreversibility, implies that there will be a increase in entropy associated with the total process (cycle), which is a measure of this difference. The following inequality, according to Clausius, must hold good for every cyclical process which is in any way possible:

\int {\frac {\delta Q}{T}}\geq 0

Therefore, anytime a system changes state, there will be an entropy change associated with this change due to intermolecular rearrangements. Later: --Sadi Carnot 03:33, 21 November 2006 (UTC)[reply]

These answers are actually all correct; entropy is a state variable, and so if you have a quantity of gas that has resumed some previous state, its entropy will be the same as it was the first time around. The trick is that the engine cannot undergo such a cycle in isolation; it must be in contact with a larger system, and it is the system as a whole — which is what allows the cycle — which must accumulate entropy in the process. Moreover, it's not automatically the case that positive entropy is accrued; Sadi Carnot quite correctly used

\geq 0

(see isentropic process). However, there are no processes which can do useful work for which

\Delta S\not >0

. --Tardis 04:04, 21 November 2006 (UTC)[reply]

Er, Sadi, I don’t see how your answer contradicts mine, or how you showed I was wrong. As Tardis mentions, if the temperature, pressure, and volume of the gas return to the same state, so will the entropy of the gas. The increase in entropy manifests in the external apparatus and such. — Knowledge Seeker দ 04:11, 21 November 2006 (UTC)[reply]

Is lightning lethal?

If you were hit by a thunderbolt, could you survive? Does the size and energy of a thunderbolt vary the survival rate?

Yes you can. It depends on lots of things thoughh. See lightning--Light current 02:31, 21 November 2006 (UTC)[reply]

Yes, yes, yes, yes, yes, yes, yes. Clarityfiend 04:30, 21 November 2006 (UTC)[reply]

A low-voltage (110 to 220 V), 50 or 60-Hz AC current travelling through the chest for a fraction of a second may induce ventricular fibrillation at currents as low as 60mA, which is usually fatal. The lightening article states that a lightening bolt is in the ten to hundreds of Amperes. That's orders of magnitude more. Normally, when it is said that people are struck by lightening, it struck near them. No living creature could take a direct hit. But wait! Lightening says the resistance of the skin is so good that lightening won't bother your heart that much. I'm kind of skeptical to that... X [Mac Davis] (DESK|How's my driving?) 04:43, 21 November 2006 (UTC)[reply]

There is an interesting book called "A Match to the Heart" by Gretel Ehrlich about her slow recovery from being struck by lightning. (Probably most of the lightning did not go through her.) In her book I found the sentence: "The overall mortality rate following lightning injury is 30% and in survivors the morbidity rate is 70%." Cardamon 05:24, 21 November 2006 (UTC)[reply]

THe trick is of course, to let it flow over the surface of your body to ground!--Light current 05:53, 21 November 2006 (UTC)[reply]

If you were wearing a completely conductive suit that wouldnt vaporise, I think you could survive a direct hit! See Equipotential suit. Unfortunately we dont have a page on it yet!--Light current 09:57, 21 November 2006 (UTC)[reply]

People who survived being struck by lightning often get mental problems. I can't remember the details, but one such person's brain was examined and it didn't operate normally. DirkvdM 08:59, 21 November 2006 (UTC)[reply]

Sorry to hear that Dirk! But you are obviously better now! Apart from the memory problem 8-)--Light current 09:51, 21 November 2006 (UTC)[reply]

I think people can survive a direct hit, LC. Apart from reports I've read recently about people struck by lightning where witnesses close by saw it hit their head, I seem to recall specific details about it often causing bleeding between the toes where the current flowed to the ground, suggesting a direct hit. Skittle 15:24, 21 November 2006 (UTC)[reply]

Hmmm difficult to tell if the full 100kA went thro them tho!--Light current 15:44, 21 November 2006 (UTC)[reply]

I believe it mostly goes through the skin and clothes. That would be consistent with a current going through a wire really going through the surface of the wire because the electrons repell each other. The skin may show burn patterns that look like lightning in the sky (searching its way). Also, metal objects worn close to the skin caused more serious burn marks. DirkvdM 08:01, 22 November 2006 (UTC)[reply]

Nevertheless, kids, do this only under strict adult supervision. --Lambiam ^Talk 13:26, 22 November 2006 (UTC)[reply]

Yes, you can survive, but it would usually have to pass through your skin, which will probably burn it to some degree or another. I remember reading about someone who received a direct hit, and survived, but whose back showed a tree-like burn (it had a particular term, but I can't find the paper in which I read it) due to the current using the skin as a conductor. Of course, it helps to be close to God too (because I do remember that the person I'm talking about was a priest)... Tito xd^(?!?) 04:54, 23 November 2006 (UTC)[reply]

Nice god. He strikes down his disciples. DirkvdM 08:00, 23 November 2006 (UTC)[reply]

remote neural monitoring

i was just a city council meeting and some concerned citizen was voicing his opinion that rnm is brainwashing america, so that raised the wuestion in me: is he for real? does this exist and what exactly is it, can smoeone please clarify this for me. thanks --69.140.210.163 02:28, 21 November 2006 (UTC)[reply]

see Remote viewing--Light current 02:34, 21 November 2006 (UTC)[reply]

Crank'll probably hold one or two answers, as well. GeeJo ^(t)⁄_(c) • 05:56, 21 November 2006 (UTC)[reply]

Scientist

This may not be a good place to ask this but does anyone know how I can get in contact with Robert Bussard?67.121.107.103 03:35, 21 November 2006 (UTC)[reply]

This is very difficult. He's mentioned all over Google but I cant seem to find an email addr for him.--Light current 10:19, 21 November 2006 (UTC)[reply]

It is definitely not the right place as Wikipeida policy would prevent disclosing of any identifying information. --Tbeatty 04:08, 22 November 2006 (UTC)[reply]

Wheeeeeeee!

What's the highest velocity a macroscopic object has been clocked at? As a subquestion, and making the fairly safe assumption that the answer to the first is probably an astronomical object, how about one that's been accelerated artificially? GeeJo ^(t)⁄_(c) • 05:52, 21 November 2006 (UTC)[reply]

Check out the Z machine from the article: 'The Z machine is now able to propel small plates at 34 kilometers a second, faster than the 30 kilometers per second that Earth travels in its orbit around the Sun, and three times Earth's escape velocity.' I couldn't tell you if that's the record though. Ariel. 07:24, 21 November 2006 (UTC)[reply]

The Helios probes are the fastest man made object, travelling at over 70 km/s. Astronomically, the solar system is travelling at 200 km/s through the galaxy. (For more info, see Orders of magnitude (speed)) La ï ka 08:13, 21 November 2006 (UTC)[reply]

Perfect link, exactly what I was hoping to find. Though going by the principle of mediocrity, I'd assume that there are probably faster-travelling large objects than our own solar system out there. Is there a list somewhere of the fastest-travelling celestial bodies discovered so far? (PS. That image in the Z machine article has to be one of the weirdest-looking pieces of equipment I've ever seen.) GeeJo ^(t)⁄_(c) • 08:40, 21 November 2006 (UTC)[reply]

For a natural object, see stars exceed 1000 km/s at pulsar kicks and [2]. I have a vague recollection of theoretical stellar encounters that left a star going relativistic but I'll leave that for you'all to google (and it's not observed, merely theoretical). Weregerbil 08:57, 21 November 2006 (UTC)[reply]

Bicycle on a slope

A bicycle is rolled down a slope. The time is measured for the bike to roll down slope. If the distance is known, hence an average velocity can be calculated; is the average velocity equivalent to the instantaneous velocity at the half way point of the slope (as indicated in the diagram) assuming constant acceleration? Oh and just to clarify, this isn't a homework question (I've finished school for the year), I'm just curious --Fir0002 05:40, 21 November 2006 (UTC)[reply]

If there's constant acceleration, the average velocity is merely the average of the starting velocity and the ending velocity. (If it starts from rest, the average is just half the final velocity. Let's suppose it does start from rest, to make this simpler.) Then see gravitational potential energy and kinetic energy (related, of course, by conservation of energy), and see whether the kinetic energy gained by the time the bike is at the half-way point is the correct amount to give it half the final velocity. (Pay attention to powers of v and h.) --Tardis 05:58, 21 November 2006 (UTC)[reply]

No. As Tardis mentioned, $v_{a}={\frac {1}{2}}(u+v)$ where u and v are the initial and final velocities and $v_{a}$ is the average velocity. using the formula $v^{2}=u^{2}+2aS$ , you can find out that the velocity of the body at half way is $v_{\frac {1}{2}}={\sqrt {{\frac {1}{2}}(u^{2}+v^{2})}}$ . Hence $v_{a}\neq v_{\frac {1}{2}}$ -- WikiCheng | Talk 08:35, 21 November 2006 (UTC)[reply]

An easy way to conceptualize this is to note, as everyone has mentioned, that acceleration is constant, so the mean velocity will be the arithmetic mean of the initial and final velocities, and will occur exactly halfway through the time interval. However, for the first half, the velocities are all less than the mean, and for the second half, they are all greater than the mean. This means that the bicicyle will travel farther during the second half of the interval, and therefore, the point at which the bicycle is at mean velocity will be closer to the start than the finish. — Knowledge Seeker দ 08:46, 21 November 2006 (UTC)[reply]

Ok am I then correct in saying that the instantaneous velocity halfway along the slope is "related" to the average speed over the whole slope, following the math below

${v_{\frac {1}{2}}}^{2}=2a{\frac {1}{2}}x$ from the formula $v^{2}=u^{2}+2ax$ , where u = 0 hence ${v_{\frac {1}{2}}}^{2}=ax$ , call this eq. 1

also

$x=ut+{\frac {1}{2}}at^{2}$ , but u = 0 hence $a={\frac {2x}{t^{2}}}$ , call this eq. 2

substituting eq. 2 int eq. 1

gives ${v_{\frac {1}{2}}}^{2}={\frac {2x^{2}}{t^{2}}}$ $v_{\frac {1}{2}}={\sqrt {2}}({\frac {x}{t}})$ and $v_{av}={\frac {x}{t}}$

hence $v_{\frac {1}{2}}={\sqrt {2}}(v_{av})$ --Fir0002 10:06, 21 November 2006 (UTC)[reply]

Correct. You can also reach the same result using conservation of energy. If final velocity is v and initial velocity is 0, then, assuming no friction

{\frac {mv^{2}}{2}}=2\left({\frac {mv_{\frac {1}{2}}^{2}}{2}}\right)

because the vertical distance between the foot of the slope and the top is twice the vertical distance between the half-way point and the top. So

v_{\frac {1}{2}}^{2}={\frac {v^{2}}{2}}

and v = 2v_av because initial velocity is 0, so

v_{\frac {1}{2}}^{2}=2v_{av}^{2}

.

Gandalf61 10:51, 21 November 2006 (UTC)[reply]

Correct. (By the way, I have corrected an error in my equation). It is $v_{\frac {1}{2}}={\sqrt {{\frac {1}{2}}(u^{2}+v^{2})}}$ -- WikiCheng | Talk 11:16, 21 November 2006 (UTC)[reply]

the body and muscles

most muscles in the body work in pairs in whaat is known as the "agonist-Antagonist" relationship. Provide three examples of muscle pairs that clearly display the agonist-antagonist relationship

Well now, this seems a bit like homework, doesnt it? Why not see our pages on muscles?--Light current 06:08, 21 November 2006 (UTC)[reply]

I don't see anything in muscle that's helpful. So perhaps we should actually try and help with the homework question. The "agonist-antagonist" relationship mentioned is, in simpler terms, just pairs of muscles that move a joint in opposite directions. The direction of movement at a joint is described in anatomical terms. See Anatomical terms of motion. Often muscles are named for the motion they produce, and you can use the naming to point you to "agonist-antagonist" pairs: flexor vs. extensor, abductor vs adductor. So for example, the "flexor pollicis longus" and "flexor pollicis brevis" oppose the "extensor pollicis longus" and the "extensor pollicis brevis" in the movement of the thumb. And the "adductor pollicis" opposes the "abductor pollicis brevis" and the "abductor pollicis longus". Other opposing pairs without such names would be biceps brachii (flexes the forearm) vs. triceps brachii (chief extensor of the forearm), or the gluteus maximus (extensor of the hip) vs iliopsoas (flexor of the hip). You may find [3] helpful in looking at the actions of muscles. - Nunh-huh 14:40, 21 November 2006 (UTC)[reply]

Help needed to identify species of Swallow

Hello, please see the photograph I took (details regarding location and circumstances are on the image description page). I need help wth identifying the species (I'm no ornithologist), and it's been posted on Wikipedia:Picture peer review to help place it in an article. Thank you!

Shock waves

Simple question that the article evades: What is the exact difference between normal compression (longitudinal) waves and shock waves in, say, a gas (like air) or indeed water?--Light current 10:37, 21 November 2006 (UTC)[reply]

What Im asking is whether the medium experiences non linear behaviour. 8-|--Light current 15:46, 21 November 2006 (UTC)[reply]

I'm not an expert in this field, but I can give the simple answer: shocks are discontinuities in variables such as pressure, temperature, etc.; normal compression waves are gradual variations. Conceptually, a shock has zero width; there is simply the fluid on one side, and the fluid on the other side, and they do not blend smoothly together in the middle. In order for this zero-width state to persist, I believe (but this is getting beyond my certain knowledge) that the shock must be travelling at at least the speed of sound in the downstream medium so that no news of its passage "prepares" the fluid for the oncoming shock, destroying the singularity. --Tardis 15:43, 21 November 2006 (UTC)[reply]

Yes, a shock can only exist at at least the speed of sound. However, the body itself does not necessarily need to be travelling that fast - in the transonic region it is possible for some bodies to be travelling at a subsonic speed, but have a localised flow around them that is supersonic. Also, a shock, while approaching a discontinuity, is not genuinely one. There are some limiting mechanisms, like viscosity for instance. Readro 21:21, 21 November 2006 (UTC)[reply]

As far as I know a compression wave is the kind you get when slap a fatty human thigh and watch it ripple. A compression wave comes from tiny pieces of the medium conducting it along, smashing into the next piece, and so on. A shock wave is what you get from C4 or a nuclear blast—the explosion pushes the air faster than it can conduct, like a sonic boom. X [Mac Davis] (DESK|How's my driving?) 17:08, 21 November 2006 (UTC)[reply]

"Watch it ripple" sounds like the fat wave has a transverse component. A shock wave sounds like a longitudinal wave, probably with a sharp rise time to the max pressure, then a damped wave train. Edison 22:41, 21 November 2006 (UTC)[reply]

Okay, okay, loudspeaker. X [Mac Davis] (DESK|How's my driving?) 01:06, 22 November 2006 (UTC)[reply]

Thallium as a prescribed drug

In light of the recent news stories regarding THALIUM as a poison, I have a query regarding its possible use as a prescribed drug.

Does anyone have any information about the use of this element, in the treatment of emotionally disturbed (depressed or hyper-anxious) patients. I have come across anecdotal evidence of it's use by INDIAN military medics, specializing in psychiatry.

Any information glady appreciated.

ROBIMARS 10:38, 21 November 2006 (UTC)[reply]

I can't offhand think of any medical use for thallium or a thallium salt other than possibly as a radionuclide during scans. A quick google search turns up this, which makes the claim that thallium was initially used as a depilatory agent, before its toxicity was fully established. GeeJo ^(t)⁄_(c) • 11:42, 21 November 2006 (UTC)[reply]

The papers claim not only that it was used in depilatory creams, but that it can still be used to treat ringworm. Skittle 15:20, 21 November 2006 (UTC)[reply]

Merck Index (10th ed) lists ²⁰¹Tl (t_1/2=73 h) as a diagnostic agent for myocardium scans, but gives no other human applications. DMacks 17:45, 21 November 2006 (UTC)[reply]

At least Merck agrees with me then :) GeeJo ^(t)⁄_(c) • 11:50, 25 November 2006 (UTC)[reply]

perhaps confusion with lithium? that's used in psych. Xcomradex 20:55, 21 November 2006 (UTC)[reply]

Is it true that people taking Lithium start feeling light headed? 8-)--Light current 23:44, 21 November 2006 (UTC)[reply]

(ec i'd say they become reduced to becoming angry after drinking, or as they say dangerous when wet ;-) Xcomradex 01:55, 22 November 2006 (UTC)[reply]

And people taking dilithium crystals want to go where no man has gone before. :-) StuRat 01:53, 22 November 2006 (UTC)[reply]

Gases

Is nitrogen heavier or lighter than oxygen?

Look at links--Light current 10:52, 21 November 2006 (UTC)[reply]

The atomic weights of the two elements should give you a hint -- WikiCheng | Talk 11:29, 21 November 2006 (UTC)[reply]

A nitrogen molecule has an atomic mass of 28 (two nitrogen atoms) while an oxygen molecule has an atomic mass of 32 (two oxygen atoms). Therefore oxygen is heavier. Prieda.

Or, if you're talking about the gas, at STP, nitrogen weighs 1.251 g/L, and oxygen weighs 1.429. Same answer, oxygen gas is heavier. -- Plutor ^talk 15:38, 21 November 2006 (UTC)[reply]

(Of course, this doesn't speak to other phases, but:) Hmm...

32/28=1.1429

, more accurately

{\frac {2\times 15.9994}{2\times 14.0067}}=1.14227

, and

1.429/1.251=1.1423

. There's gotta be some sort of connection here! --Tardis 17:16, 21 November 2006 (UTC)[reply]