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April 9

low blood pressure and i. v. heroin use

can low blood pressure be caused from heroin abuse? what can the outcome be? — Preceding unsigned comment added by 174.19.186.92 (talk) 02:20, 9 April 2012 (UTC)[reply]

I wouldn't expect it as a direct result, but perhaps it could be an indirect result, if while shooting up the addict damages a blood vessel and bleeds out. Death could result. This is most likely when mainlining into major arteries. StuRat (talk) 03:50, 9 April 2012 (UTC)[reply]

Here are three sites [1], [2], and [3] that show an association but usually in cases of overdose. The causes of low blood pressure can be widely variable and the subsequent effects can be unpredictable depending on the particular physical circumstances of the person at the time. Richard Avery (talk) 07:25, 9 April 2012 (UTC)[reply]

Perfectly elastic collisions for equal masses

From the article on Collisions:

"Collisions play an important role in cue sports. Because the collisions between billiard balls are nearly elastic, and the balls roll on a surface that produces low rolling friction, their behavior is often used to illustrate Newton's laws of motion. After a low-friction collision of a moving ball with a stationary one of equal mass, the angle between the directions of the two balls is 90 degrees. This is an important fact that professional billiard players take into account.[1] Consider an elastic collision in 2 dimensions of any 2 masses m1 and m2, with respective initial velocities u1 in the x-direction, and u2 = 0, and final velocities V1 and V2. Conservation of momentum: m1u1 = m1V1+ m2V2. Conservation of energy for elastic collision: (1/2)m1|u1|2 = (1/2)m1|V1|2 + (1/2)m2|V2|2 Now consider the case m1 = m2, we then obtain u1=V1+V2 and |u1|2 = |V1|2+|V2|2 Using the dot product, |u1|2 = u1•u1 = |V1|2+|V2|2+2V1•V2 So V1•V2 = 0, so they are perpendicular."

Although I see no problem with the math, I've been having a hard time trying to believe the result that the final velocities are perpendicular. Can anyone provide evidence for this statement (A video, if possible)? Also, if correct, what mechanism is directly responsible for the change in direction of the initial velocity? If the final velocities are indeed perpendicular, that would imply a force acting on a different direction than that of the moving ball's trajectory. However, the only forces I can see are the respective normals of each ball, which, as far as I know, are perpendicular to the contacting surfaces, that is, parallel to the initial velocity of the moving ball. Is there something I am missing?186.28.49.19 (talk) 02:57, 9 April 2012 (UTC)[reply]

The cueball need not hit the target ball straight-on (square with the direction of travel)--they are not point particles. Here's a handwaving explanation for if not colliding straight-on: if the cue makes a glancing blow on the target, you have to decompose the initial velocity as one vector normal to the contact and one orthogonal to it. The contact-normal component results in energy transfer to the target in the direction of that vector. The orthogonal component is not affected by the collision, so the cue keeps moving in that direction. DMacks (talk) 03:23, 9 April 2012 (UTC)[reply]

In the real world, there is an additional complication in that when the collision occurs, the cue ball will usually have some spin, which may include spin imparted by the cue tip (in any direction) and/or a previous bounce from a cushion (in a horizontal direction), and usually (sometimes solely) a component caused by its roll along the baize: after the collision, this spin (which is not greatly affected by the near-frictionless contact between the balls) operating through friction with the baize, will contribute to the subsequent direction of the cue ball, changing it from the Newtonian ideal. If the ball is spinning and/or travelling with any appreciable speed, it "skids" on the baize, and only when it slows down sufficiently in either sense does it fully interact with the baize, so the ball may well travel in a curve, or change direction abruptly after a brief interval, following the collision.

Rather than videos, may I suggest that you go to the miniclip.com website and try playing the 8 Ball Pool QFP game (in its solo version via the "Play Instead" button rather than the multiplayer mode reached via the "Play" button). This gives, for an online version, quite a good simulation of most of the parameters of a billiards-type game. (Confession time – I'm an addict.)

Pertinent to your particular query, you will find that a hard cue shot played from any position against a ball on the centre spot (a situation that is automatically set up at some re-racks, and always if one contrives to leave the last ball of a frame in the rack area when potting the penultimate ball) so as to pot the latter into a side pocket will always send the (skidding) cue ball at right angles, parallel to the length of the table, whereas if you play the cue ball more gently, its rolling friction-induced spin will be able to cause it to deviate from this ideal direction. You will also see that the helpful post-strike direction indicator (absent if you play in "Expert" mode) always indicates the "ideal" (spinless) post-collision directions of cue and object balls to lie at right angles.

Should you be so minded, you may through repeated playing be able to develop both a reasoned and an intuitive grasp of the balls' behaviours in various situations of speeds and spins. {The poster formerly known as 87.81.230.195} 90.197.66.4 (talk) 05:42, 9 April 2012 (UTC)[reply]

To answer your question directly it seems there is indeed something you are missing. Any spin ("English") applied to the Que ball will be imparted onto the object ball in some proportion. HominidMachinae (talk) 06:04, 9 April 2012 (UTC)[reply]

But how would that spin be imparted in such a way that the resulting velocities are perpendicular? Also, since the formulation given in the article makes no reference whatsoever to spin or to the fact that they are balls (it only considers momentum and kinetic energy), it should be possible to change the problem to two cubes of equal mass on a low friction surface. In such a case, in which there is clearly no spin, what would be the mechanism that makes the final velocities turn out perpendicular to each other? By: 186.28.49.19 (the OP) 157.253.197.112 (talk) 21:13, 9 April 2012 (UTC)[reply]

I agree that the result that they end up perpendicular regardless of the initial angle is surprising, but that the math is right. And zero spin is assumed, so spin has nothing to do with it. You say "If the final velocities are indeed perpendicular, that would imply a force acting on a different direction than that of the moving ball's trajectory." But it's bouncing off -- things going in a straight line bounce off at an angle -- this happens if a cue ball bounces off an angled wall, so it's not surprising that it happens when the cue ball bounces off something that puts up less resistance than a wall does. So to me the only surprising part is the invariance of the result to the initial angle -- this somehow results from the assumption that the two balls have equal mass. Duoduoduo (talk) 14:47, 10 April 2012 (UTC)[reply]

Maybe the reason the 90° angle result seems counterintuitive is that we're used to seeing balls rolling with topspin continue in more or less the same direction after hitting another ball at an angle. But presumably the fact that it is rolling with topspin means that the cue ball, at the moment of impact, uses friction with the table to impart more momentum in the original direction, something assumed away in the math. Duoduoduo (talk) 15:17, 10 April 2012 (UTC)[reply]

CARDIOLOGY

Sorry, Wikipedia cannot give any medical guidance. For guidance on medical conditions talk to a physician. -RunningOnBrains^(talk) 16:22, 10 April 2012 (UTC)[reply]

The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

SIR PLS CAN U TELL ME WHAT IS THE GENERAL MEANING OF WHEEZING IS IT RELATED TO HEART DISEASE BUT THE GENERAL MEANING WHAT I KNOW IS THAT WHEEZING IS JUST DUE TO AILMENTS IN RESPIRATORY CAVITY BECAUSE OF INSUFFICIENT OXYGEN THE PERSON SUFFER FROM BREATHING TROUBLE IS THERENT ANY MEDICINE TO ERADICATE IT PLEASE GIVE ME SUGGESTIONS WHY CANT LADIESBECOME A CARTDIOLOGIST WHAT IS THE PROBLEM — Preceding unsigned comment added by Akshayaz (talk • contribs) 12:24, 9 April 2012 (UTC)[reply]

The reference desk will not answer (and will usually remove) questions that require medical diagnosis or request medical opinions. AndyTheGrump (talk) 13:20, 9 April 2012 (UTC)[reply]

Typing all in capital letters is bad manners too. Wickwack121.221.31.226 (talk) 13:22, 9 April 2012 (UTC)[reply]

Please re-write your question using correct capitalisation and punctuation. It is very difficult to understand it at the moment. --Tango (talk) 13:33, 9 April 2012 (UTC)[reply]

Um, there are female cardiologists; there's no reason that ladies can't become cardiologists. Nyttend (talk) 15:33, 9 April 2012 (UTC)[reply]

I can't see how this question is asking for medical diagnosis or opinion. It is asking for facts about a medical phenomenon. If he was to ask (for example)'why are bruises blue' the answers would fly in. Since when have we been only answering questions in perfect English? Please, Tango! it is not difficult to understand the three questions being asked. I trust all your comments will be flawless in future. Now, Akshayaz, the answer to your first question can be found by clicking on wheeze. The answer to your second question is not lack of oxygen but a narrowing or partial obsruction of the airways in the lungs (it may of course cause a lack of oxygen). There is medicine available to improve this problem but it must be prescribed by a doctor. Finally, in many countries in the world there are lady cardiologists. Perhaps it is your experience that makes you think that cardiologists are always men. I hope that helps. Richard Avery (talk) 07:37, 10 April 2012 (UTC)[reply]

The article Wheeze will tell you the general meaning of wheezing. You will have to consult a doctor to find out what is causing wheezing in a particular case. --Colapeninsula (talk) 08:38, 10 April 2012 (UTC)[reply]

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Seahorse gender

So one one of all time favorite factoids - seahorse females have penis and males get pregnant. Why do we think that males are males, not the other way around? (Besides genetics - I hear this assumption was around before gene testing) ~~Xil (talk) 13:19, 9 April 2012 (UTC)[reply]

The females produce eggs - that's what makes them female. Males produce sperm - that's what make them male. Male seahorses do not get pregnant - they merely have a carrying pouch to protect the young. Wickwack121.221.31.226 (talk) 13:28, 9 April 2012 (UTC)[reply]

And what female seahorses have isn't a penis, it's an ovipositor. Penises transmit semen; ovipositors transmit eggs. Red Act (talk) 13:41, 9 April 2012 (UTC)[reply]

The question, though, was how did science first arrive to that conclusion. I am pretty sure it was discovered before you could genetically test seahorses and also their eggs probably are microscopic enough for it to not be all that obvious. So how seeing an individual with penis-like appendage and another that gets pregnant you conclude that first is female and the second is male? ~~Xil (talk) 16:41, 9 April 2012 (UTC)[reply]

Fish don't get pregnant, so if you see a fish doing something analogous to pregnancy (and only loosely analogous at that), there is no reason to assume that is the female. The female's ovipositor isn't really penis-like. While the reproductive method of the seahorse seems superficially like the reproductive method of mammals with the roles reversed, that isn't how scientists would determine their sexes. They would compare them to other fish, and the female seahorse has far more in common with other female fish than other male fish, and vice versa. --Tango (talk) 17:28, 9 April 2012 (UTC)[reply]

In a strict sense, that is correct, as no fish nourishes offspring internally in the process of placental vivipary. However, a number of fish retain their fertilised eggs internally in the process of ovoviviparity until they hatch and thus exhibit viviparous birth. I even caught such a "pregnant" female fish once, a Viviparous eelpout, whose condition was not apparent until it was prepared for dinner. Some species of snakes and other reptiles, and some amphibians, do the same.

The seahorses' method has parallels in other Classes, such as the Gastric-brooding frog. {The poster formerly known as 87.81.230.195} 90.197.66.34 (talk) 19:53, 9 April 2012 (UTC)[reply]

You are right that scientists made the male/female identification long before genetic tests came along. Even today, it will be hard to tell the sex of an individual of an arbitrary new species just using a genetic test without knowing a little about the genetics of a related species. As 121.221.31.226 pointed out above, the scientists distinguished the sexes by looking for eggs and sperm. A microscopic / histlogical characterization was common very long ago. Also (as tango mentioned) when compared to other fish, the internal organs - testes, ovaries and accessory sexual organs are quite distinguishable, both macroscopically and under a microscope. So given the choice between calling an individual with eggs, ovaries, and a "penis" a male with eggs or a female with an ovipositor the choice was kinda obvious with enough details. Same goes for a female with testes and swimming sperm / male who has a pouch. Hope this clarifies things. mark as solved if you think it is.Staticd (talk) 18:39, 9 April 2012 (UTC)[reply]

What is the essential minimum difference that enables one to declare "this cell is an egg and that one is a sperm" if you know nothing about the animal that produced it? Is it "sperm have tails that they use to swim but eggs do not actively move"? Roger (talk) 07:53, 10 April 2012 (UTC)[reply]

Size (see sperm). --Colapeninsula (talk) 08:40, 10 April 2012 (UTC)[reply]

Note that isogamy does exist ancestrally. I suppose that it is possible that in some lineage, males and females eventually lost the differences between eggs and sperm and returned to isogamy, then later reacquired the differences with the opposite pattern, thus inverting male and female roles at the gamete level. But I certainly do not know of such a case. Wnt (talk) 16:59, 10 April 2012 (UTC)[reply]

white hole

a white hole sounds alot like a super nova. what is the proposed life expectancy of a white hole? if like some other phenomena, it can only last for a short period of time why couldnt this be what the formulas are seeing? wouldnt this also follow occams razor? — Preceding unsigned comment added by 165.212.189.187 (talk) 14:44, 9 April 2012 (UTC)[reply]

Please read White hole. It is not related to a super nova in any way at all. What formulae are you refering to? Plasmic Physics (talk) 14:48, 9 April 2012 (UTC)[reply]

Whatever formulae physicists use to interpret the universe that come up with these conclusions.165.212.189.187 (talk) 15:23, 9 April 2012 (UTC)[reply]

By their articles it is difficult for me to see how they are as different as you say. Could you name some differences? Also the white ~~dwarf~~ hole article says nothing can enter it does that include light? — Preceding unsigned comment added by 165.212.189.187 (talk) 15:03, 9 April 2012 (UTC)[reply]

I think in the original post when the OP typed white hole he meant white dwarf. When a moderate size star ages and eventually shrinks, it doesn't have enough mass to go supernova, so it becomes a white dwarf. When a high-mass star ages and collapses, the collapse results in it going supernova and then becoming either a neutron star or, in the very high mass case, a black hole. Duoduoduo (talk) 15:37, 9 April 2012 (UTC)[reply]

No he didn't165.212.189.187 (talk) 15:47, 9 April 2012 (UTC)[reply]

A white hole is a type of singularity. A supernova is an exploding star. They have nothing in common. White holes aren't actually very interesting - it turns out when you do the maths that they are indistinguishable from black holes. --Tango (talk) 18:25, 9 April 2012 (UTC)[reply]

ThHis "In addition to a black hole region in the future, such a solution of the Einstein equations has a white hole region in its past. However, this region does not exist for black holes that have formed through gravitational collapse." and this "objects falling towards a white hole would never actually reach the white hole's event horizon the white hole event horizon in the past becomes a black hole event horizon in the future" sound to me like the explaination of a supernova which results in a black hole. in the same relative terms could you explain an actual supernova-to-black hole event? 165.212.189.187 (talk) 19:26, 9 April 2012 (UTC)[reply]

More relevantly, note the nearby bit in white hole about there being no known physical processes that could result in a white hole. Conversely, we are entirely certain that supernovae and black holes exist and we're well-versed in the processes surrounding them. Consider using those articles as a basis for understanding the phenomena in question. — Lomn 19:41, 9 April 2012 (UTC)[reply]

I did. why cant someone list some differences in the properties of the two events in question instead of remanding me to the articles?165.212.189.187 (talk) 19:52, 9 April 2012 (UTC)[reply]

People have pointed out that supernovae exist and white holes don't. What further needs to be added on the question of whether one is really the other? --Sean 20:39, 9 April 2012 (UTC)[reply]

A supernova does not have singularity, or a event horizon for one. Plasmic Physics (talk) 23:20, 9 April 2012 (UTC)[reply]

If there is a white hole, it has always existed (by definition as far as I know, but at least in the commonly regarded case of the Schwarzschild metric), and a black hole exists at the same location in space. If what we see as a supernova happens at the location of such a pair of white hole and black hole, then the supernova would not "be" a white hole, but it would just be a particle ejection event happening once in the history of the white hole.

Look at the diagram in alternative coordinates: The hyperbolic curves are curves of constant r (the radial Schwarzschild coordinate). There are lines of constant t (the Schwarzschild time coordinate) which are not shown in the diagram, but they are simply straight lines passing through the origin. But only in the "exterior" regions (left and right quarters of the diagram) these lines of constant t actually correspond to lines in spacetime which can only be traversed in one direction (from past to future). In the interior regions (top and bottom quarters of the diagram) the hyperbolic curves of constant r are such boundaries between past and future (more technically called spacelike curves). That is because the Schwarzschild r and t coordinates sort of switch roles at the event horizon.

So a particle (or call it light and matter) ejection event would only be a pretty confined event happening at the boundary of the lower quarter and the right quarter of the diagram, at one small part of the event horizons many particles would cross to the outside.

I don't know why you would think that a supernova would be something like that. I would think that it is a product of stellar evolution rather than some white hole event. In case of the white hole event, I have no clue why there would be an ejection at a particular time, and the particles would come more or less from nowhere (from the singularity).

Icek (talk) 23:51, 9 April 2012 (UTC)[reply]

What do you mean? Wouldn't they come from the star? — Preceding unsigned comment added by 76.117.202.67 (talk) 05:27, 10 April 2012 (UTC)[reply]

You do not always know whether there was a star at the position where you see a supernova. If you want to explain even the observation of a star which becomes a supernova by a white hole, then make the white hole mysteriously eject particles at a lower rate for a longer time and at a higher rate for a shorter time... but again, a white hole hypothesis doesn't seem to explain anything. Icek (talk) 05:59, 10 April 2012 (UTC)[reply]

To summarize, the answer to your question is that white holes do not eject particles, and are not localized in time (they are extremely long-lived). Those are two reasons they can not be supernovae. -RunningOnBrains^(talk) 16:28, 10 April 2012 (UTC)[reply]

OK, I get it. but why does the first sentence of the article say that particles are emitted? — Preceding unsigned comment added by 165.212.189.187 (talk) 17:44, 10 April 2012 (UTC)[reply]

It doesn't, and it didn't before you edited it (today at 17:46). Icek (talk) 18:18, 10 April 2012 (UTC)[reply]

The wiktionary entry did! (until my edit).165.212.189.187 (talk) 17:51, 12 April 2012 (UTC)[reply]

Huh? It says "A white hole, in general relativity, is a hypothetical region of spacetime which cannot be entered from the outside, but from which matter and light have the ability to escape." And his edit of 17:46 merely changed "may" to the synonym "has the ability to". So does the lead sentence need to be changed to be consistent with the above assertion of Runningonbrains? Duoduoduo (talk) 16:35, 11 April 2012 (UTC)[reply]

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