Wikipedia:Reference desk/Archives/Science/2008 August 4

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August 4[edit]

Heparin lock[edit]

I've been undergoing minor testing in the hospital for several days, during which I've had a Heparin Lock in my arm. I've gone looking online for information about it, which has led me to conclude that it's potentially notable, even though there's no article about it. Does anyone know much about such a device? The medical staff have answered all my medical questions, so I'm not asking for medical advice for my specific situation: I'm simply more interested in information about the device in general. I've learned somewhat about it from the websites I've looked at, but as a distinctly non-medical person, I don't understand too much about it, except as it applies to my testing. Could someone explain the basics of this device: what it was originally meant for, what it's used for (in general today), information like that? Nyttend (talk) 01:32, 4 August 2008 (UTC)[reply]

A heplock provides access to the bloodstream for easy injection or sampling. Heparin is an anticoagulant, and is used to prevent the port from clotting. The alternatives would be to poke you with a new needle each time or to require you to have an IV drip to keep the line open. DMacks (talk) 05:17, 4 August 2008 (UTC)[reply]
It looks like there's a passing mention in port (medical), but it could use some beefing up. It'd be dandy if someone with suitable expertise could have a look, and perhaps create appropriate redirects. TenOfAllTrades(talk) 13:46, 4 August 2008 (UTC)[reply]

five kindoms[edit]

tigar which filam? —Preceding unsigned comment added by 121.247.108.75 (talk) 02:43, 4 August 2008 (UTC)[reply]

Your question is so full of typos and missing grammar that it is not possible to answer. It appears that you are asking which Five Kingdoms movie has the Tiger - but that is only a guess. Try spelling actual words and using complete grammar and someone will probably be able to give you an answer. -- kainaw 02:53, 4 August 2008 (UTC)[reply]
You can learn which phylum the tiger is in by looking at: Tiger. Dragons flight (talk) 02:55, 4 August 2008 (UTC)[reply]
Look at the box on the right. You don't need to read the article. --Bowlhover (talk) 03:42, 4 August 2008 (UTC)[reply]
BTW, the Five kingdoms model, while still taught in some schools is generally considered a rather poor model nowadays. One of the most obvious flaws is it puts Archaea and Bacteria together into Monera. Nil Einne (talk) 13:27, 4 August 2008 (UTC)[reply]

chordee[edit]

can steroid help cure chordee? —Preceding unsigned comment added by 116.71.150.79 (talk) 06:52, 4 August 2008 (UTC)[reply]

If you look at chordee it says treatment is best carried out in infancy by surgical means. If there was a treatment involving steroids I guess that would be the preferred treatment. Richard Avery (talk) 07:38, 4 August 2008 (UTC)[reply]

A lot of quasi-related questions[edit]

Sorry, I know these questions are going to be odd, but I couldn't find the answers on the net, so please bear with me. First, when we submerge something in water, there will be an upward force (buoyancy) acting on the object, right. The magnitude of this force (as per the wikipedia article) will equal the weight of the water displace, but why, intuitively, should this be? Secondly, an object that floats at the surface will also float at the bottom of the ocean, but how is this possible considering that there's the weight of the entire ocean above it that it has to contend against? The only explanation for this that I can think of is that water is a liquid, and so because the object can pass through it, it's weight is meaningless. But if this was the case, why do people keep mentioning the weight of the atmosphere above us as important (i.e. puts stress on objects, causes pressure, etc.) And why does air pressure drop as we travel farther from the surface of the earth is the weight of the atmosphere above us isn't significant? Finally, is bouyance responsible for convection (why hot air rises)? I would think so (hot air expands, thus less dense), but I wasn't sure. Thanks in advance! —Preceding unsigned comment added by 76.68.246.7 (talk) 09:16, 4 August 2008 (UTC)[reply]

By pushing an object into water, you displace water, and raise the water level slightly, so you're work done to push the object into the water, is stored as gravitational potential energy. The water level, acted on by gravity, is pulled down, and this forces the object up, providing that it is lighter than the water it displaces, this is the origin of buoyancy. Secondly yes, the water above an object does provide pressure in the same way air above us does, but it does not pin us down does it? gravity does that, and in the same respect, objects underwater are not 'pinned down' by the weight of the water above them.An object that floats placed at a greater depth will in fact experience a greater up thrust, as the higher pressure means the weight of the water it is displacing is greater. Philc 0780 11:28, 4 August 2008 (UTC)
A correction - The weight of the water above a submerged object is realized as pressure, which acts equally in all directions on an object, thus canceling out, so the only relevant forces are the weight of the object and the buoyant force. Also, the thrust does not increase with depth because the weight of displaced water does not increase with depth, because water is incompressible. To answer the other questions, you can say air pressure drops as you climb in altitude for the same reason water pressure drops as you approach the surface - just less weight piled on top - but air pressure also drops because its weight decreases with height due to the declining pull of Earth's gravity (relevant well above the troposphere). The weight of the atmosphere is certainly significant as atmospheric pressure. Buoyancy, in a sense, is responsible for convection, but since convection is a well-defined statistical mechanical effect, it might be more accurate to say that convection is responsible for buoyancy (but I'm not sure on this). SamuelRiv (talk) 13:54, 4 August 2008 (UTC)[reply]
I think this is splitting hairs, but yes, I would prefer to say convection causes buoyancy. Buoyancy is the simple approximation of the net fluid motion effects acting as a single equivalent upward force; it also assumes the object is sufficiently small that there is not a significant pressure gradient between its top and bottom parts, etc. The more rigorous treatment yields a similar effect via much more horrible integral equations. Nimur (talk) 15:50, 4 August 2008 (UTC)[reply]
Thanks a lot guys, really. But sorry, just one last thing, why is the pressure increase caused by the weight of water/air above us distributed evenly?
Because in the case of a liquid or gas, the molecules are free to move in all directions. They are under pressure (the weight of all the molecules above them) so they will push whichever way they can go to escape the pressure. If they can get a little more room by pushing your body inwards a little bit, that's what they'll do. They're pushing back just as hard on the molecules above them, beside them, and below them. (That is the highly unscientific explanation, but it works for me :) Franamax (talk) 02:14, 5 August 2008 (UTC)[reply]
"it also assumes the object is sufficiently small that there is not a significant pressure gradient between its top and bottom parts, etc" Buoyancy is caused by the pressure gradient between an object's top and bottom. The pressure increases as depth increases, so it pushes on an object's top with less force than on the object's bottom. The buoyancy article elegantly shows why this pressure difference is equal to the object's weight in water. A volume of water of any shape is not moving, so water must be pushing it upwards with a force equal to the water's weight. When the water is replaced by a solid, this force remains.
Convection is caused by buoyancy, which is in this case the pressure difference between a mass of air's "top" and "bottom". --Bowlhover (talk) 07:58, 9 August 2008 (UTC)[reply]

I was washing dishes...[edit]

And I noticed two things happen that I can't fully explain. First, if I were to plunge a glass into the water (with the open side facing it), the inside won't get wet. I'm not sure why this, and my only explanation is that the air pressure inside the glass is stopping it from entering (and if the water level were to rise, the pressure would increase and stop it eventually). Second, if I quickely remove the drain, a huge bubble will erupt. I not more or less why, but the details seem to be lacking. The water enters some sort of secondary container, and the air needs to escape. But what gives the column of air so much force? Thanks. —Preceding unsigned comment added by 76.68.246.7 (talk) 09:21, 4 August 2008 (UTC)[reply]

In response to the first, the air in the glass takes up the volume of the glass. We'd normally say that the glass is empty but it's not. It's full of air. The air would rise out of the glass if there were a hole in the bottom (now the top) of the glass. This would of course cause the glass to fill with water. You can demonstrate this with a straw though it may be hard to see. Plug one end of the straw with your finger and plunge it into water the same way as your glass into the sink. Then remove your finger. The water will push the air up the straw and out the top. Dismas|(talk) 10:05, 4 August 2008 (UTC)[reply]
To answer your second question, the air that is in the drain beneath the plug has to go somewhere when the water in the sink tries to go down. It can either follow the pipe or it can attempt to rise through the water. Water will attempt to fall because gravity is pulling it into the drain, forcing the air to attempt to rise because it will be displaced by the falling water. This the bubble you see. The reason the air doesn't just move down the drain is because the water falling into the sink does not form a solid uniform wall to prevent the air from escaping. EagleFalconn (talk) 15:38, 4 August 2008 (UTC)[reply]
This is more or less what I thought, but when I say huge, I really mean it i.e. water went everywhere.
If you release it quickly, there is a large amount of air that is rising. It has a bit of momentum when it reaches the water's surface and therefore lots of water gets spritzed everywhere. Dismas|(talk) 20:29, 4 August 2008 (UTC)[reply]
If you ever go SCUBA diving, you can take this experience to a whole new level. Just under the surface the water level in the up-ended glass will be near the lip. As you descend and pressure builds the water will force it's way further up into the glass. When you hit 33 feet under, the water will be exactly (by volume) half-way up the glass. At 66 feet it will be 3/4 and at 99feet it would be at 7/8. As you ascend, it will slowly "empty" in exactly the same manner. ---J.S (T/C/WRE) 23:44, 4 August 2008 (UTC)[reply]

Drinking water vs. pouring it over your head to cool down[edit]

If you had a small amount of room-temperature water, say 30 cL, on a hot, dry summer day (say 28 °C), would it help cool you down more/faster to drink the water, or to pour it over your head? What if it was hot or cold water? Or what if it was a humid day? --58.37.187.141 (talk) 10:11, 4 August 2008 (UTC)[reply]

I don't expect drinking the water will actually help to cool you down directly much, but it will help recover a small amount of the lost fluids from sweating. It doesn't matter much what the temperature of the water is (just don't burn your mouth!). Pouring it over you, provided the water is at a lower temperature then your surface temperature is obviously going to cool you down somewhat but it won't help you with the water lose due to sweating and given the small amount, isn't going to help that much. Obviously the colder the water, the more effective. However particularly if you have fairly cold water, I would suggest pouring it on a cloth and holding it over your head (or wherever you want to cool down) will be much more effective. Also, while 28 °C isn't exactly an extremely hot day whatever the case, you may feel there is greater benefit in humid conditions to using the water to cool you down as your sweating is less effective Nil Einne (talk) 13:15, 4 August 2008 (UTC)[reply]
If you drink it, 100% of the temp of the water will go into changing the temp of your body, while if you pour it on yourself, only a portion will, and the rest will go into changing the temp of the air. However, there is no evaporative cooling if you drink it, while there is if you pour it on yourself. Evaporative cooling works best on hot, dry, windy days. If the water is hotter than the air and you, I would expect it would be best to leave it to cool to the air temp before using it for cooling. Also, if you are already sweaty, pouring water on yourself won't help much, as you already have the benefit of evaporative cooling from the sweat. StuRat (talk) 13:23, 4 August 2008 (UTC)[reply]
I've heard that drinking 500 ml cold water will make your body burn 100 kcal to get back at its normal temperature. Regardless of the correctness of it, it implies that any cooling of the body is useless in itself unless you have elevated temperature because cooling when having normal temperature will be reversed by your body. So, the correct answer would be; it depends on the body's temperature. --Ayacop (talk) 13:32, 4 August 2008 (UTC)[reply]
The math: 1 kcal raises a L of water by 1C. 0.5L of cold water raised by 35C (that's from near-freezing to body temp) = 17.5kcal.
Whether the water is cold or warm will certainly affect the comfort of drinking or wearing it. If it's too warm, it could exacerbate the problem of overheating...humidity will reduce the ability of evaporative cooling to effectively remove heat from the body by sweating or evaporating doused water.
Furthermore...one may assume that drinking the water allows one to capture 100% of the coldness and it will be used for further cooling as sweat. Pouring the water on one's self will waste some of it as it falls off the body. In general, drinking the water would be more effective as a cooling mechanism, and to stay properly hydrated. — Scientizzle 15:59, 4 August 2008 (UTC)[reply]
Extra drinking water will only lead to extra sweating if it's hot and you were in a dehydrated state to beign with. Otherwise, water you consume will just be eliminated as urine. StuRat (talk) 20:21, 4 August 2008 (UTC)[reply]
Let's do the math a bit differently. Imagine that you drink 1% of your body mass as water which is 60°F cooler than your body temp. That would result in lowering your body temp by about 60/100 or 0.6°F. That might not seem like much, but the body temp is very closely regulated, and that can be the diff between sweating profusely and being quite comfortable. (And, of course, if that takes your body temp too low, your body would react and burn some calories to bring it back up.) StuRat (talk) 20:28, 4 August 2008 (UTC)[reply]
Intuitively, it seems to me that if you only have a small amount of water you should drink it, assuming it is cooler than your body temperature. As noted above, this way you capture 100% of the coldness, and it helps supply your body with water for sweating. If you have plenty of water, you should drink enough to become well hydrated, and then pour some over yourself.
Partly, this question is hard because you've asked the wrong question. Body temperature is tightly controlled, so unless you are experiencing hyperthermia you can't actually "cool yourself down" with water. What you really mean is "which will make you feel cooler?" There, the answer is less clear, because it involves questions about how the rate of sweating and the sensation of being cool depend on your state of hydration.--Srleffler (talk) 02:46, 5 August 2008 (UTC)[reply]
Well, body temp is controlled, but it's not constant, it varies a bit. And even a few tenths of a degree in body temp is what makes the diff between you feeling hot or cold, so actually the question is dead on, as lowering body temp (a tiny bit) will also make you feel cool. StuRat (talk) 03:20, 5 August 2008 (UTC)[reply]
See [1] for previous responses to the same question.
The rate of cooling determines apparent temperature, not the amount of heat removed. Although drinking water removes more heat than pouring it on the skin, the latter feels much more effective in the short term because a thin film of water heats and evaporates very quickly. --Bowlhover (talk) 07:39, 5 August 2008 (UTC)[reply]
But, as already stated, pouring water on yourself will only help by evaporative cooling if it's hot and dry out, and yet you aren't already sweating, which is an unusual combo. One reason for this might be dehydration, in which case drinking the water becomes more important. Also, if you lower your temp too quickly then the body will react by generating heat, via shivering and other means. So, lowering body temp steadily may be the best way to go, especially if the effect lasts longer. StuRat (talk) 17:15, 6 August 2008 (UTC)[reply]
Evaporative cooling helps whenever the relative humidity is less than 100%. Poured water cools the body via conduction while itself heating up, and after reaching body temperature, the water would cool the body as sweat would do. As for reflexes, shivering and other reactions to cooling are trigged when the core body temperature drops; a drop in skin temperature would not decrease the core temperature because the body is already generating too much heat. --Bowlhover (talk) 11:08, 7 August 2008 (UTC)[reply]
Yes, evaporative cooling helps when less than 100% relative humdity, but adding water won't help in that way if the person was already completely coated with sweat. And cooling from conduction is never as effective as if the person drinks the water, as a portion of the coolness of the water goes into cooling the air, and much of the water drops off the person before it can cool anything down. I don't agree on shivering, however, as people shiver instantly when exposed to something cold, like diving into cold water, long before the core body temp could have had time to drop significantly. StuRat (talk) 14:58, 7 August 2008 (UTC)[reply]

Existance of God[edit]

Is there some connection between finding life on Mars and the existence of God? Does finding life on Mars refute the existence of God or prove it? —Preceding unsigned comment added by 71.100.5.89 (talk) 12:23, 4 August 2008 (UTC)[reply]

No. Algebraist 12:24, 4 August 2008 (UTC)[reply]
At worst, it would imply that the Judeo-Christian account of the creation of life was leaving out a few things. (But it's not like that was exactly on the line from a scientific point of view.) --98.217.8.46 (talk) 12:26, 4 August 2008 (UTC)[reply]
If you carefully read the Bible's account of the creation of extraterrestrial bodies, you'll notice that only "greater light", "lesser light", and "stars" are mentioned as being created: "sun" and "moon" are not used, and nothing is said specifically about the planets at all. I doubt anyone believes that this account includes everything. Nyttend (talk) 13:44, 4 August 2008 (UTC)[reply]
The Bible was written from the understanding of the universe they had at that time. Similar events occur in science where completely new ways of thinking or seeing things develop. These are called paradigm shifts. --Russoc4 (talk) 13:47, 4 August 2008 (UTC)[reply]
Neither. If/when we start discovering life on other planets, for the religious it will simply be a matter of saying "well, God created the entire universe and *all* the life in it". As far as I am aware, the Christian religion (I'll assume for the sake of discussion that you are referring to this) has never claimed that their God is only the god of *our* planet. Nothing will really change at all. It will still be impossible to prove or disprove the existence of an 'entity' which presumably would exist (if he/she/it exists) somewhere beyond/outside the universe as we know it. --Kurt Shaped Box (talk) 14:24, 4 August 2008 (UTC)[reply]
Quite the opposite, really. Christians believe God created everything, not just the stuff on Earth. —CycloneNimrodTalk? 16:01, 4 August 2008 (UTC)[reply]
That was the point I was making in the first place. Or at least I was trying to... ;) --Kurt Shaped Box (talk) 22:59, 4 August 2008 (UTC)[reply]
The existence of God might be strengthened by similar life on different planets (i.e. similar species). Life in itself doesn't really say much, it could have formed the same way we think it did here. —CycloneNimrodTalk? 15:26, 4 August 2008 (UTC)[reply]
The discovery of intelligent extraterrestrial life with very similar religious beliefs to those of humans would certainly provoke heated debate. I don't know how much the discovery of (for example) a fish-shaped creature in an extraterrestrial ocean really would prove, other than that that a fish shape is a good shape for swimming in the ocean and that convergent evolution works. --Kurt Shaped Box (talk) 15:56, 4 August 2008 (UTC)[reply]

Whew... I have to say that is certainly a relief. I would call this consensus that the discovery of life on Mars would not have any impact on the question of whether God exists or not so there is no reason for believers to panic or for non-believers to put them down. —Preceding unsigned comment added by 71.100.7.91 (talk) 19:17, 4 August 2008 (UTC)[reply]

There are always the same reasons as before ;) —CycloneNimrodT@lk? 19:48, 4 August 2008 (UTC)[reply]
God of the gaps is an article generally related to this topic. It discusses the interaction of theism and science over the long-term. ---J.S (T/C/WRE) 23:30, 4 August 2008 (UTC)[reply]
The question of whether or not there is a God in the abstract sense cannot and will not be verified or disproved by science. That is fairly clear. The question of specific religious ideas though can vary—if religions make concrete claims about the natural order of the world, then they can (and do) easily overlap into the domain of science, and conflict can (and does) occur. But if things are discussed in abstract, non-testable, entirely "unnatural" (that is, not pertaining to operations of the natural world) ideas, then there is not even the possibility of conflict. --98.217.8.46 (talk) 03:00, 5 August 2008 (UTC)[reply]
The abstract and the concrete, however, are quickly proving themselves to be reliant upon each other and inseparable for that reason and being only different aspects of the same thing. For instance, I can have an abstract concept which a computer program can emulate such as logical equation reduction. Although an abstract concept the results can be implemented directly by computer hardware, just as people can think and believe a certain abstract thing and control their actions accordingly to comply with or to implement the abstraction they are thinking and believe. Conversely, they can adapt their abstract thoughts and beliefs to accommodate feedback they get from the concrete world just a computer sensors can collect information from the concrete world which software can use to modify or to update an abstraction. In other words if you belief that God did not create life on any other planet and life was discovered on another planet then you would have to update your abstract belief with the concrete information or suffer abstract invalidation. —Preceding unsigned comment added by 71.100.7.91 (talk) 07:40, 5 August 2008 (UTC)[reply]

Calculation of pipe dia. Formula Needed?[edit]

Flow of water at standard pressure of 3bar, Calculation of pipe dia. to be used? When water needs to be pumped for a distance of 3 kilometers, and a constant pressure of 3 bar is supplied what is the optimal diameter(mm) of pipe that can be used? What formula can be used to calculate the diameter of pipe to be used when the pressure ,substance (water) and distance are known. Can anyone provide me with information? Arjaa (talk) 12:53, 4 August 2008 (UTC)[reply]

There's some missing info:
1) What is the desired flow rate through the pipe ?
2) Is the pipe straight and level or some other configuration ?
3) Is the pressure at the far end important ? StuRat (talk) 13:28, 4 August 2008 (UTC)[reply]


If the pipe is horizontal, and without leaks or taps, there should be no loss of pressure, and the diameter will only affect the flow rate. You can calculate the flow rate from the exit-velocity; density times velocity times flow area will give you mass flux. Mass flow rate has some equations that may put you in the right direction. If you use a pump or something else to induce a pressure differential across the pipe, you can use that to calculate the fluid velocity. Nimur (talk) 16:52, 4 August 2008 (UTC)[reply]
Is your first statement actually true in a real pipe? I would have thought that in an actual pipe there would be a pressure drop with distance due to the drag of the fluid on the walls of the pipe. The pressure drop should decrease with increasing pipe diameter (since there is less wall per unit volume), and also should increase with increasing flow velocity.--Srleffler (talk) 02:58, 5 August 2008 (UTC)[reply]
I'd say it's a suitable approximation; it would really depend on the fluid velocity (a stationary fluid would exhibit no drag!) The discharge coefficient is the quantity which would measure such a loss, but that's not typically applied to pipes, it's usually applied to the entry or exit orifice. In a real pipe, that quantity should be very close to 1.0. Nimur (talk) 04:48, 5 August 2008 (UTC)[reply]

Medical appliances (?), early/mid-20th C.[edit]

This picture shows prostheses and similar devices taken from victims of the Auschwitz II (Birkenau) camp and exhibited in a museum on the grounds of the Auschwitz camp. I haven't succeeded in identifying these other objects and need to know how to describe or otherwise refer to them. Would "medical appliances" be appropriate, or is there a more suitable term you'd suggest? -- Thanks, Deborahjay (talk) 13:50, 4 August 2008 (UTC)[reply]

My understanding, based on many years as registered nurse in the UK is that 'prosthetic' refers to some sort of replacement of a body part, sometimes to restore function sometimes for aesthetic reasons. Examples would be a glass eye or false hand for aesthetic reasons and wooden leg as a functional prosthesis. There are nowadays many more much more complex prostheses available. A 'medical appliance' I have always understood to mean a contraption, device or gizmo to assist some failing part of the body such as a spinal support for someone with scoliosis, a metal sprung calliper for drop-foot, a hearing aid for deafness or the good ol' truss for a hernia, as three examples. I hope that helps. Richard Avery (talk) 14:23, 4 August 2008 (UTC)[reply]
"Medical appliance" is quite vague, and includes such things as catheters and stoma appliances. The picture appears to be of orthopaedic-type equipment. It would perhaps be appropriate to refer to them as prostheses and orthoses, or prosthetic and orthotic equipment. Gwinva (talk) 22:14, 4 August 2008 (UTC)[reply]
I have two hesitations in accepting this solution en bloc: (a) some may indeed be hernia trusses (not clear in the photo, but used in this population) which are not orthopedic, and (b) the readership is mainstream (including non-native speakers and "orthoses/orthotic" are probably unfamiliar. Is there another comprehensive term rather than "medical appliance"?-- Deborahjay (talk) 06:39, 5 August 2008 (UTC)[reply]
If you're writing for a general readership, then "appliance" might be misunderstood; most people probably imagine an appliance is something you plug in. (WP doesn't even have an article for "medical appliance", although appliance mentions the concept briefly.) How about "prosthetics and other medical equipment" or "medical/health care aids"? Gwinva (talk) 22:52, 6 August 2008 (UTC)[reply]

Post thunderstorm ionization[edit]

Is it true that after thunderstorms people and animals feel better or get a 'high' from the change in ions in the air? Is this ionization? —Preceding unsigned comment added by 72.171.0.146 (talk) 14:42, 4 August 2008 (UTC)[reply]

No. In fact, some evidence[2] suggests the opposite, due to ozone production from the lightning. You may want to read ozone, though. Nimur (talk) 15:54, 4 August 2008 (UTC)[reply]

Tree of life[edit]

Does the DNA of currently-living organisms provide enough information to construct a full phylogenetic tree back to the most recent common ancestor of all such life? Are we likely to have such a tree in the next 50 years? Thanks. --Sean 14:51, 4 August 2008 (UTC)[reply]

No. We can make educated guesses and probably create a pretty good approximation, but we will never have a complete tree free from error. Several problems prevent this. First, many ancestral species no longer exist, which means one has to estimate the ancestral relationships between modern species. This can be done fairly well for recent speciation events, but as one looks further back in time, these ancestral relationship become far more speculative and there isn't enough information in currently existing DNA to be definitive. Keep in mind, more than 99% of all species that have ever lived are extinct. Secondly, with bacteria (and very infrequently some higher species) lateral gene transfer violates the assumptions generally used to construct such trees. As one gets closer to the most recent common ancestor, the more likely it becomes that non-descendent mixing will make it impossible confidently define descendent relationships. Dragons flight (talk) 15:42, 4 August 2008 (UTC)[reply]
But you might be interested in Richard Dawkins' book The Ancestor's Tale. It follows human evolution back and discusses the splits (or, in that direction, merges) encountered. He describes about 35 or so common ancestors, and also makes clear how much is hypothesis and how much is fairly solid. I found it in an airport bookshop on sale before I ever heard of Dawkins, and I was hooked by the beautiful writing and clear but not oversimplified exposition. --Stephan Schulz (talk) 00:23, 5 August 2008 (UTC)[reply]
Some clues to ancient life may be found in meteorites on the moon, or floating in the asteroid belt. Some would be derived from the earth, particularly in it's early stages. It is possible that vacuum dried bacteria may still be present. This is not a whole tree, just another node preserved from the past. This sort of thing may be found in the next 50 years. Graeme Bartlett (talk) 00:46, 5 August 2008 (UTC)[reply]

Blood plasma[edit]

If you take blood and just centrifuge it you get blood plasma. Plasma still has the clotting factors in it (as opposed to blood serum). My question is will blood plasma clot or does clotting only work when the cells and or platelets are in there? ike9898 (talk) 16:09, 4 August 2008 (UTC)[reply]

Yes, you need the platelets (and some factor proteins, read coagulation) which are in the heavy part of the centrifuged blood. --Ayacop (talk) 17:59, 4 August 2008 (UTC)[reply]

Minor correction; If you simply centrifuge whole blood you'll end up with a pellet of cells at the bottom and serum on top; If you centrifuge blood treated with anticoagulant you'll have a pellet of cells with plasma on top. When you have a blood sample taken you may notice that different colour tubes are used. The colours indicate the additive in the tube. If you put blood into a blue (Lithium citrate) tube the coagulation cascade is inhibited such that after centrifuging the coagulation proteins are still present in solution and the yellow fluid on top of the pellet is Plasma(or strictly speaking Plasma+Artificial anticoagulant). Various diagnostic tests/assays test different blood components, hence the need for different tubes. [3]Od6600 (talk) 16:56, 6 August 2008 (UTC)[reply]

Help diagnose my ear deficiency![edit]

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.
This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis or prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page. --~~~~

Earths core[edit]

This is a schematic diagram of earth's interior. Nimur (talk) 16:57, 4 August 2008 (UTC)[reply]

How do we know the core of the earth is molten? Has someone drilled right down. Is it possible there could be another planet at the center like in journey to the center of the earth? —Preceding unsigned comment added by 79.76.204.221 (talk) 16:47, 4 August 2008 (UTC)[reply]

This is what we use to determine the above layers. Nimur (talk) 16:57, 4 August 2008 (UTC)[reply]
We measure earthquakes and compare the data from many different places on the planet. Then we use that data to solve the wave equation to estimate what kind of material the wave traveled through. In some cases the waves appear to travel through solid rock, and other cases the wave appears to have traveled through liquid rock. Close data analysis lets us estimate the density, and thus the temperature and pressure, and other parameters. The deepest drilling humans have completed is probably deep petroleum exploration, which may go as far as 10 or 12 kilometers in some extreme cases; this does not even penetrate the crust. It is very unlikely there is another planet or anything like Journey to the Center of the Earth speculates about. Nimur (talk) 16:58, 4 August 2008 (UTC)[reply]
The Bertha Rogers well was the deepest petroleum exploratory well and reached 31,441 feet (9,583 m) encountering pressures approaching 25,000 psi (172,369 kPa). In 1974 the hole was abandoned after encountering molten sulpher that melted the drill pipe!
The Kola Superdeep Borehole made it to 12,262 metres (40,230 ft or 7.62 mi) deep in 1989. Saintrain (talk) 23:01, 5 August 2008 (UTC)[reply]
(edit conflict) We know because earthquakes travel differently through the different layers, and refract at the boundaries. As for drilling, see Mantle_(geology)#Exploration. Note that that's for drilling to the mantle, not the outer core (the liquid layer). Anything in the center of the earth wouldn't fulfill the definition of a planet. If you're asking about having a gas or vacuum layer (so the next layer down would look like a planet), then no. It wouldn't conduct earthquakes as well as it does. If having a liquid layer with a solid layer under it counts, then it's correct. The outer core is liquid, and the inner core is solid. If you could survive the temperature and pressure, the inner core-outer core discontinuity would seem similar to a liquid covered planet. — DanielLC 17:16, 4 August 2008 (UTC)[reply]
But the inner core would be covered by an ocean of liquid metal, and that would be opaque and probably red-hot. It's not quite the typical "liquid-covered planet" you might envision. Nimur (talk) 17:43, 4 August 2008 (UTC)[reply]
What inner core made of? If iron why isnt it molten? —Preceding unsigned comment added by OpticalSphincter (talkcontribs) 17:48, 4 August 2008 (UTC)[reply]

Who said the core was molten? The core is solid. In fact most of the layers of the Earth are solid. The pressure would force any matter near the core into a solid. Of course if you were to bring that hot matter to the surface, it becomes a viscous liquid. ScienceApe (talk) 19:46, 4 August 2008 (UTC)[reply]

The outer core is liquid. Dragons flight (talk) 19:49, 4 August 2008 (UTC)[reply]
Actually, OpticalSphincter asked why the core isn't molten. It's made of mainly iron and nickel, and is solid because the pressure on it prevents molecules from moving freely. The Sun`s core is even more interesting; it comprises mainly hydrogen and helium at 15 million degrees Celsius, yet the density is 150 g/cm^3. --Bowlhover (talk) 19:52, 4 August 2008 (UTC)[reply]
How do we know that?
Pretty much the same way we know about the Earth's core, but rather than studying earthquakes, you study sunquakes. See Helioseismology. --Tango (talk) 03:41, 5 August 2008 (UTC)[reply]
Fun (aside) fact—in his memoirs, Edward Teller says that working with George Gamov on questions relating to the core of the earth gave him the idea that you could use a solid sphere of plutonium inside an implosion design nuclear weapon, that under the pressures of a few hundred megabars even an apparently dense solid could get a LOT denser than it would be at a single atmosphere. --98.217.8.46 (talk) 02:55, 5 August 2008 (UTC)[reply]

Galactic Center of the Milky Way[edit]

Hypothetically, if there was no cosmic dust and such in the way, how bright would our galactic center would be to us earthlings? If anything, how would this extra light or whatever, affect us on earth? --Anilmanohar (talk) 18:57, 4 August 2008 (UTC)[reply]

Well there should be a super massive black hole at the galatic core of our galaxy, and it's surrounded by an immense accretion disk. This disk should be very hot and larger than any other object in our galaxy, so it should be very bright. ScienceApe (talk) 19:52, 4 August 2008 (UTC)[reply]
Actually by itself the black hole, official known as Sagittarius A*, is very dim (as galactic nuclei go). Only about 1000 times the luminosity of the sun, or so. For comparison, the neutron star at the heart of the Crab Nebula and some individual stars, such as Rigel, are at least 10 times more intrinsically luminous than this black hole. Since the black hole is 27000 light years away (give or take), it would not be visible to the naked eye (even if you could distinguish it from all the stars clustered around the galactic center). As a mature black hole Sgr A* has already consumed most everything in its immediate vicinity and so now it accretes matter only very slowly. Young black holes, which have lots of matter to eat, can power active galactic nuclei which are among the brightest objects in the universe, but our black hole thankfully became quiet billions of years ago. Also, the bulk of the accretion disk is only about the size of a solar system. The black hole itself is not huge on astrophysical scales. Dragons flight (talk) 20:40, 4 August 2008 (UTC)[reply]
No not necessarily. I guess you are talking about the black hole at the center of our galaxy. But the it depends on which supermassive black hole we are talking about because they vary in size by quite a lot. The one in our galaxy is pretty small at only about 3.7 million solar masses. But Q0906+6930 is much larger at 16 billion solar masses and has a event horizon volume that's 1000 times our solar system. And that's just the event horizon. The accretion disk is much much larger. ScienceApe (talk) 03:28, 7 August 2008 (UTC)[reply]
(Edit conflict) Okay, let's work through this problem - it's an interesting one that requires a bunch of assumptions and formulas. First, Galactic Center gives us relevant information on the area immediately around the supermassive black hole, and it tells us that within 1 parsec (3.something light years) you "thousands of stars... most of them are old red main sequence stars." Great, so we can ignore most of them, because such stars are very very dim compared to the "more than 100 OB (blue giant)... stars that have been identified so far". We'll use Rigel as a model for this type of star, which is at 40,000 apparent solar luminosity. One more thing: this central parsec is 7600 parsecs from us, which, using trig, gives us 27 seconds of arc, which is about 1/60th the diameter of the moon in the sky, so roughly the angular size of a planet.
So now to crunch numbers to find apparent magnitude of this galactic center from Earth. We can add luminosities since it's just a measure of power and get about 4*10^6 solar luminosities, which is an absolute magnitude of -11.7, which is an apparent magnitude at 7600 parsecs of 2.7, which would make it one of the brighter lights in the sky. With all that done, it would probably be more relevant to get a number for the whole galactic bulge, since the combined luminosity could probably outshine the moon. But I'm tired now. SamuelRiv (talk) 20:16, 4 August 2008 (UTC)[reply]
This webpage has luminosity data on bright galaxies. Taking the spiral galaxies M31, M77, M81, NGC 3521 and averaging the brightnesses of their inner 2 arcminutes, we find that galactic nuclei are around magnitude 8.7 per 4pi square arcminutes or magnitude 20.3 per square arcsecond. This corresponds to class 5 on the Bortle dark sky scale, described as "Milkyway washed out at zenith and invisible at horizon. Many light domes. Clouds are brighter than sky. M31 easily visible. Limiting magnitude about 5.6 to 5.9." The galactic nuclei would be conspicuous unless light pollution is heavy. --Bowlhover (talk) 10:28, 5 August 2008 (UTC)[reply]

No one ha explained how this will appear to us earthlings or the effect it would have on earth...... --Anilmanohar (talk) 14:06, 5 August 2008 (UTC)[reply]

Using SamuelRiv's calculations, we would see something about the same size as a planet (i.e. showing features when viewed through binoculars or a telescope) and just about bright enough to be seen in an urban night sky. Gandalf61 (talk) 14:24, 5 August 2008 (UTC)[reply]
SamuelRiv only calculated the size of the parsec around the supermassive black hole; his purpose was to give an indication of brightness. Anilmanohar: see this excellent photo of the Milky Way. Look at the bright region on the right and imagine the dust lane does not exist. The galaxy would continue to get brighter as one approaches the center of where the dust lane is now, and the region around Sagittarius A*, as SamuelRiv calculated, would be magnitude brighter than magnitude 2.7. On average, the bright region would be as bright as a suburban night sky, but it would be extremely luminous at the center and comparatively extremely dim at the edges. --Bowlhover (talk) 20:56, 5 August 2008 (UTC)[reply]
So, in summary, an unobscured galactic centre would not be at all spectacular when viewed from Earth - we are just too far out in the galactic 'burbs. And, to answer the other part of the original question, I don't see how it would have any effect on life on Earth. Gandalf61 (talk) 09:00, 6 August 2008 (UTC)[reply]
It wouldn't be particularly bright, but I still think it would be quite spectacular - it's pretty impressive with the dust cloud in the way, without it it would be even better! --Tango (talk) 09:12, 6 August 2008 (UTC)[reply]

Why can't bullets fly ?[edit]

(fly in the sense that its form causes a little lift to compensate the free fall) —Preceding unsigned comment added by 77.224.230.72 (talk) 19:47, 4 August 2008 (UTC)[reply]

It does. A little. It would need constant thrust like a rocket though. ScienceApe (talk) 19:53, 4 August 2008 (UTC)[reply]

A form that would create lift would also increase drag, decreasing the effective range of the weapon. You also have to consider stabilisation; how would a lift-creating shape work with the spinning motion that most modern firearms use to increase accuracy? FiggyBee (talk) 20:02, 4 August 2008 (UTC)[reply]
Back in the olden days of SciRefDesk, you could get a whole treatise about how angle of attack, not asymmetric air-foil shape, can be key to lift. So have its center of mass displaced relative to the center of lift: essentially weight down the rear, increasing the angle of attack. The bullet flies rotating on its axis (spin-stabilized) but the axis is pointed nose-up. And has enough surface area on the "forward-facing bottom" so that the air pushes the bullet up, but not so much that you lose all your forward motion. And still maintain stability in the forward (as well as axial) direction without starting to tumble. Pretty soon you're just firing a normal rifle nearly straight up:) DMacks (talk) 21:50, 4 August 2008 (UTC)[reply]
Now that sounds interesting. Can you point us to an article that says all that/ —Preceding unsigned comment added by MisterRSole (talkcontribs) 01:10, 5 August 2008 (UTC)[reply]
Indeed, I would like to see that treatise. It seems to me that after the bullet exits the barrel in a straight line, the extra weight at the rear would cause the bullet to (gyrate? pronate? precess?) go into a speed wobble. Franamax (talk) 01:54, 5 August 2008 (UTC)[reply]
I had the same reaction. Weighting the rear would precess the spinning bullet and it would wobble its way dramatically off course. I'm skeptical of any angle-of-attack-related lift for a spinning, axially-symmetric shape. Nimur (talk) 04:53, 5 August 2008 (UTC)[reply]
Bullets can and do wobble to some degree—a really wobbly bullet produces horrible forensic wounds if it hits someone. I believe the degree to wobbling depends on how the weight is distributed in the bullet itself. --98.217.8.46 (talk) 02:49, 5 August 2008 (UTC)[reply]
Note my conclusion: "Pretty soon you're just firing a normal rifle nearly straight up". I'm completely doing thought experiments, speaking from first-principles physics and not any specific training in balistics design. The angle-of-attack issue is real for airplane wings though. But back to the bullet...by the time you fix all the problems and have enough tilt and enough spin to maintain stability and a steep enough angle of attack, your gonna tumble or something else rather than "fly". I don't think you could get a stable design that gave enough push "up" enough to generate real lift without pushing "up and over backwards" and creating back-spin. Wobble really does occur (and pretty much makes hamburger of whatever it hits). An easy way to get it is to have the weight distributed off-center from the axis...rotation about the center of mass thus has the bullet itself spinning off-center. Dunno how badly pushing center of gravity forward or backward would lead to wobble (lift on one end of the axis of rotation would produce torque at right angles, sending the bullet wobbling or curving sideways?). Maybe could also off-axis weight to counteract it? DMacks (talk) 04:25, 5 August 2008 (UTC)[reply]
Yeah, if I remember my gyroscopes right (which I don't) changing the direction of the axis of rotation produces precession (hah, I got to use the word :). Since the direction of flight isn't changing, that would add up to a wobble. Now I don't think adding an off-axis weight could counteract that, because - where would you put the weight? The bullet is spinning, so the off-axis weight is spinning around too, seems to me you could only increase the wobble. And now to complicate things, bullets are pointy-shaped, so aren't they already heavier at the back end? Franamax (talk) 04:57, 5 August 2008 (UTC)[reply]
It's not "heavier at the back" (CoG relative to object's overall form) but (dim memory) where the CoG is compared to center of lift or something like that. There was something in an Estes catalog about finding CoG of a model rocket, then tying a string there (wherever it was front-to-back) and swinging it around. If it flew level (not pitching up or down) then you were all set and you wouldn't wind up launching at through the neighbor's window by accident. Well I'll be damned, I managed to learn some science from the toys I played with as a kid! (learned about torque from Capsela too:) DMacks (talk) 05:50, 5 August 2008 (UTC)[reply]
OK, I think I understand you, the rocket on a string analogy is good, I thought they were only good for poking your eye out. Now the rocket over its various shape and contours will have areas of greater or lesser positive and negative lift. Adding up these vectors will give you the centre of lift. But how do you do that with the spinning bullet? If a patch of the surface has positive lift, it will have negative lift in one-half a revolution. Spinning the bullet will put one centroid of lift onto the rotation axis, now how do we move the fore-aft centroid away from the CG? Add aerodynamic drag at the back end of the bullet with little extra furrows? Would that add net negative lift at the back? Franamax (talk) 06:38, 5 August 2008 (UTC)[reply]
If one could stabilize the spinning bullet with an angle of attack, the Coanda effect would cause it to curve way off course (see curveball).
A (perfect) flying bullet would have about the same range as a ballistic one but, when it got to that range, it would just drop out of the air with no impact speed, all its umph having been used up as drag.
(Side note:if you fire a bullet horizontally and drop one from the same height they'll both hit the ground at the same time! (Unless one of them is "flying" :-) How cool is that!?) Saintrain (talk) 22:23, 5 August 2008 (UTC)[reply]
  • 'Cause the distance the air has to travel on the top of the bullet is the same as the distance the air has to travel on the bottom of the bullet. Hence, no vacuum. For an airplane wing, it is curved on the top and straight on the bottom. Comparted to the wing bottom length, the longer air travel distance on the top results in vacuum, which sucks the airplane right off the ground. Of course, if you laid a bullet on the ground and put the end of a vacuum nozzle over it, the bullet would fly upward. And if you shoot a small airplane wing out of a gun, you might miss what ever you are aiming at. Suntag (talk) 17:28, 7 August 2008 (UTC)[reply]
    • "Back in the olden days of SciRefDesk, you could get a whole treatise about how angle of attack, not asymmetric air-foil shape, can be key to lift." DMacks (talk) 21:50, 4 August 2008 (UTC)[reply]
      • Ah, good times. Good times. Practical experiments were recommended and everything. 79.66.38.215 (talk) 23:28, 7 August 2008 (UTC)[reply]
      • Alas, the dumbing down of society reaches everywhere. Suntag (talk) 19:26, 8 August 2008 (UTC)[reply]
      • You mean we're not going to do practical experiments? What am I going to do with all these bullets I've filed down? Especially the ones I've already tied string to? :) Franamax (talk) 19:40, 8 August 2008 (UTC)[reply]

Plant identification: red berries[edit]

I saw these berries in High Park Toronto yesterday while I was out for a walk. I was wondering what kind of berries they are and if they are edible. They looked like red currants, but I'm not sure if that's what they are. Please see the photo I've linked bellow.

http://img.photobucket.com/albums/v244/shigil/berries.jpg —Preceding unsigned comment added by 99.254.23.238 (talk) 21:31, 4 August 2008 (UTC)[reply]

I don't live in Ontario anymore, otherwise I'd be running out to my little backyard paradise to look at the Honeysuckle bushes I planted. That was my immediate thought when I brought up the photo you link. Unfortunately, I no longer have the record of the exact species and my Audubon guide is not helping out, so maybe someone else can chime in here.
Our red currant article states "3-10 berries on each raceme" and I can't quite get to three when I count from your pic, so I'd say red currant is unlikely.
And of course, unless you're sure, assume they're not edible!
And looking a little more, a lot of people love High Park [4] [5], so I'm sure you can find some locals to ask if we can't pin it down. Franamax (talk) 01:35, 5 August 2008 (UTC)[reply]
They give me the impression of being from the dogwood family. This family is quite large and more expert knowledge will be needed to make an accurate identification. I certainly wouldn't put them on my pizza. Richard Avery (talk) 07:32, 5 August 2008 (UTC)[reply]
Lonicera morrowii I think. See http://www.ppws.vt.edu/scott/weed_id/lonmo.htm and http://www.illinoiswildflowers.info/weeds/plants/morrow_hs.htm William Avery (talk) 19:02, 6 August 2008 (UTC)[reply]
Aw crap - looks like I'm right about honeysuckle, but I always (tried) to take care to only plant native species in my yard. Maybe I wasn't clear with Dan at the garden centre that year. Any chance there's an equivalent native Ontario species? (I agree with your diagnosis, just trying to salvage my principles. You may have nailed the OP's plant right, it's my own I'm worried about now :) Franamax (talk) 09:41, 7 August 2008 (UTC)[reply]

Why did Newton think light speeded up in glass?[edit]

The Principle of least action was already a well known principle that explained Snell's law when Newton was investigating optics, so why did he come to the conclusion that light went faster when it entered glass rather than slower? —Preceding unsigned comment added by Dmcq (talkcontribs) 21:42, 4 August 2008 (UTC)[reply]

Did he? I hadn't heard of that, but it's a reasonable layman's interpretation to infer from observation that sound waves travel faster in denser materials, so perhaps electromagnetic waves would do the same. It's wrong, but understandable. ~Amatulić (talk) 22:29, 4 August 2008 (UTC)[reply]
Yes, it's how he explained refraction (thus accurate measurements of the speed of light in different materials were the first good evidence against Newton's ideas). I don't know how he came to this position, but it's probably in his Opticks. Algebraist 23:20, 4 August 2008 (UTC)[reply]
It is explained in [6]. Newton imagined that as light moved through air, the net attractive force on it from all air particles equaled zero. But at the interface between air and water, there was a net force pulling it towards the water, speeding it up. As its velocity along the surface remained unchanged, light would bend. this was Newton's explanation of snell's law and he was quite happy about it. :-) The different refractions of different colours were explained by the different colours being of different masses, so that the same force gave different changes in velocity. It wasn't until the 1850's that Léon Foucault got to compare the speed of light in air and water and discovered that light slows down in water.
As an apropos, [7] tries to speculate on what Newton would have made of simple experiments where the refraction of colours are opposite to the usual. EverGreg (talk) 07:12, 5 August 2008 (UTC)[reply]

Thanks very much. Very interesting. I must say I find his argument a bit unconvincing though and can see why he would be worried about Hooke! By his argument light of a single colour even if all the corpuscles were going a the same speed would have a different speed after entering glass depending on the angle they entered it at. But all the different speeds would have their vertical speeds leaving the glass changed by exactly the same ratio, that's a very strange idea for a particle theory - you'd expect the slower ones to change more if anything. I can see why he'd want a corpuscular theory with different shaped particles to try and explain polarization though. Dmcq (talk) 07:56, 5 August 2008 (UTC)[reply]

Sorry I misunderstood. He said the vertical attraction would change the velocity by the ratio rather than the vertical component. That makes a lot more sense. I guess he might have modeled it as two flat surfaces separated by a slope and the light particle going faster after it goes down the slope, the speed change wouldn't depend on the angle the slope was approached at. If the particles had differing speeds though they would bend differently so I guess he thought all light of a particular colour was going at the same speed - the obvious way then would be to have the different colours going at different speeds rather than have different masses. I must check if that is what he thought. Also I guess he liked having different particles for the different colours because Fourier was far in the future - who would have though different wavelengths could practically ignore each other like particles. Dmcq (talk) 18:27, 5 August 2008 (UTC)[reply]

Nope it looks like Newton believed all light moved at the same speed in space and he actually had quit a good estimate. I guess his reasonng was different speeds for different colours would give funny effects when looking at the moons of Jupiter. So he thought different colours were different masses. All a bit strange but it loks like he wasn't altogether sure of his theory himself because of Newton's rings Dmcq (talk) 19:13, 5 August 2008 (UTC)[reply]

Gas[edit]

Let's say I had a closed vessel containing only two liquid chemicals (no air), and this vessel had a known weight. If the two chemicals started a reaction which produced a gas, and I were to weigh the vessel afterwards, would the recorded weight constant (i.e. can a balance record the mass of a gas)? A second and unrelated question, how does water behave in a vacuum? —Preceding unsigned comment added by 76.68.246.7 (talk) 22:12, 4 August 2008 (UTC)[reply]

Assuming the vessel itself has mass, you can certainly weigh it "full of" something (liquid, gas, whatever), and then subtract the weight of the vessel itself to see what the effect of its contents are. If your reaction produces a gas, remember gas is still matter, and that matter has to come from somewhere--the liquid! See Conservation of matter. DMacks (talk) 22:15, 4 August 2008 (UTC)[reply]
As to how water behaves in a vacuum, the phase diagram for water shows that at zero pressure and some positive temperature, water will be in a gaseous state. The phase diagram for water also shows an anomaly in that for a certain range of temperatures, increasing the pressure will cause a transition from gas to solid, and then to liquid. Most other substances don't behave that way. ~Amatulić (talk) 22:27, 4 August 2008 (UTC)[reply]
The mass of the liquid, solid and gaseous contents of a sealed container would not change measurable as it went through chemical reactions or phase changes. In theory, it should weigh a tiny bit less (too small a change to detect on existing scales) if it emitted energy by heat or by radiation, and it should weigh more if it absorbed energy through the container, due to E=MC2. Edison (talk) 23:57, 4 August 2008 (UTC)[reply]
You'd need a very accurate scale to measure that! Even nuclear fissions yield mass discrepancies of tiny fractions of an atomic mass unit (relatively, on the order of hundredths of a percent). A chemical reaction has a far smaller energy change; as Edison said, no conventional mass measuring equipment could detect it (I am imagining some sort of emission spectrum fine structure analysis might reveal some indirect measurement which could reveal mass changes on this order of magnitude, but I've never seen anything like that in any lab I've been around). Nimur (talk) 05:01, 5 August 2008 (UTC)[reply]
I think you misunderstood my question. I know the total mass in the vessel will remain the same. I was wondering if a balance can detect the weight of a gas (afterall, it's not really touching the balance).
Yes, a balance can weigh a gas. Ultimately, the balance is reacting to pressure exerted, rather than "touch". Similarly, a truck full of birds weighs the same if all the birds begin flying within the truck. — Lomn 18:17, 5 August 2008 (UTC)[reply]
Whether the gas touches the balance or not, gravity is still acting on it, and it in turn presses down on the other substances in the vessel. If you stand on a scale, carrying a backpack would increase the weight reading even though the backpack is not in contact with the scale.
Yes, but if you let go of the backback (i.e. it's in freefall), then the scale won't take it into account. From my knowledge of gasses, molecules are touching each other and thus spend a considerable portion of their time in freefall, and so the balance shouldn't be able to detect their weight. So wouldn't the total weight measured be slightly less, because some of the atoms of gas are in freefall? —Preceding unsigned comment added by 76.68.246.7 (talk) 13:04, 7 August 2008 (UTC)[reply]
If you're trying to actually weigh a gas, however, you must take buoyancy into account. It has a significant effect on apparent weight when dealing such low-density substances. --Bowlhover (talk) 20:29, 5 August 2008 (UTC)[reply]
Buoyancy shouldn't have a meaningful impact on this example, though, with the gas in question enclosed in a vessel. — Lomn 23:20, 5 August 2008 (UTC)[reply]

Viscious liquid[edit]

i read lot of science pages. I notice lot of mention of vicious liquid. Why is it vicious. will it attack me/ —Preceding unsigned comment added by MisterRSole (talkcontribs) 23:07, 4 August 2008 (UTC)[reply]

Yes. It is a a very dangerous liquid. Simply vicious. It will tear your heart out if you give it half a chance. On the other hand a viscous liquid is a very thick, slow flowing liquid, such as maple syrup. Though they may lead to sticky situations, most viscous liquids are not plotting to kill you. Dragons flight (talk) 23:16, 4 August 2008 (UTC)[reply]

Viscosity is a measure of the resistance of a fluid which is being deformed by either shear stress or extensional stress. In general terms it is the resistance of a liquid to flow, or its "thickness". - Viscosity

(vicious edit conflictx2)Does that help? ---J.S (T/C/WRE) 23:17, 4 August 2008 (UTC)[reply]
Of course, viscosity is definitely at its finest with Non-Newtonian fluids. -- Captain Disdain (talk) 01:38, 5 August 2008 (UTC)[reply]
...Funny you should say that, because one of the definitions of a non-Newtonian fluid is one that doesn't have a single, well-defined viscosity. —Keenan Pepper 02:15, 5 August 2008 (UTC)[reply]
Yeah, and that's what makes them cool. -- Captain Disdain (talk) 10:02, 5 August 2008 (UTC)[reply]
Like cold custard? —Preceding unsigned comment added by 79.76.159.153 (talk) 20:47, 5 August 2008 (UTC)[reply]
According to this[8] the danger is the sharks. Julia Rossi (talk) 02:16, 6 August 2008 (UTC)[reply]

Expansion on shaking container of hot water[edit]

If I take a plastic food storage container and partially fill it with cold water, snap the lid on and shake it, nothing happens. If I repeat the procedure with hot water, very shortly after I begin shaking, the lid will vigourously pop off due to pressure buildup inside the container. If I do the experiment with a mason jar, there's a "Pfff!" of escaping air if I loosen the lid after shaking, but again, only with hot water. Why? The addition of dishwashing liquid seems, if anything, to amplify the effect, though I'm guessing this is due to the volume of suds produced. —Scheinwerfermann (talk) 23:20, 4 August 2008 (UTC)[reply]

Hot water heats the remaining air in the container. Warm air takes up more space than cold air. Shaking the container mixes the water with the air, warming it more quickly. Pressure builds up until the pressure difference between the contained warm air inside and the free, cool air outside is enough to pop the lid. 79.66.32.107 (talk) 23:30, 4 August 2008 (UTC)[reply]
Also, if the air is quite warm and the water is cold, you might notice the container collapse a bit. ---J.S (T/C/WRE) 23:45, 4 August 2008 (UTC)[reply]
"As the temperature is raised, gases usually become less soluble in water" <-- Maybe the heated water is releasing dissolved air. --JWSchmidt (talk) 03:55, 5 August 2008 (UTC)[reply]
Today's safety tip: Be very careful when blending hot liquids, basically for the reasons 79 gave above. -- Coneslayer (talk) 12:10, 5 August 2008 (UTC)[reply]
@JWSchmidt: Not likely, for two reasons. One; the water isn't being heated in the container and two; the dissolved gasses take up very little volume. ---J.S (T/C/WRE) 17:27, 5 August 2008 (UTC)[reply]

Here's a nice simple experiment to demonstrate the expansion of air with heat. All you need is a sink with hot water and an empty lightweight plastic bottle (like a big soda pop bottle). If your hot tap water is not very hot, you might start by heating some with a stove, microwave oven, or electric kettle instead of using it straight from the tap, in which case a funnel might be handy for pouring safely. Okay, first fill the bottle with hot tap water to warm the bottle, then dump out the water. Next put in just a little fresh hot water from your hottest supply in the bottom. Cap the bottle and shake it vigorously so the water heats the air, then uncap it and drain the water. Repeat with a little fresh hot water again, to make sure the air is really hot. After draining the water this time, quickly recap the bottle and close tightly. The heated air will now cool to room temperature and you will see how much it shrinks. --Anonymous, 18:36, August 5, 2008.