Wikipedia:Reference desk/Archives/Science/2006 December 23

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How long does it take to build a motorway from scratch? Forget about greenbelt land and planning rejections and pretend that it's 100 miles long. Living in the 21st Century, I seem to take motorways for granted, there seems to be a lot of roads built in very little time and I can't comprehend how it's happened so quickly. How many people would you need to employ? How much tarmac would you need. It just doesn't make sense! ImbalancedZero 23:17, 22 December 2006 (UTC)[reply]

Without trying to come up with exact numbers, I'll just observe that the time required depends critically on how many people (and machines) are used. At one extreme you could have one crew starting at one end of the road, building one carriageway, working toward the other end, stopping work on the main road whenever they had to build a bridge or a slip road, and finally working back building the other carriageway. At the other extreme, if there are 25 intersecting roads, you could send out 25 bridge-building teams and 100 road-building teams, building both carriageways in both directions from each cross road until they met. In practice something in between would apply, of course. But there would be a real difference between a road needed as a top-priority military project and one being built because there was a bit of money left in this year's budget.

As to "how much tarmac", I think modern freeways are normally paved mainly with concrete, and an optional layer of asphalt (tarmac) on top. The total thickness of the two materials, supposing that to be what you really wanted, is probably something like 18 inches or say 50 cm. Four lanes (say 18 feet wide) and four paved shoulders (say 10 feet) makes a total width of 112 feet (say 35 m). 100 miles is 160 km. So that's 0.5 x 35 x 160,000 = 2,800,000 cubic meters of asphalt and concrete. If it's say 2.5 times as dense as water, that'd be 7 million tons. Plus a bit more for slip roads (ramps) and a fair bit more bridges. Understand that this is just a rough estimate -- I don't know the actual width or thicknes. It shouldn't be too far off, though. --Anonymous, December 23, 00:04 (UTC)

In the United States, we never build roadways out of concrete anymore. Instead, we put about two plies of tarmac down over 12" or so of crushed stone. It's terrible construction practice, and in the North, with winter freeze-thaw cycles, the roadway is guaranteed to start falling apart within five years or so, but it sure does keep highway construction gangs employed.

Atlant 20:23, 23 December 2006 (UTC)[reply]

Anonymous is very close, probably close enough for an acceptable answer. If I was at work I'd have access to precise figures, but, sadly, I am not at work right now. In any case, a rule of thumb from the local department of highways here states that a mile of highway (2 lanes wide) can be paved per day. Actually estimating daily production rates and total construction time is an art form, and thousands of variables are involved. You can see some of what goes into it in this document. The local terrain, of course, has a big effect if there is a lot of grading required. Generally, though, the construction company has a strong incentive to complete the job on time, and usually will get a bonus for completing the job early. 192.168.1.1 10:52am, Dec 23 '06

I see lots of roads made out of thick concrete slabs with rebar in them. It is on a gravel base with asphalt on top in some cases. Edison 19:19, 25 December 2006 (UTC)[reply]

Say I was constructing an electrical device to be battery powered and I needed both 12v and 9v supplies. Would it be acceptable to create a 12v battery from eight 1.5v cells and by inserting an additional connection in between the sixth and seventh cells, get a 9v and 12v connection from the same battery? --Username132 (talk) 23:45, 22 December 2006 (UTC)[reply]

This will work, but the cells supplying 9 V will be discharged sooner. Consider: if the 12 volt load is 0.3 amps, and the 9 V load is .1 amps, then the 6 cells making up the 9 V section will have a load of.4 amps, which would discharge them long before the additional 2 cells supplying only the .3 amp load go flat. Because of this, the 9 volt load would see its supply voltage drop a little faster than the 12 Volt load. If you had a charger charging the entire battery at 12 volts, the cells would be unevenly charged and the whole arrangement would give unsatisfactory service. If the cells were promary cells or if they were removed from the device and all charged individually, you would have better operation. If the tapped 9 V circuit drew only a much smaller current tham the 12 volt load, or if it were intermittent, you would also get better service. You could always change the cells around periodically to even out the load. Edison 00:29, 23 December 2006 (UTC)[reply]

But using a suitable resistor would allow me to get about 9v from 12c supply, supposing that the reistance of the 9v device was constant? --Username132 (talk) 01:33, 23 December 2006 (UTC)[reply]

If the 9 volt load is reasonably constant, a voltage divider circuit can produce the 9 volts, at the expense of a constant flow of wasted current through the two resistors comprising the voltage divider. If the 9 volt load is a constant resistance a single dropping resistor might work, at the cost of the energy dissipated in the dropping resistor. A three terminal regulator circuit could be used to produce a constant 9 volts over wide variations of load, once again at some small constant flow of wasted current, but could certainly be a sound choice. Integrated circuits are also available which could step up a lower battery voltage to a higher voltage for some part of the circuit. Edison 08:10, 23 December 2006 (UTC)[reply]

My math's teacher had a "boil-off" with some friends in which they used their kettles to boil a certain amount of water, hoping that their kettle would be the fastest. He was asking us (his A level students) for advice. It's all over now, but I have wondered many times since, how could one optimise one's chances of winning a boil off? How difficult would it be to modify a kettle to include two heating elements, powered by seperate plug sockets? --Username132 (talk) 23:55, 22 December 2006 (UTC)[reply]

Electricity, water, and red hot heating elements are a recipe for accident or injury, so modification of the electric kettle is not recommended. Edison 00:31, 23 December 2006 (UTC)[reply]

Given the constraint that you have no post consumer modification of the electric kettle, the question is which kettle should you bring to the competition. (i.e. which kettle on the store shelves should you buy. Since you are looking into converting electrical energy into thermal energy, you aren't really concerned with "efficiency", as due to the law of conservation of energy and the laws of thermodynamics, any energy which you "lose" in inefficiency will be converted to heat (which is what you want anyway). Thus, what you want is the kettle capable of drawing the most electrical energy, or the one with the highest wattage. -- 01:27, 23 December 2006 (UTC)

And one with good insulation, to make sure that the heat generated by the element is retained in the water. Maelin (Talk | Contribs) 09:25, 23 December 2006 (UTC)[reply]

Lowering the pressure can cause water to boil sooner, even at room temperature. Now, how you can lower the pressure of the water without making it obvious, that's the trick. StuRat 15:06, 23 December 2006 (UTC)[reply]

I want to make a device the flashes as brightly as possible, so it might be seen from 50+ meters away in daylight. Device must operate from 12v and be (as always) as cheap as possible. What kind of bulb might do such a thing? --Username132 (talk) 23:59, 22 December 2006 (UTC)[reply]

Does the flash unit on cameras count? --Bowlhover 01:30, 23 December 2006 (UTC)[reply]

It does! There are xenon flash blubs available on ebay - suppose I want to power such a thing from a 12v car battery - what circuitry am I going to need? It needs a "trigger" apparently, but these have three prongs - how does it wire up? --Username132 (talk) 04:47, 23 December 2006 (UTC)[reply]

Could you adapt a disposable camera? It has all the circuitry in place. - CHAIRBOY (☎) 05:55, 23 December 2006 (UTC)[reply]

I'm no expert in electronics, but I think all you need are a xenon flash bulb, a capacitor, and a battery. When charging the capacitor, connect the battery's terminals to the capacitor's terminals and disconnect the bulb from the capacitor. When making the flash, disconnect the battery from the capacitor and reconnect the bulb. --Bowlhover 17:46, 25 December 2006 (UTC)[reply]

I'm hoping for something higher powered. I failed to mention that I want the flash to be seen at 50+ meters in _all_ directions. --Username132 (talk) 05:59, 23 December 2006 (UTC)[reply]

Another possibility: visit the local car spares and accessories shop, and examine one of the rotating roof lights that plug into the accessory power plug (c******** lighter) of your car, being attached to the roof by a permanent magnet. Those that emergency workers use are 12V, easily visible in daylight, and should not be too difficult to emulate using scraps. They are, however not as bright as camera flash lights. --Seejyb 06:04, 23 December 2006 (UTC)[reply]

Please do not tinker with electronic flash circuitry, since it steps up the battery to hundreds of volts and stores the energy in capacitors. This can cause electricution, similar to the units hospitals use to defibrillate. People have died tinkering with electronic flashes. Perhaps find a pre-assembled strobe unit which just plugs into the 12 volts. Edison 08:05, 23 December 2006 (UTC)[reply]

But those beacons aren't bright enough for my application. One watt simply wont do. I probably wont electrocute myself because the circuit will be enclosed when I switch it on. --Username132 (talk) 09:01, 23 December 2006 (UTC)[reply]

I'm no expert but it seems to me that you can in theory make a one-off flash as bright as you like, no matter how limited your power source is, as long as you have a large enough capacitor and a quick and efficient means of discharging it. However, storing a large amount of energy in a capacitor and discharging it rapidly sounds like a risky thing to do, especially if you are inexperienced - read this hazards and safety section. So I would second the views of other responders - find a ready-made solution or find an expert to help, but do not attempt a DIY solution. This sounds like it could turn into a Darwin Awards situation ! Gandalf61 09:34, 23 December 2006 (UTC)[reply]

Perhaps some old-fashioned magnesium flash bulbs would be in order. They generate their light from combustion, so need very little energy to trigger the rapid oxidation of magnesium. Of course, they are one-time use, as well, so you could only use them for a limited number of flashes, depending on how many you buy. Eventually your car battery would go dead, but that would be after hundreds or thousands of flashes. StuRat 14:55, 23 December 2006 (UTC)[reply]

Can you not pick up a slave flash unit from your local photography shop. Theyre not too expensive.--Light current 18:11, 25 December 2006 (UTC)[reply]

1) Is there ANY correlation between dirty/sweaty hair and acne?

2) Is technology something that can be progressed gradually, or can we make a huge leap with what we have now?

3) I've read somewhere around here, I think in "Senescence" that one day we'll have the technology to stop aging. According to that, can there be any guesstimate as of how long it will take for us to achieve that? PitchBlack 00:23, 23 December 2006 (UTC)[reply]

1. In my experience, yes. They are both symptoms of an underlying conditon, though, so washing your hair will generally do little to improve complexion.
2. In my observance, technology does indeed improve in fits and starts. Right now, there seems to be a lot of rapid improvement across all areas of science.
3. I question the assumption that technology is the key to immortality. Why not just a good diet and outdoor exercise? ←Vranak→

1. No. See Acne#Causes of acne.

2. It's not necessarily a case of either-or. Both can occur. See innovation and invention.

3. See wishful thinking. BenC7 01:17, 23 December 2006 (UTC)[reply]

1) I would say both are symptoms of excessive fat consumption, and, in the case of acne, trans-fats in particular.

2) Both. World wars seem to particularly spur on technology leaps.

3) I think we can lengthen the lives of biological bodies, but not extend them forever. The weak link is the brain. We might eventually get to the point where we can replace every other organ when the old one fails, but replacing your brain means replacing "you". However, if the brain can be copied into a computer, then, with backups and redundancy, it would be possible for that "brain" to live forever. Of course, many would argue that the brain in the computer isn't really "you", but rather a soulless copy of you. I would expect these type of advancements over the course of centuries, not decades. StuRat 14:49, 23 December 2006 (UTC)[reply]

the title of the book is: Futures, Options, and Swaps, by R. Kolb, Blackwell publishing, 4th ed. 2003 —The preceding unsigned comment was added by 150.210.171.79 (talk) 00:10, 23 December 2006 (UTC).[reply]

Wouldn't that be an illegal copy? Then, we cannot help you. — Kieff 02:14, 23 December 2006 (UTC)[reply]

Not necessarily; some websites that sell books also sell them in electronic format. After a Google search, however, I couldn't find any PDF copy of "Futures, Options, and Swaps". --Bowlhover 04:02, 23 December 2006 (UTC)[reply]

In that case if you must have one the only option seems to be to contact the publishers adn ask if they can do that for you.87.102.4.34 11:47, 23 December 2006 (UTC)[reply]

{{spoiler}} In the latest James Bond movie, he suffers a cardiac arrest after drinking a spiked martini. Is there really such a poison that can create that effect in that little time? --The Dark Side 02:10, 23 December 2006 (UTC)[reply]

A google search for "fastest-acting poison" turns up this forum page which suggests that nicotine taken orally in sufficient quantities could cause unconciousness within a few seconds and death within a minute. --Robert Merkel 03:41, 23 December 2006 (UTC)[reply]

If I remember correctly, the poison in the move was identified as either digitonin or digitoxin. They're related compounds, both of which are used in small amounts as heart medications; overdoses can cause arrhythmia, tachycardia, fibrillation, and cardiac arrest. As far as that goes, the movie gets it mostly 'right'. On the other hand, as far as I know neither one acts as quickly as the movie suggests. In real life, oral doses take tens of minutes to hours to get into the bloodstream and take effect. They are occasionally administered intravenously if more rapid action is required, but obviously the movie martini was taken orally and not IV. (This is the same school of movie science that lets investigators on CSI run a sample through a gas chromatograph in thirty seconds.)

Something that might fit the bill is batrachotoxin: one of the very potent toxins secreted by certain species of poison dart frog. It is efficiently absorbed through the mucous membranes and generates cardiac symptoms similar to those seen with digitonin overdose. On the other hand, it may also cause paralysis (batrachotoxin is a neurotoxin); perhaps we could write that off as the explanation for James' loss of coordination.

Our article doesn't identify an antidote, though it does offer suggestions on what treatments might work if you're desperate enough. Perhaps MI6 has some very good medical researchers on staff. TenOfAllTrades(talk) 06:40, 23 December 2006 (UTC)[reply]

I believe high doses of caffeine can also cause a heart attack (much higher levels than you can get from drinking coffee). StuRat 14:41, 23 December 2006 (UTC)[reply]

When he drank the salt water, was that to slow the poison or to force himself to vomit up the poison? --The Dark Side 01:23, 25 December 2006 (UTC)[reply]

I'd say to vomit, as that is a commonly used emetic. Drinking lots of water is sometimes also used to dilute certain poisons, but there's no reason for it to be salt water, in that case. StuRat 12:29, 25 December 2006 (UTC)[reply]

Is it possible to estimate the length of time an electric device can be powered by a battery-based uninterruptible power supply (UPS)? I was thinking one could use a device such as the Kill-A-Watt[1] to measure how much electric power was needed for a device. But how would one use that information along with the specifications of a UPS to know how long the UPS could power the device? Thanks for any ideas! --71.171.1.236 02:27, 23 December 2006 (UTC)[reply]

Typical UPS specifications include: Watts, VA and time. The VA is volts times amps, which is greater than the watts because anything other than a resistive load draws more volt-amps than it does watts. With a resistive load they would be the same. I tested early personal computer UPS units over 20 years ago with a resistive load box. When new, it would supply the rated volt amps for the specified time. Examining the current from the batteries, it was clear that it was stressing the batteries toward the end of the time. As their voltage dropped, the invertor circuit drew more and more current from the batteries to maintain the rated output to the bitter end. This was a somewhat destructive test, since the batteries were being drained at about a 1 hour rate (the rated time) instead of a more normal 10 hour rate. If the load were, say, half of the rated load, the carry time would more than double. Thus there is no simple mathematical answer to the question posed, except to say that a well designed UPS, when new, should supply the stated watts (or volt*amps) for the stated time, and should supply a lower load for a correspondingly longer time, but to a longer time than simple reciprocity would suggest. On the other hand, if the battery is beyond its service life, or has been repeatedly discharged to shutoff, the carry time at full load will be less than when it was new. I do not recommend loading it beyond its rated load. Edison 07:55, 23 December 2006 (UTC)[reply]

The pharmaceutical production plant where I work is enclosed in a Faraday cage but I want to know why? There are no radio communications in there that I'm aware off - does it protect again interference that may damage equipment? --Username132 (talk) 05:57, 23 December 2006 (UTC)[reply]

Do you mean there is a metal or wire mesh structure surrounding the building, or than each room is separately in such a cage. or that all internal walls are covered in sheet metal or wire mesh? It would be more common to enclose certain labs in a Faraday Cage. If the entire building were externally shielded, cell phones radios and pagers would not work, but when someone tried to make a cel call or used a Walkie Talkie it could affect the delicate equipment being shielded. There are various industry standards for the level of electromagnetic interference a piece of equipment should withstand. Edison 08:01, 23 December 2006 (UTC)[reply]

WOW! Edison is answering a question on Faraday! ;-) -- WikiCheng | Talk 18:30, 23 December 2006 (UTC)[reply]

Is is possible that the whole metal cage is grounded to earth - a way of preventing sparks?87.102.4.34 10:44, 23 December 2006 (UTC)[reply]

Thermocouples are sensitive to EM radiation (among other devices). I suspect it's a way to guarantee that temperature sensitive operations aren't altered by spurious radiation. --Tbeatty 18:54, 23 December 2006 (UTC)[reply]

Respected Reader, I would like to ask you this question, "Why are astronauts quarantined after they come from space. --""""

There are various reason. Radiation, disease, etc. Read the article quarantine. Sign your posts with four tildes, rather than quotations by the way. :) --Proficient 08:46, 23 December 2006 (UTC)[reply]

Until we are not sure that no extraterrestrial viruses or bacteria exists against which we have no immunity, it seems logical to impose a quarantine.--V. Szabolcs 10:19, 23 December 2006 (UTC)[reply]

I'm not sure that "quarantine" is quite the right word, as that implies the public is being protected from infectious diseases they may have. I think it's more that they need time to recuperate, especially from the negative effects changes in gravity have on the body. StuRat 14:22, 23 December 2006 (UTC)[reply]

Historically, the early Apollo astronauts who went to the Moon were quarantined on their return because, at the time, scientists weren't absolutely certain that the Moon was free of disease-causing organisms. This precaution was dropped after it was determined that the lunar surface is sterile.

As far as I know, astronauts since are not quarantined after their return from space. However, NASA does keep a watchful eye on their returning astronauts. Though not quarantined, a returning astronaut might as well be—there will be a post-flight debriefing and any number of medical tests. As StuRat says, an astronaut who has spent an extended length of time in freefall will take some time to re-adapt to live in one gee.

Before launch, astronauts are quarantined to prevent them from being exposed to any nasty communicable diseases, and to allow any infections they do have to surface. In addition to being kept isolated, their diet is also carefully monitored. (You don't want to send someone to space with salmonella poisoning or the flu, right?) Access to medical care is very limited in space, and a sick crewmember can't work very hard (if at all). TenOfAllTrades(talk) 14:48, 23 December 2006 (UTC)[reply]

Please be aware that Wikipedia cannot give medical advice to astronauts so if you are seeking medical advice you must see a physician. -THB 18:15, 23 December 2006 (UTC)[reply]

Where can I find the exhaustive and extensive list of ranges of values of the criteria necessary for human life to evolve from the Big Bang to the present, i.e., size of the sun, the distance from the sun of the life supporting planet, formation of an atmosphere, evolution of amino acids and certain proteins and cycles such as the Krebs cycle, etc.? I am looking for things that are absolutely critical to the evolution (existence and then maintenance) of human life anywhere in the cosmos. Adaptron 10:01, 23 December 2006 (UTC)[reply]

For a general overview see rare Earth hypothesis, anthropic principle and fine-tuned universe - although I doubt that you will find the precise quantitative data you ask for, as many parts of the argument involve the extrapolation of probabilties from very few examples. For the other side of the argument see mediocrity principle, Texas sharpshooter fallacy and survivor bias. Gandalf61 11:30, 23 December 2006 (UTC)[reply]

I also think it's a mistake to assume that all other life must be identical to humans and thus require identical environments. Just on Earth, we see an amazing variety of life, including life that isn't dependent on the Sun for energy, but feeds off deep ocean sulfur vents. There's no reason to think that such organisms couldn't eventually, or on some other planet, evolve intelligence comparable or beyond our own. StuRat 14:13, 23 December 2006 (UTC)[reply]

Its quite possible that even more "exotic" life forms exist but my inspiration for asking is not so much to consider the other life forms that could have evolved but rather but rather because of the realization that of how many, many thing had to fall and stay within certain range values, time frames and sequences of events in order for human life to exist at all. I mean think about it. The universe is basically the inside of a vacuum chamber and yet here I am able to make these comments today. Adaptron 16:31, 23 December 2006 (UTC)[reply]

That's where "life as we know it" comes into play... --Proficient 19:34, 23 December 2006 (UTC)[reply]

What is the pressure inside an inflated party balloon? 1.1atm,2atm,1000atm????87.102.4.34 11:45, 23 December 2006 (UTC)[reply]

If the graph on page 4 of this Word document is accurate and for a typical balloon, it is about 110 to 112 kPa, or 1.09 to 1.11 atm. I hope you don't think human lungs could supply a pressure of 1000 atm: that is about the pressure when you dive to a depth of 10 km, well beyond the crush depth of even an Alfa class submarine. That is about the depth of the Challenger Deep, the deepest known point in the oceans.  --Lambiam^Talk 14:41, 23 December 2006 (UTC)[reply]

Excellent thanks.

My back of envolope tells me that the energy stored (and released when it pops), for a smallish party balloon is (1 litre volume) of the order of 10 Joules..87.102.4.34 15:51, 23 December 2006 (UTC)[reply]

That's the equivalent of 0.000000000002 kilotons of TNT (or 0.002grams of TNT [bang!]).87.102.4.34 15:56, 23 December 2006 (UTC)[reply]

Just to avoid any confusion I used the approximation change in Energy =(pressure x volume)_initial - (pressure x volume)_final to find the actual work done is more complex and approx. 1/10th of this.87.102.4.34 17:33, 23 December 2006 (UTC)[reply]

A water manometer could be a long piece of clear aquarium tubing filled with water. See how tall a column of water the balloon can support when blown up and connected to the tube of water. (Not: do it outside, since water may spray around.) Then note the pressure equivalent of that column of water. I do not remember the exact equivalent, but isn't it something like 32 feet of water per 1 atmospherre? So a tenth of an atmosphere would be around 3.2 feet? There should be a water reservoir between the balloon and the water column. Consider a clamp to hold the water tube while the blown up balloon is connected. Open the clamp with the water level close to the final one so the balloon pressure does not drop while it is pushing up water, but not so high that the descending water column pops the balloon. (again, this sounds like an outdoor experiment, use adult supervision, wear eye protection, etc etc.. Edison 20:12, 24 December 2006 (UTC)[reply]

Is it true that the resultant of a gravitational force is also gravitational,that of an electromagnetic force is electromagnetic and so on?

-Sruthi

There may be a problem with terminology here. The term "resultant" in mathematics and physics is usually applied to the net result of adding two or more forces (or other vector quantities). You could have the resultant of a single force, but it would just be a very indirect way of referring to the force itself, so it would be unusual to use the word "resultant" in this sense. But in this rather odd sense, then yes, the "resultant" of a gravitational force is a gravitational force. On the other hand, if you mean "result" rather than "resultant", then the "result" of applying one or more forces to a body is an acceleration - see Newton's second law. Accelerations don't come in different types - they are just accelerations. So it would be incorrect to say that the "result" of a gravitational force is a "gravitional acceleration" - it is just an acceleration (or, alternatively, a change in momentum). Gandalf61 13:27, 23 December 2006 (UTC)[reply]

You might have meant "for an object in equilibrium, (zero resultant forces), is it necessary that all forces be of the same type ?". In that case, the answer is no, with a boat being a simple example. Gravity pulls it down and water pressure/buoyant force pushes it up. StuRat 14:08, 23 December 2006 (UTC)[reply]

A force is a force. It doesn't matter if it is gravitational or magnetic. Adding to StuRat's answer above, another example is that you can keep a magnet 'floating' on top of another one if you make the like poles face each other. Here, the weight of the magnet (gravitational force) acting downwards and the repulsion between the like poles (magnetic force) acting upwards nullify each other to keep the upper magnet in equilibrium. Usually, it is sufficient to say that a force of x Newton acts on a body in a perticular direction. It does not matter whether the force is caused by gravity or magnetism or a spring or just plain pulling by hand. In effect, the answer to your question (is it true that ...) is No. It is not true -- WikiCheng | Talk 17:36, 23 December 2006 (UTC)[reply]

Having been subjected to a firecracker explosion within close proximity of my private hearing space, I was wondering what I could do to somehow attentuate any bad effects that may arise form this occurance. There is a bluntness in my right ear, closest to describing it would be a clogged ear sensation.

-:( -—Preceding unsigned comment added by 83.5.227.70 (talk • contribs)

Medical advice and diagnosis comment deleted. -- TenOfAllTrades(talk) 16:42, 23 December 2006 (UTC)[reply]

Suggested diagnosis comment deleted. -- TenOfAllTrades(talk) 17:13, 23 December 2006 (UTC)[reply]

You might have a perforated eardrum. If so, that requires medical attention. In any case, avoid any further loud noises. (Loomis 12:34, 27 December 2006 (UTC))[reply]

Ten Of All Trades, are you objecting to the first or second part of the sentence? Tragic romance 06:04, 24 December 2006 (UTC)[reply]

If you have a perforated eardrum, that requires medical attention. In any case, avoid any further loud noises. (Tragic romance 00:30, 24 December 2006 (UTC))[reply]

A firecracker exploding near the ear results in "acoustic trauma" much as being hit by a car results in a "motor vehicle accident". Try googling that term to find information on what has happened to you. There is also an article in Wikipedia on "noise induced hearing loss" although it focuses on chronic trauma.

The answer to your actual question is that, no, you can't attenuate the damage as it has already been caused, only time can heal it, just like in any other injury to your body.

Of course, there is no way to get diagnosis or treatment recommendations without being examined by a health care provider. Unfortunately, because of the holiday, if you want to see a physician you will have to go to an emergency room or, more appropriately, an urgent care center. That might not be a bad idea as it is possible that the force was sufficient to rupture the eardrum and if that has happened there are precautions you need to take to prevent infection. -THB 18:07, 23 December 2006 (UTC)[reply]

Says who? I just called my doctor and her office is seeing patients through the whole 'holiday'. Both celebratory practices and health care practices vary through the English-speaking world; it is important for us (question answerers) to try to remember this when answering questions. Anchoress 18:14, 23 December 2006 (UTC)[reply]

In that case, please call Anchoress's physician to schedule an appointment. -THB 18:16, 23 December 2006 (UTC)[reply]

Questioner is in Poland. -THB 18:22, 23 December 2006 (UTC)[reply]

Unfortunately, we can't diagnose your condition over the internet. You should seek the advice of a qualified medical professional, who can discuss the incident with you in detail, analyze your symptoms, and – if necessary – call you in for a physical exam or to prescribe tests. More briefly—call your doctor with any questions. TenOfAllTrades(talk) 16:42, 23 December 2006 (UTC)[reply]

Only a doctor can really evaluate damage that has been done to your ear. S/he will tell you what treatment (if any) is going to improve the condition. It is important to get this looked at as it may be a perforation.--Light current 18:10, 23 December 2006 (UTC)[reply]

Uh, it's kinda stupid to delete a comment, but link to the now-deleted comment, innit? In any case, the comment was not an attempt at a definite diagnosis with a recommended course of action; it was more of a suggestion that may better arm the person asking the original question when they do contact a medical provider. I don't see how deleting it served any good except to pretent to police the reference desk rules. 192.168.1.1 whatever.

Frankly, if someone has some information to give that may be of use, they should be able to give it. This nonsense is causing a disruption of Wikipedia. People's posts should not be getting deleted. This disclaimer at the top of the page is plenty enough without a big to-do every time someone asks a medically oriented question. I should have nipped this in the bud when I saw people's email addresses being removed when they almost certainly wouldn't come back to view the answers but rather expect them in their inbox. Gosh darn it, if I want my email address displayed, I'd gosh diddly darn display it; somewh(email edited to protecdt the user from spam) ail.com. I'm 21 years old, an adult and should be allowed to make my own decisions without molly coddling to protect me from spam. --Username132 (talk) 20:04, 23 December 2006 (UTC)[reply]

Doesn't seem like medical advice; it seems like common sense. If you have a perforated eardrum, it requires medical attention. Avoid any further loud noises -- common sense. In order to be "medical advice," the comment would have to say "If you avoid further loud noises, then you will be cured of/ will be treating/ will be at risk of, etc. But it doesn't. All it does is tell you to avoid further loud noises, and does not make any suggestion of what benefits or risks that entails. 71.97.74.56 00:23, 24 December 2006 (UTC)[reply]

What specification of capacitor is required to deliver 300 J at 150 V over 5 seconds and what to deliver 1 J at 12 V over 0.01 seconds? This is not for homework (I'm not a students and when I was (and hopefully will again be in a few months), it's biologically orentated!). I would like to know how you arrive at these specifications, so I can work things out again when I get more precise info. --Username132 (talk) 15:29, 23 December 2006 (UTC)[reply]

See our article on capacitance for the relationship between energy, capacitance and voltage. Note that the discharge time does not affect the capacitance, although it may affect the type of capacitor that is best for the job. Note also that the voltage across the capacitor does not stay constant as it is discharged, but decreases linearly with the charge left on the capacitor. Gandalf61 16:47, 23 December 2006 (UTC)[reply]

You're going to have to be a bit more specific about your needs. Bear in mind that the voltage and current from a capacitor (discharged across a constant resistance load) fall off with a first-order exponential decay. To get something close to a constant voltage, you either need to have a capacitor that has much higher capacitance and won't discharge fully, or employ some electronics.

The time that a capacitor takes to discharge (for moderate rates of discharge) is largely dependent on the load to which you attach it. (Lower resistance = higher current = faster discharge, and vice versa.) If you're planning on rapidly and repeatedly charging and discharging, be aware that you may need to provide cooling for the capacitor.

The total amount of energy stored in a capacitor is given by

${\displaystyle E_{\mathrm {stored} }={1 \over 2}CV^{2}.}$

For example, if you want to have a starting voltage of 150 V and total stored energy 300 J, then you need at least a 0.027 Farad (27 mF) capacitor. Note that such a capacitor will only start out delivering 150 V. The voltage will decay to zero as the capacitor discharges. (For reference, the leftmost capacitor in the back row of Image:Capacitors Various.jpg is rated at 5 mF and 100 V.)

If instead you had a 100 mF capacitor at 150 V (this would be a very substantial capacitor), you would only be down to about 130 V by the time you had drawn off 300 J. TenOfAllTrades(talk) 17:09, 23 December 2006 (UTC)[reply]

Right! The questioner doesnt say what waveform he requires at the load. In problems of this sort, it is essential to know this. Or is he just interested in energy transfer? for instance, there are ways to deliver square pulses from an energy staorage circuit, but Im not sure if that is what the OP wants 8-)--Light current 17:39, 23 December 2006 (UTC)[reply]

Thanks for all your interest. Maybe a capacitor is the wrong way to go about this. Here's the dealio - I'm considering a game called 'laser skirmish', like laser tag but outdoors, like airsoft but not illegal (yet!) in the Netherlands. What I want to do is construct a simulation of a small [nuclear?] bomb in which after switching the circuit on, a random time is set between say 20 and 60 minutes. After completion of the time, another circuit is acivated which counts down 30 seconds and plays a 30 second clip of an air-raid siren (cheaper to do with electronics than an actual electromedchanical siren) before "detonating". The detonation should consist of an intense flash of light, shortly after which, 100 (470 mW) IR LEDs arranged in a 2x50 circle around the device are simultaneously lit with the desired signal (58 kHz carrier signal with another 1.8 kHz signal embedded into it (I'm unacquainted with this topic)) that activates the sensors and "kills" anyone caught in the blast which may last a few seconds. The device then plays an all-clear siren and game-play continues. It should run from a 12v car/motorbike battery.

The strobe will need some kind of capacitor and trigger unit. The LEDs are quoted with a surge current of 3A - specs are here http://pdf1.alldatasheet.com/datasheet-pdf/view/45652/SIEMENS/SFH400-3.html

You can probably see a connection now, between some of the questions I've been asking recently. --Username132 (talk) 19:34, 23 December 2006 (UTC)[reply]

Regarding using 12V power, I would suggest instead buying a 12V -> 110V power inverter like this one for $65USD, then using off-the-shelf 110V products to piece it together with minimal need to build your own circuits. A couple time delay relays for another $50USD or so, and a few of these inexpensive mini strobe lights for $12USD and you have all of the parts needed for the timing and strobe light circuit. No need to solder anything! You may want to disassemble part of the housing around the strobe light head, so that the bulb has a wider field of view, and with minimal handyman skills that shouldn't be difficult to do without compromising the integrity and safety of the high-voltage circuits inside. Hmm... I used to build stuff like this all the time. Can I build it and sell it to you? 192.168.1.1 2:25pm, 12/23/06

Given your intended use had you considered connecting a capacitor and electrical relay in parallel to a smaller battery - and using the relay to switch the 'psuedo bomb' - as the capacitor charges the effective resistance of the capacitor is low - so most of the current goes into the capacitor - as it becomes fully charged the resitance rises and so more current will flow into the relay - eventually there should be sufficient current to trip your relay. It's a simple and relatively inaccurate timing method..87.102.4.126 11:22, 24 December 2006 (UTC)[reply]

If grasshoppers have white blood and lobsters have blue blood, does this mean they lack red blood cells (made red by the presence of iron in hemoglobin) ? If so, how do they transport oxygen ? StuRat 16:47, 23 December 2006 (UTC)[reply]

There is a pretty good list of alternatives to iron-based hemoglobin for the transport of oxygen at Hemoglobin#Other_biological_oxygen-binding_proteins. I believe lobsters have Hemocyanin which is copper based; not sure about grasshoppers. --TeaDrinker 16:56, 23 December 2006 (UTC)[reply]

Insects do not transport oxygen in their blood, but rather rely on a network of tracheal tubes to deliver air directly to their tissues. For more details, see Insect#Morphology and development. —Ilmari Karonen (talk) 18:28, 23 December 2006 (UTC)[reply]

Not quite; there's always some tissue the tracheae cannot reach. In any case, in insects (and lobsters, and squid) there is no closed blood-vessel system; the body cavity fluid is not blood in the ture sense but hemolymph which does what blood does and more. But your guess that these animals are lacking red blood cells is correct.

Some Antarctic fish also have colorless blood (or rather blood the color of uncooked egg-white). Their hemoglobin IIRC has evolved into some freeze-protect protein; they need so little oxygen and have so much available by comparison that they simply transport it dissolved in their blood.

(Also, the greenish skin color of Star Trek's Vulcans was explained due to their home planed having a lower partial pressure of oxygen than Earth, forcing them to evolve a copper-based protein instead of hemoglobin. This they use in "red" blood cells like humans, so the theory goes, instead of having it swim around freely in a hemolympgh as crustaceans do with their hemocyanin) Dysmorodrepanis 04:28, 28 December 2006 (UTC)[reply]

Can somebody, please, tell me a list of books typically needed for a medicine degree? Thank you very much.

This would depend greatly on which university you went to. It would also depend on what you wanted to specialize in. BenC7 01:55, 24 December 2006 (UTC)[reply]

It would depend especially on which country you are speaking of. In America, it's possible to give a general outline for all (specialties really only come into play in postgraduate training). In the first year, there are basic sciences: Medical Biochemistry; Human Anatomy; Clinical Ethics in Medicine; Medical Embryology; Medical Genetics. In the second year, the study of disease and treatment: Microbiology; Clinical Medicine; Pharmacology; Pathology. In the third year, clinical rotations in Psychiatry; Obstetrics and Gynecology; Internal Medicine; Family Medicine; Surgery; Pediatrics. In the fourth year, clinical rotations (including about four elective clerkships, and Acute Care Clerkship; Ambulatory Care Clerkship; a Subinternship in Internal Medicine, Surgery, or Pediatrics; Neurology Clerkship; Advanced Cardiac Life Support. And of course each course requires its own textbook. I can list some standards, and anyone should feel free to add to the list. But these really are useful only in the context of concurrent coursework; I don't know that you'd get much benefit from them otherwise. Many courses are taught from handouts, with the texts as supplementary reading and references. - Nunh-huh 03:08, 24 December 2006 (UTC)[reply]

• Anatomy:
• Biochemistry: Stryer, Biochemistry
• Embryology:
• Genetics:
• Pathology: Robbins and Cotran's Pathologic Basis of Disease
• Microbiology:
• Pharmacology: Goodman & Gilman's The Pharmacological Basis of Therapeutics
• Psychiatry:
• Obstetrics & Gynecology:
• Internal Medicine: Cecil Textbook of Medicine; Harrison's Principles of Internal Medicine
• Surgery:
• Family Medicine:
• Pediatrics: Nelson Textbook of Pediatrics
• Clinical handbooks: Washington Manual of Therapeutics; the Harriet Lane Handbook: A Manual for Pediatric House Officers; Bates' Guide to Physical Examination and History Taking

The basic books in the first two years of American medical school typically include one standard text on each of the following: biochemistry, embryology, anatomy, genetics/cell biology, physiology, histology, microbiology, neuroscience, pathology, pharmacology, behavioral science, physical diagnosis. The specific books are usually specified by the professors, and typical US prices are $50-150 each. Books for the clinical rotations are more variable, as students have more choices: many buy an abbreviated or synopsis text for areas of lesser interest but a full standard textbook for their areas of greatest interest out of the following: internal medicine, general surgery, psychiatry, pediatrics, obstetrics ($50-200 each). Most students buy brief clinical handbooks for use on the wards ($20-50 each). Most students buy additional texts on areas of specific interest such as cardiology, or infectious diseases, or fluid & electrolytes, and there are many different topics like this ($50-300 each). Finally, many students buy study guides of various types ($30-50 each). Lotsa books lotsa money. alteripse 02:51, 24 December 2006 (UTC)[reply]

I am searching for the address to contact Yasargil and his team ( supposedly the best in the world) because my friend, a 41 years old mother of three, desperatley needs a second opinion concerning breastancer with metastasis to the brain. Norwegian Healthcare does not have any more treatment to offer her. Is there anyone who knows the address of this fabulous team? If so I will be for ever grateful! Merry Chistmas.

Best regards

Vigdis Bjørnøy Totengata 10 0658 Oslo Norway

I shall say no more than quote the entry from the second google entry for Yasargil:"Ask your physician to discuss your case with Dr. Yasargil. For a consult, please call Dianne Yasargil, RN at (501) 686-5270, or send an e-mail to Niketa Webb at the address WebbNiketa at uams.edu." --87.74.20.6 18:01, 23 December 2006 (UTC)[reply]

What makes flowers blossom vigorously?Or otherwise, how can flowers be made to blossom vigorously?

What does "vigorously" mean here ? Quickly ? StuRat 21:57, 23 December 2006 (UTC)[reply]

I mean to say in great amount or in great number with large bunches.

Generally it depends on the specific type of flower, can you be more specific? Seraphimblade 10:16, 25 December 2006 (UTC)[reply]

My father was subjected, many years ago, to an explosion at close range that perforated his eardrum. His hearing as far as I can tell is ok for his age. So, is it possible to have a return to 'normal hearing', having had a drum perforated by a loud explosion.--Light current 18:32, 23 December 2006 (UTC)[reply]

Yes, a ruptured tympanum can heal without permanent hearing loss. Here's more info: [2] from the U.S. gubmint because Wikipedia has nothing about it as it's incomplete. -THB 18:38, 23 December 2006 (UTC)[reply]

Of course, you should check with your father's doctor, to be sure! We can't answer medical questions. :P 192.168.1.1 11:05am, 23 December, 2006, Merry X-mas, etc.

There is a huge difference betweeen health education and diagnosis & treatment. Health care professionals have a duty to educate. -THB 22:26, 23 December 2006 (UTC)[reply]

So are you saying we should be offering medical advice to the readers?--Light current 22:38, 23 December 2006 (UTC)[reply]

I think there's some misunderstanding here. We're not supposed to diagnose people's medical problems ("this is what's wrong with you") or prescribe treatment ("this is what you should do") but giving general medical knowledge is perfectly fine. THB's answer was excellent. —Keenan Pepper 23:25, 23 December 2006 (UTC)[reply]

The seller of these lenses does not know the focal length. I sent instruction of how to estimate it using paper and a window but am still awaiting a reply. I'm impatient. Can someone give the focal length from this picture [3] and the 64 mm diameter and 9 mm depth? --Username132 (talk) 18:46, 23 December 2006 (UTC)[reply]

Your question asks for an opinion or a guess. Sorry, we are only allowed by the administrators to refer you to Wikipedia articles. This is a Reference Desk, not a chat room. Please ask your question in a chat room. Sorry. -THB 18:55, 23 December 2006 (UTC)[reply]

I've reworded my question to demand a more assertive answer. --Username132 (talk) 19:12, 23 December 2006 (UTC)[reply]

Thanks! -THB 19:32, 23 December 2006 (UTC)[reply]

I may be wrong, but I dont think you can tell just from the dimensions. I think you need to know the refractive index of the material as well. THe method you outlined above would seem to be the simplest method:

Use parallel light beams from a distant object and focus the image on a piece of card. Since

1/u + 1/v =1/f where 'u' is the object distance, 'v' is the image distance (both from the lens) anf f is the focal length. Since the object can be considered to be at infinity , u= inf, the distance from the lens to the card is its focal length. If the seller cannot do this 10 second experiment, is he worth dealing with?--Light current 20:42, 23 December 2006 (UTC)[reply]

I figured someone would bad-mouth the seller ;) Indeed they are worth dealing with since, just like I don't have much choice about using Easyjet, there aren't many people selling lenses on ebay. I need it to turn an infrared beam with a 6 degree half-angle from an LED into a parallel beam that will range about 200+ meters (for a laser tag gun). Apparently I want a short focal length but a small magnification - I'm not sure why a large maginification should disperse the light more, and figure the larger the magnification, the smaller the focal length. Optics was one of my least liked aspects of A level physics... blurgh --Username132 (talk) 22:09, 23 December 2006 (UTC)[reply]

OK I didnt bad mouth him- I asked if you wanted to deal with him. Ok you do. Now the bad news. THis is not as simple as it may first appear. Are these lenses made of glass? If so you may have a problem getting sufficient transmission of IR through it. You may in fact need a plastic lens. Also have you considered using a conical plastic reflector from a torch instead? You may want to see our article on Infra red for more details on the transmissivity of different materials 8-)--Light current 22:30, 23 December 2006 (UTC)[reply]

Sorry; I wasn't attacking you. I just figured someone would question whether I ought to deal with the seller on the basis stated. I had no idea that glass absorbed IR but I can't find it in the infrared article either (although I do note that it specifically mentions the use of plastic lenses). Thanks for the tip. I've reported back to the laser tag forum that I've also been using, so they don't make the same mistake. --Username132 (talk) 23:29, 23 December 2006 (UTC)[reply]

Do you know more about the lens - is it convex? is one side flat? Also if you can describe it's type that would be helpful - is it described as spherical, aspherical etc.? Perhaps most importantly what is is made from glass, plastic etc.? The focal length could be estimated given the inner and outer thickness, the width and the refractive index of the lens, as well as the shape of the lens.87.102.4.34 21:26, 23 December 2006 (UTC)[reply]

I posted all the scanty details they gave, however, the provide a picture [4] from which it appears to be a fat little glass biconvex thing. Could you give a ball-park figure or could it really be anything? Ideally, it should be no longer than 30 cm but less is more for accurate 'lasering'. --Username132 (talk) 22:09, 23 December 2006 (UTC)[reply]

Well it should easily work for your 6 degree half angle - a big guess at the focal length would be 10-20cm - most likely closer to the 10 but I really can't say - if you want ir you should use a parabolic reflector - just buy a torch with the reflector at the front - cut off the front plastic part of the torch (use a hot knife or a saw) - the reflectors are usually metallised on the inside and should reflect ir into a beam really well - if you are interested - ask another question.87.102.4.126 11:28, 24 December 2006 (UTC)[reply]

by the way torch is called flashlight in some backward parts of the world - look a the link - the green one - second picture would do - put the led where the bulb is - cut off the front 'glass' (it'll be plastic)(cut the back of the torch off as well i guess) and hopefully that's what you wanted???87.102.4.126 11:32, 24 December 2006 (UTC)[reply]

Sometimes it is yellowish green. What is the name of it and what is the chemical composition? What is its funcshun?-THB 19:36, 23 December 2006 (UTC)[reply]

Error, grasshopper disassemble. (bonus points for catching the reference)

Short Circuit, of course! Seraphimblade 10:19, 25 December 2006 (UTC)[reply]

"Hemolymph". Leave those grasshoppers alone and read their article instead. Femto 20:14, 23 December 2006 (UTC)[reply]

"What is that red stuff inside humans when you squish them?" — The answer to both questions is the same. — Kieff 20:27, 23 December 2006 (UTC)[reply]

All that shit is blood? It's like 80%???? It doesn't taste salty. -THB 20:32, 23 December 2006 (UTC)[reply]

No, some of that "blood" is shit, which is no doubt why it doesnt taste salty. alteripse 20:49, 23 December 2006 (UTC)[reply]

Err... I was thinking of "their guts mixed with blood"... — Kieff 20:57, 23 December 2006 (UTC)[reply]

Thank you to all of the people who gave valid, well, reasoned answers. Meta-comments belong on the talk page and have been moved there. -THB 21:14, 23 December 2006 (UTC)[reply]

(moved from Miscellaneous Reference Desk)

On my homework, I have to write out "LiCl". I know this is Lithium and Chlorine, so do I just write "Lithium Chlorine", or does it have to be "Lithium Chloride" or "Lithium Chlorate". I just need someone to explain to me how that all works. Thanks, Musli Miester 18:47, 23 December 2006 (UTC).[reply]

The term, LiCl, does not simply represent the two chemical elements independently, but actually represents a compound of the two. See Lithium chloride for more info. Lithium perchlorate, LiClO₄, also contains oxygen and Lithium Chlorine doesn't exists as a single entity. Rockpocket 18:54, 23 December 2006 (UTC)[reply]

I know they are bonded ions. I just want to know how to properly write them. Musli Miester 18:56, 23 December 2006 (UTC)[reply]

You write out the name of the molecule. In this case, lithium chloride. See chloride to see why that is so as opposed to, say, chlorate. --Justanother 19:03, 23 December 2006 (UTC)[reply]

chloride refers to a chemical compound in which one or more chlorine atoms are covalently bonded in the molecule. Chlorates are compounds that contains the chlorate ion (ClO₃⁻) with chlorine in oxidation state +5. Rockpocket 19:12, 23 December 2006 (UTC)[reply]

Chloride is a monatomic ion, it's not covalently bonded to anything. —Keenan Pepper 23:12, 23 December 2006 (UTC)[reply]

You need to know that the Lithium is thought to be Li⁺ and the chlorine is thought to be Cl^-, the charge on the ion changes the name. If you knew that and just need the names then Ions#Common_ions would be a good place to start looking - note negatively charged ions usually end in "ide" (eg S^2- is called "sulphide"), positively charged ions often don't change the name (eg Ca²⁺ is still called "calcium").

If you didn't know the charge on the atoms then say so and someone will try to explain that. Hope this helped you.87.102.4.34 19:19, 23 December 2006 (UTC)[reply]

Okay... Does it work the same when with polyatomic ions? EG: Mg(HCO₃)₂ - Magnesium and hydrogen sulfite? Musli Miester 19:21, 23 December 2006 (UTC)[reply]

See Dictionary of chemical formulas --Justanother 19:33, 23 December 2006 (UTC)[reply]

ps Bicarbonate not hydrogen sulfite --Justanother 19:43, 23 December 2006 (UTC)[reply]

Taking into acount the above corrections - yes (HSO₃^- would be hydrogen sulphite).

The name is "Magnesium hydrogencarbonate" (oxygen containing ions often have the ending -ate.) The charged HCO₃^- ion is called hydrogencarbonate.

Just to confuse things hydrogencarbonate is also called bicarbonate. You don't need to say di-hydrogencarbonate because the formula is already known - the magnesium always has a plus 2 charge so there must be two of the hydrogencarbonate ions.

Most people learn by using the names - and the endings (ide, ate etc) are consistent eg all monoatomic ions are ended in "ide" Iodine becomes iodide I^-, Oxygen becomes oxide O^2-, Phosphorus becomes phosphide P^3-.87.102.4.34 20:02, 23 December 2006 (UTC)[reply]

if two virgins have sex, i.e. no previous intercourse, within reason (i.e. no 1 in a million chance ofone of them having caught aids of a rusty nail or something) what protection does using a condom offer that the female going on the pill wouldnt? -172.159.187.49

A condom offers the same protection to everyone who uses it correctly. If you're asking for us to give spitball estimates of the chances of disease transmission in the circumstances above, I'm not sure that's a good thing to ask. Many STDs are transmitted in other ways, so the "1 in a million" shot you're describing above isn't very accurate. But, this question is really asking for medical advice, which should come from a doctor. Ned Wilbury 21:04, 23 December 2006 (UTC)[reply]

No you completely misunderstood the question. Im asking if there are any diseases which can be transmitted via intercourse that virgins could possibly have within reason, as no doubt they could have aids, but living in a western country, the chances are less than one in a million (that one of them has aids) so they neednt worry. So do you understand what im asking now?

The chance that a virgin will have AIDS is a lot better than 1 in a million; you can also get it from sharing needles, from birth (if your mother had it), from blood transfusions, etc. StuRat 21:55, 23 December 2006 (UTC)[reply]

The pill is better at preventing pregancy than condoms. However, you should never assume your partner is disease free. Anyway, you could get sick from analingus, for instance, even if your partner is disease-free. Frankly, it's not worth the risk outside of an established relationship to have sexual intercourse without barrier protection. If you are Catholic, you should talk to your priest about birth control. -THB 21:13, 23 December 2006 (UTC)[reply]

Sure, you could get herpes just from kissing or through oral sex. Hepatitis, too. There are diseases that you can get without having sex that can then be transmitted via sex, in addition to natural stuff like e. coli infection. -THB 21:21, 23 December 2006 (UTC)[reply]

Ok cheers, that answers my question pretty conclusively. Thanks!

I'd like to add, and I'll probably get told off for it, but I'm still a believer in nature! If two people are virgins, and in love and been in a relationship for six months and have seen a doctor to get tested, I can't say I think condoms are mandatory, if you want to have kids you'll eventually have to stop using them anyway;) Yes there was probably better then one in a million chance that I got a STD when I lost my virginity without a condom, but I'm glad I took that chance. Maybe I was lucky, maybe if 10 people do it one of them does catch something, it is safer to use a condom, don't listen to me, make up your own mind, weigh up the risks and live with the consequences. Vespine 22:06, 26 December 2006 (UTC)[reply]

Rider question

If one person has normal herpes, is it possible to induce genital hepers in your partner by means of oral sex?--Light current 00:39, 24 December 2006 (UTC)[reply]

Yes. I believe they're caused by one and the same virus. Don't play with herpes! --Username132 (talk) 00:51, 24 December 2006 (UTC)[reply]

So when you ve got cold sores, you shouldnt have oral sex with you bf/gf. is that right?--Light current 00:59, 24 December 2006 (UTC)[reply]

The herpes virus is present even in the absence of visible sores. If you have oral or genital herpes, a Dental dam and/or condom should be used at all times. Hipocrite - «Talk» 01:07, 24 December 2006 (UTC) Hipocrite[reply]

So any one I had unprotected oral sex with could now have genital herpes?--Light current 01:11, 24 December 2006 (UTC)[reply]

That is a correct interpretation of the information provided above. - CHAIRBOY (☎) 01:13, 24 December 2006 (UTC)[reply]

Oh Shit!--Light current 01:19, 24 December 2006 (UTC)[reply]

The herpes simplex article indicates that generally the oral and genital manifestations are the result of difficult viruses (HSV-1 and HSV-2). --24.147.86.187 03:25, 24 December 2006 (UTC)[reply]

Meaning?--Light current 04:05, 24 December 2006 (UTC)[reply]

He meant to say 'different' viruses, meaning the tranmissibility of oral to gential and versa must be lower than oral-oral or genital-gential. He meant the chance of infection of your partners is lower than you may have originally concieved, although we don't know how low. At university, we we're told, "don't do it, man!"!

Without having had the other person under observation for their lifetime, it would be impossible to know that a claimed virgin is an actual virgin. People have been known to lie about their sexual activity. Therefore the same safe sex practices should apply to relations between purported virgin as for relations between non-virgins. Edison 20:25, 24 December 2006 (UTC)[reply]

What various ways of there of creating a loud 'bang' sound, similar to 'party poppers'? Basically for scaring my friends! --Chachu207 ::: Talk to me 21:42, 23 December 2006 (UTC)[reply]

See the article on Explosive material. You must promise not to be a terrorist. -THB 22:13, 23 December 2006 (UTC)[reply]

He said terrorizing his "friends" was the whole point. alteripse 02:55, 24 December 2006 (UTC)[reply]

Depends how you want to use it. A paper bag or balloon might do the job, but you could just as easily say BANG, which would actually be the more environmentally-friendly option. Do you have any scenario in particular in mind? --Username132 (talk) 22:13, 23 December 2006 (UTC)[reply]

Newspaper Poppers are surprisingly loud, if you make them right. It's a little like cracking a whip. —Keenan Pepper 23:09, 23 December 2006 (UTC)[reply]

Haha, I remember those from middle school. X [Mac Davis] (DESK|How's my driving?) 00:17, 26 December 2006 (UTC)[reply]

Some people seem to think that now that the national helium reserve will be liquidated, we will run out of helium, or at least experience a severe spike in its price due to the scarcity after squandering such a resource. So, how much He is left to be extracted from the Earth before it starts to run out? Are there any estimates?--Deglr6328 01:05, 24 December 2006 (UTC)[reply]

The US Geological Survey, in this PDF indicates the US produces about 84 million cubic meters of helium per year, and also has been withdrawing about 50 million cubic meters per year from the reserve. Domestic consumption is about equal to production, at 84 million cu m per year, and the US is a net exporter of helium (one of few mineral commodities for which this is the case) to the tune of about 50 million cu m per year. So it looks like we'd just stop exporting it. The report says the US reserves of helium amount to 3,600 million cu m, or 43 years at current consumption rates. Cheers Geologyguy 03:24, 24 December 2006 (UTC)[reply]

hm! interesting. that's quite a lot! *scratches off of list of trivial things to worry about*...--Deglr6328 03:54, 24 December 2006 (UTC)[reply]

So why did the US ever decide to store so much ? Do Congressmen just really like to give helium balloons to their mistresses/pages ? StuRat 04:48, 24 December 2006 (UTC)[reply]

For various space age stuff and airships when they were previously all the rage, aparently. Oh and by the way, if I ever inexplicably develop a musical talent I am TOTALLY naming my band The Helium Crisis so don't you dare even think about it. :) --Deglr6328 06:24, 24 December 2006 (UTC)[reply]

Will they all sing will high falsetto voices, Chipmunk style ? StuRat 12:21, 25 December 2006 (UTC)[reply]

So the US doesn't need to worry; they just stop exporting it. The rest of us however are in deep trouble! The end is nigh! *Starts stock-piling helium canisters in the attic* --Username132 (talk) 10:52, 24 December 2006 (UTC)[reply]

Don't worry, I'd be willing to bet that if and when international demand justifies it, US private sector producers will increase production of helium to meet that demand. Cheers Geologyguy 15:26, 24 December 2006 (UTC)[reply]

See our articles National Helium Reserve and Helium#Applications for reasons why the U.S. government might want to have helium handy. TenOfAllTrades(talk) 16:10, 24 December 2006 (UTC)[reply]

Once used, most helium escapes into the atmosphere and ultimately bleeds into space because it is too light to be trapped by the Earth's gravity. Hence all the helium we have is left over from the formation of the Earth. Absent someone inventing cheap fusion, helium is basically irreplacable. So what is the value of dwindling and irreplacable natural resource? Apparently not as high as one might think. Dragons flight 18:26, 24 December 2006 (UTC)[reply]

But, depending on the quantity on Earth, it may take millions of years for us to use it all up. If so, this really isn't much of a concern. StuRat 12:15, 25 December 2006 (UTC)[reply]

I would say that liquid helium is actually tremendously important, which means squandering so much of it is really kind of silly--71.247.120.5 23:14, 25 December 2006 (UTC)[reply]

The atmosphere still contains only about 24 trillion cubic metres of the stuff. This is indeed worrisome; what are we going to do when it runs out? Since the rate of depletion does not depend on consumption but only on the bleeding off into space, I propose that NASA envelop the Earth with a big balloon.  --Lambiam^Talk 00:48, 26 December 2006 (UTC)[reply]

Don't be silly, there isn't enough rubber to make a balloon that size--71.247.120.5 00:52, 26 December 2006 (UTC)[reply]