Wikipedia:Reference desk/Archives/Science/2006 October 10

Super Cannon of Doom
I asked a long time ago if it was possible to send an object wieghing 500 tons hurtiling through the air at 7.2 miles per second and the answer was yes but with the amount of energy that little boy the atomic bomb released. So To further the question is it possible, I'll even be happy if you give me the equations to lead me in the right direction, to build a cannon with a ten foot bore capable of sending an object of this wieght? If so how long would the cannon be and what are some Ideas on what compound to use that would create the required energy? Thanks I know this is really random and of course it was inspired by Jules Vernes.


 * Project HARP--Light current 02:23, 10 October 2006 (UTC)

So could Harp be put on some steroids for my purposes or it this totally and uterlly impossible.
 * I think if Gerald Bull could have made it bigger, he would have.--Light current 14:02, 10 October 2006 (UTC)


 * To start, here's the article on kinetic energy, to calculate how much energy is required to accelerate the payload to the required speed. The way I crunch the numbers, the amount of TNT required to provide that amount of energy (assuming 100% efficiency, which of course is not possible) would be a cube 16 metres (roughly 50 feet) to a side, and weighing 7,000-odd tonnes.  That should give you an idea of the gargantuan scale of any such cannon.  --Robert Merkel 05:43, 10 October 2006 (UTC)


 * Some points of reference: Jules Verne's From the Earth to the Moon(1865) had a giant cannon fire a heavy projectile from the Earth to the Moon. Skeptics say "bad science": it would have killed the passengers and fallen back to the ground. Big Bertha (Howitzer) a WWI siege mortar fired shells weighing 820 kg each to a maximum range of 12 km. The Paris Gun, another WWI artillery piece, built for long range bombardment, fired 94-kilogram (210 lb) shell to a range of 130 kilometres (81 mi) and a maximum altitude of 40 km (20 mi). The 16 inch guns on the WWII battleship USS Missouri were 16 inch (406 mm)/50 caliber Mark 7 naval gun. The article does not give range, projectile velocity, or muzzle velocity, but an earlier gun is described at http://www.geocities.com/fort_tilden/16ingun.html For a 16" Mk2 they state 2750 feet/sec, projectile weight 2100 pounds, The largest bore naval guns of the 20th centyury were the Japanese WWII 40 cm/45 Type 94 which had 18.1 inch bore and fired a projectile of 1460 kg or 3220 lb at a velocity of 780 m/s or 2560 ft/s to a range of  42,000 meters or 26 miles.  The WWII Schwerer Gustav could fire a projectile weighing 4.8 ton (4,800 kg) at a muzzle velocity of 820 m/s, using 2,500 lb (1134 kg) of propellant. Now you can figure the kinetic energy of your proposal and compare it to that achieved by practical artillery. Graph the KE against the weight of the entire gun, and extrapolate to what the approximate required weight and propellant charge of your gun would be. You specified 7.2 miles/sec = 38,000 feet/sec or 11,600 meters/sec. This is way higher than the achievements of artillery designers, so your projectile would likely melt and lose velocity very rapidly, like a meteor reentering the atmosphere. Certainly you would want to consider having the muzzle be at the top of Mt. Everest, to take advantage of the thinner atmosphere. Consider a lower muzzle velocity with rocket power to boost the projectile to the ultimate velocity when it is up to a region of thin atmosphere. Then consider rail guns and mass drivers, or perhaps acquire some old blueprints and build a Saturn V, which could boost 260,000 pounds (130 tons)(118,000 kg) to low earth orbit. You could get your payload into orbit with 4 such rockets, assemble it there, then send up more boosters to launch it on it way to wherever. Edison 16:08, 10 October 2006 (UTC)

Awesome thanks for the ideas they are exactly what I was looking for.

Is there a way to calculate the arc or when then the projectile will start going down instead of up?


 * I'm no good with "miles" so 7.2 m is 11.5872768 km, call it 11.6 km/s, or 11600 m/s. Gravity is close to 10 m/s/s, if you divide initial velocity by the acceleration of gravity you get the 'velocity = 0' point, or the time when the projectile starts going down = 1160 seconds = 19.33 minutes. Because it is accaleration constant, the projectile would travel as far as if it had been going half its initial speed for the full length of the journey, = 5800m/s for 1160 s = 6728000 mhad meters, 6.7 thousand km, about 4 thousand miles. Of course wind resistance and orbital motion screw all that up, the above would only be true in a frictionless environment and shooting straight up. Vespine 06:40, 16 October 2006 (UTC)

An estimaste of the thermal power output and specific heat capacity of a human being
I am curious how long it would take someone like me to overheat if wrapped up in a ton of blankets. Assume an average sized human in a fairly sedentary state. This human is then put in a perfect insulator. About how long would it take for the human's temperature to be raised by approx. one degree celcius? I originally guessed the specific heat capacity for human tissue would be close to that of water but looking at a table in specific heat capacity and noting that the specific heat capacity for wood was about ten times lower, I am not so sure. The equation should be (energy in)/ (mass * specific heat capacity) = (change in temperature). Sifaka  talk  02:06, 10 October 2006 (UTC)


 * One average human being produces as much heat as a 100W bulb (ie 100W) when sedentary.--Light current 02:20, 10 October 2006 (UTC)


 * Since we're mostly made of water, I suppose using the SH of water would give you a good estimate of the temperature rise per hour etc. Energy = mass*specific heat*temperature rise (if I remember correctly).


 * This is one of those rare occurrences where the calorie comes in handy (although it is not necessary). 100W is 100 J/s. The article says it takes 4.185 kJ to increase the temperature of 1 kg of water by 1 °C. So if we take a 70 kg person, it takes about 30 kJ. So it would take 30,000/100 = 300 s or 5 minutes. So you'd be dead within half an hour. In other words, don't try this at home (if you'd have such a perfect insulator, that is). Or anywhere else, for that matter. DirkvdM 08:42, 10 October 2006 (UTC)


 * THis analysis of course excludes the effects of the body's temperature regulation mechanism for which at present I cant find the page (cos I havent looked!). I believe some tests were done on how many blankets to put over a young baby without overheating it. Cot death?--Light current 08:50, 10 October 2006 (UTC)

All of this stuff is relevant only to dead bodies. Live human beings are homeothermic and will expend considerable energy to maintain a healthy body temperature. alteripse 09:35, 10 October 2006 (UTC)


 * I wasnt aware that dead people could generate any heat 8-)--Light current 14:05, 10 October 2006 (UTC)


 * Presumably a corpse would generate some heat, since decomposition is a form of oxidation. If the dead person were named Ernest, then heat might be generated by maggots fighting in dead Ernest over tasty morsels. And I've seen 60 watts as the approximate heat output of a person at rest. (Some have slower metabolism than others) People doing sustained work can burn 500 kcal/hour, or if very athletic up to 800 kcal per hour. Another rule of thumb is that a man can produce about 1/8 horsepower of sustained work, while an athlete can produce perhaps 1/2 horsepower for a short while. These figures are off the cuff and totally unreferenced. The stated problem is extremely complex, because an overheated person wrapped in blankets would sweat profusely, and presumably some cooling would occur as the moisture evaporated with the blankets acting as a wick. Respiration can also produce cooling. Per Specific heat Q = n C ΔT where delta T=1 degree, m=100kg for a good sized person,, and c= 4.18, (for a person like water> it would take 4.18 *10^5 Joules, which at 60 watts of heat produced would take 1.9 minutes. At http://www.engineeringtoolbox.com/human-body-specific-heat-d_393.html they give the specific heat of the human body as 3.47 J/*(g*degK), and at various sites the 100 watt output figure is commonly bandied about, so do the math.Edison 16:58, 10 October 2006 (UTC)


 * Calories and horsepowers? What century are you from? Also, there is no such thing as degK. There is K, though. DirkvdM 08:32, 11 October 2006 (UTC)


 * From the mid-2oth century, and proud of it. Still have my sliderule. Per K (disambiguation), K represents 21 different things, so perhaps you could edit that page and add degrees Kelvin as the 22nd. Deg K= deg C so far as delta T is concerned. Motors are still rated in HP, so it is an interesting comparison.  In many parts of the world, human or animal labor still pumps water for irrigation and cultivates fields. A champion cyclist can put out 1/2 HP for 75 minutes, but in competition such cyclists only produce an average of 1/3 HP continuously. The average person can only produce 1/10 to 1/5 HP continuously, per http://qlipoth.blogspot.com/2006/08/our-slave.html The net result is that one barrel of oil represents one year's hard labor for a person. Such labor consumes food energy, usually measured in kcal. Metabolism produces heat. See http://www.ieer.org/reports/energy/3-power.html for an input-output analysis of bullocks in Joules, and HP. A bullock is about 6% efficient. Are you? Edison 14:22, 11 October 2006 (UTC)


 * Edison - I think Dirk is just pointing out that the kelvin scale isn't measured in degrees. Matt Deres 18:46, 11 October 2006 (UTC)

I appreciate the Earnest joke, which should tell you I am no engineer, but I am relying on the definition of specific heat as simply the heat needed to raise the temperature of a substance 1 degree. It takes a heckuva lot more heat to overcome the homeothermic mechanisms of a live human being and raise the body temperature from 37 to 38 degrees than it would to raise the temperature of a corpse by 1 degree (and before you ask, no I have not tested this personally). Are the engineers at the site you linked to that simple, or are you and they assuming a different definition of specific heat? alteripse 21:15, 10 October 2006 (UTC)

UG Nuclear tests
In films of American UG nuclear tests, you see a large area of ground in the test area sinking rather than being pushed up. Why is this?--Light current 03:29, 10 October 2006 (UTC)


 * I expect it's the collapse of the mine or tunnel or cave they put the bomb in, but I'm just guessing. --Allen 04:32, 10 October 2006 (UTC)


 * See nuclear testing and Subsidence crater. Dismas|(talk) 04:57, 10 October 2006 (UTC)

THanks. But I would still expect the ground to bulge before it sinks back, and it doesnt seem to. 8-|--Light current 13:59, 10 October 2006 (UTC)


 * Sometimes it does bulge. I imagine it depends on how large the blast is and how close it is to the actual surface. This file shows a graphic in which it indicates some bulge. But the bulging and the sinking are two different things — one is a shock wave, the other is the result of the molten rock having cooled and falled into the newly created hole. --Fastfission 21:22, 10 October 2006 (UTC)

A question that relates to UG nuclear tests. How are the effects of Electromagnetic pulse affect above ground? Or do the EM pulses never reach the surface? --Agester 12:26, 10 October 2006 (UTC)


 * You shouldn't get any real EMP effects unless you are in the atmosphere. If you read our EMP article it discusses why this is (they dissipate fairly quickly). --Fastfission 21:22, 10 October 2006 (UTC)

fainting at the sight of blood
My boyfriend faints when he see's blood. He always has but to different degrees. He has a vagal responce. He gets pale then cold and sweaty. A doctor told him the name of a syndrome he though start with a V. I am trying to find out info. Thanks Pattibeach


 * Vasovagal syncope? Sifaka   talk  04:01, 10 October 2006 (UTC)


 * Vampirism? They are reputed to be pale and cold. Edison 17:05, 10 October 2006 (UTC)


 * I think a Vampire would be in trouble if he fainted when he saw blood. That's one thirsty Vampire... Benbread 20:19, 10 October 2006 (UTC)

weapon testing
North Korean explosion caused earthquake of magnitude 3.58 on ricter scale..Any estimation of weapon power??
 * see 2006_North_Korean_nuclear_test =--GangofOne 06:25, 10 October 2006 (UTC)

Lead acid storage batteries
Conventional Lead storage Acid batteries have water as the electrolyte, which has the risk of electrolysis resulting in battery explosion.Can any substitute be used for water?


 * i guess another system with good solubility for sulfuric acid would work, as long as the sulfuric acid could dissociate. i can't find too much on it though. Xcomradex 07:27, 10 October 2006 (UTC)

This is the best article I could find. The fanciest lead acid batteries use a glass mat in a tight roll. They do not discharge explosive hydrogen gas under normal conditions. --Zeizmic 12:11, 10 October 2006 (UTC)


 * Someone from a battery company once pointed out that if a battery is maintained, it is full nearly to the top with liquid, leaving very little space for enough hydrogen to accumulate to produce an explosion capable of rupturing the battery case. He said cases of batteries blowing up were probably cases where the electrolyte level had been allowed to drop very low.Edison 17:08, 10 October 2006 (UTC)

Critical MAss
After any Mass goes critical,from where does that one neutron come to create the reaction?Even in refining, what i dont understand is that when the percentage of enrichment goes too high it gets critical and undergoes reaction.But how does that happen without the first neutron to start the process?


 * it comes from spontaneous fission. Xcomradex 07:18, 10 October 2006 (UTC)


 * That's fine for a reactor, where the critical mass is assembled statically and you can wait a fraction of a second before a spontaneous fission supplies the neutron that starts the chain reaction. In a bomb, the critical mass may only exist momentarily.  In that case an initiator is provided, which is a combination of substances that will emit plenty of neutrons.  The Fat Man bomb used a beryllium-polonium initiator with the two substances kept separate until detonation, so the neutrons wouldn't cause trouble eralier.  On mixing them, alpha particles from spontaneous decay of the polonium hit the beryllium atoms and the debris includes neutrons.  See section 8.1.1.2 on this page.  --Anonymous, 04:30 UTC, October 15.

Transition Metal Chemistry question
Hi I need info and examples to do with, Transition Metal chemistry. I need info on, 1. properties, trends, and oxidation states of the first row of transition elements.

2. Uses in volumetric analysis.

3. coordination compounds, coordination number, chelates, isomers, optical activity, substitution reactions.

4. complex ions in medicine: cisplatin, chelation therapy.

5. bonding in complex ions, d-orbitals, colour in complex ions.

6. transition metals in qualitative and quantitive analysis.

7. ellingham diagrams, free energy of oxide formation and carbon as a reducing agent, the thermite reaction.

I also need info on:

1. Thermodynamics: Gibbs free energy and spontaneous change.

2. Redox reactions and electrochemical cells.

3. Info on the non-metallic elements:

i.e properties, trends, oxidation states of the non-metals

i.e the nature o covalent bonding


 * (ec)try clicking on some of the blue words. and get a good textbook, i recommend atkins, inorganic chemistry. Xcomradex 08:48, 10 October 2006 (UTC)


 * This is a reasonably good site for the basic stuff. Also try some of the periodic tables available from www.download.com, some of which have details on each element. For a detailed answer on the reference desk, you will need to be more specific. BenC7 08:35, 10 October 2006 (UTC)

Question about Kinetic Theory of Gases and the Physical Meaning of Temperature
I have some (perhaps very stupid) confusions about kinetic theory. From the theory we can show that the internal energy of the system is related to the temperature: $$ U = \frac{3}{2} \cdot N k  T. $$

So it encourages me to interpret the real meaning of temperature as internal energy of gases. But what will be happen if we evacuate all the gases from the system (i.e. make a vacuum closed black box)? Can we say anything about the temperature of a system with no gas at all? (does a vacuum box has a temperature?)

What is the physical nature of temperature? particle or something? -- 131.111.164.229 14:39, 10 October 2006 (UTC)


 * Hm, I have not thought of this lately. My guess is that if you had a pure vacuum with nothing in it, then you'd in principle reach absolute zero. However, in our world, this vacuum is inside a vessel of some sort, and because the vessel has a vapour pressure, you'll have residual gas atoms inside, no matter how hard you try to pump the vacuum. The internal energy is not as simple as that, for example, there are also contributions from rotational, vibrational, and eletronic transitions. Wait for another Wikipedian to chime in. --HappyCamper 15:07, 10 October 2006 (UTC)
 * Anyway, because of vacuum fluctuations, it a "box with nothing in it" is a non-physical concept. Batmanand | Talk 15:41, 10 October 2006 (UTC)
 * I guess that would be the quantum explanation. --HappyCamper 16:13, 10 October 2006 (UTC)
 * Your original conclusion hits the nail. You cannot ascribe a temperature to a vacuum. Maybe I add a few pints to make things clearer. (i) Heat is defined as the internal energy of matter associated with its temperature. The difference between heat and all other kinds of energy is that heat can never be fully used again. See the second law of thermodynamics for details. An intuitive explanation is this: If you have a lot of particles of a gas travelling in the same direction, you have wind. By meand of a windmill you can make use of the kinetic energy of this movement. But if all molecules speed in different directions, so that their movement cancels on average, it is heat. You cannot use the energy easily -- actually, not at all, unless you have another piece of matter with less temperature. (ii) As Happy Camper points out, there can also be energy be stored in internal degrees of freedom. See here for the textbook example. A gas with diatomic molecules as depicted there has twice as much internal energy as a single-atom gas (for which your formula is valid only) at the same temperature and hence it has twice as much heat capacity, with heat capacity being defined as the the internal energy that the matter stores per degree of temperature. Vacuum cannot store any heat energy, and hence has heat capacity 0. As just stated, temperature is internal energy devided hy heat capacity, and 0/0 is undefined, so it does not make any sense to assign a temperature to ideal vacuum. Less mathemematically: A body is called warmer than a reference body at temperature $T$ if it lets heat flow to the reference body, and colder, if it sucks heat out of it. Now, putting vacuum "next to" the reference body does neither, hence it has the same temperature as the reference body, no matter what this body's temperature is. You see, this does not make sense, vacuum does not have a temperature. Simon A. 17:01, 10 October 2006 (UTC)


 * A vacuum most definitely does have a temperature. An evacuated space would still contain a photon "gas" which, at equilibrium, is at the temperature of the walls of the container. So the container also has a heat capacity. If you put a hot object in a vacuum at a colder temperature, it will "suck heat" out of it. For an excellent explanation, see "The Heat Capacity of a Vacuum" and surrounding pages at . And then ask any followup questions. --GangofOne 04:28, 11 October 2006 (UTC)


 * Ok, I should have expected that somebody brings that up, but I had already written a long answer. Now, there are some subtleties. First, we did not mention blackbody radiation yet. In my example, the two bodies exchanged heat because they touch. But every body also continuously emits and absorbs electromagnetic radiation, in the case of room temperature mainly infrared radiation. If a cool body is standing next to a warm one, even with an evacuated gap in between, they will exchage heat because the hotter one send more radiation to the cooler one than it absorbs from the cooler one. Now imagine a Hohlraum, i.e. a hollow body of matter. The inner walls exchange heat by the means of radiation, and the hollow, evacuated interior is filled with a bath of photons with a frequency distribution corresponding to the walls' temperature (as explained in the notes cites by GangofOne). In this sense, this so-called photon bath has indeed a temperature. But in a way, it is just the temperature of the walls. To see why this is a problem, imagine one wall is continuously heated and the opposite one cooled. In the Hohlraum, we get a mixture of the Planck spectra correspoding to the two temperatures. If the vacuum were replaced by some material filling, we would get a temperature gradient, i.e. the temperature would change smoothly from hot to cold. This is because matter is able to equilibrate: If the atoms vibrate strongly at one place and weakly at the adjacent place, they will exchange energy and end up both vibrating equally strong. Photons cannot equilibrate because they do not "feel" each other. All equilibration can only happen via the material walls. This is why one can ascribe a temperature to en evacuated space if it is surrounded by walls in equilibrium (i.e. all walls at the same temperature) but not in other cases. Another point is: Let's hold a thermometer into outer space. The photon bath there is in equilibrium, more or less and despite what I just said, namely at the famous 3 Kelvin of the microwave background radiation. How long would it take for the matter of the thermometer to cool down to these 3 K. Within the solar system, it would never get as cold, because the sun keeps heating it up. You would have to go really far out to see the 3 K as actual temperature. Simon A. 07:13, 11 October 2006 (UTC)


 * Some minor clarifications: for many materials, the heat capacity is not a constant and $$T\not=U/c$$; what is true is that $$dT=dU/c$$. Also, it's often the case that certain degrees of freedom are "frozen out" &mdash; even their first excited state is all but unpopulated &mdash; and do not count for internal energy calculations; in particular, at normal temperatures diatomic gasses get 5/3 the heat capacity of monatomic, not twice it.  --Tardis 15:44, 11 October 2006 (UTC)

more
Thanks everyone, here is more refined question. Suppose the system is just inside the container (do not regard the container as a system), and contains only one (ideal gas) atom. Then the internal energy of the system is equivalent to the total energy of that atom (correct me if I'm wrong). What is the temperature of the system? Can I calculate the temperature from this formula? $$ T= \frac{2U}{3Nk}. $$

Does it make sense to say about the temperature of an atom? Or would it be better to think about the temperature as macroscopic quantity, and try not to relate it to the microscopic world? -- 131.111.164.110 13:25, 11 October 2006 (UTC)


 * The fundamental issue is that a temperature is only ascribable to a system in equilibrium; this is why the interior of Simon's hohlraum has no temperature. (When we speak of temperature gradients, we are saying that "locally" there is equilibrium, or perhaps only a small deviation from it, with a different equilibrium at every point.)  As far as a single atom goes, it's not useful to call it an ideal gas because it isn't a collection of particles that interact with each other and their surroundings in any particular fashion (for an ideal gas, this fashion is "not at all" and "only by collision with walls or so").  You can give the atom a temperature with that formula, but remember that in the atom's frame of reference it's not moving at all.  The problem here is that a single particle (or even a collection of very few) doesn't have a well-defined average velocity which we can identify to separate a bulk material velocity from a collection of random velocities (whose corresponding kinetic energies count as heat).  The problem of average velocity is exascerbated by interactions with the walls; only for a large number of particles will those interactions average to (very near) 0 momentum exchange over a reasonable time period.  What you can say without complication or inaccuracy is that there is a particular temperature T at which this particular particle's velocity (in this particular reference frame) would be most typical of a system (e.g., an ideal gas of particles like the one in question): the characteristic temperature associated with that (hypothetical) system and velocity.  When applying that equation, be sure to keep in mind the distinction between (even random) kinetic energy and internal energy, which also includes the non-translational degrees of freedom to which Simon referred.  Does that help?  --Tardis 15:44, 11 October 2006 (UTC)


 * That's really help. Thanks very much! -- 131.111.164.228 08:28, 12 October 2006 (UTC)

Symmetry in technical drawing
I'm not sure whether this should go in Science or Miscellaneous, so feel free to move it if you want. (If there was a Technology section, I would have put it there.)

How does one properly indicate circular symmetry in a technical drawing? I have seen an example here at Wikipedia, but I don't know if that is the best way to do it. Actually, in my case it's a hexagon rather than a circle, so it's not truly circular symmetric. Should I draw it as such anyway or is there another way for my case? —Bromskloss 17:10, 10 October 2006 (UTC)


 * To indicate an equilateral hexagon, draw a small line (dash)crossing the mid point of each side (not quite perpendicular) and a small semi circle indicating the inside angle of each corner. The effect you are looking for is to indicate that each side and angle are the same. If you had a hexagon where there were 2 different lengths to the sides and 2 different angles to the corners, you would mark one set of sides with one dash and the other with two dashes, one set of angles with one semicircle and the other with two. I unfortunately don't have an example I could link, in this case a picture would be worth about 100 words ;) Vespine 06:00, 16 October 2006 (UTC)

WEIGHT OF HUMAN BEING IN THE MOON
WHAT WILL BE THE WEIGHT OF A MAN IN MOON, IF HE IS WEIGHING 60KGS ON EARTH.


 * ~9.9&frac14; kg B00P 17:30, 10 October 2006 (UTC)


 * Why are you using only capital letters? —Bromskloss 17:34, 10 October 2006 (UTC)


 * Well, the Kilogram is not a measure of weight, but mass. So the mass on the moon would be 60Kg.  The weight would be about 22lb.


 * This is the relevant calculation (Newtonian-style), resulting in a measurement of 97.62 N. On a kilogram scale calibrated for Earth's gravity, the person would seem to have a mass of 9.96 kg, though that measurement isn't correct (see Mass vs. Weight). -- Consumed Crustacean (talk) 19:05, 10 October 2006 (UTC)
 * Lbs also measure Mass, a unit called pounds-weight (equilavent to the weight in Lbs time 9.8) is used for weight. however if you are using metric i.e. kg, then you should measure weight in N --Englishnerd 19:50, 10 October 2006 (UTC)
 * No, you shouldn't. Kilograms are the proper units for body weight, in the normal and proper usage in the medical sciences and in sports.  Furthermore, the pounds we use for those purposes are also the normal pounds, units of mass equal to exactly 0.45359237 kg by definition, and not pounds-force.
 * If you weigh 60 kg (not only not KGS, but not Kg either, and with a space between the number and the unit) on Earth, you would weigh 60 kg on the moon. If you weigh 130 lb on Earth, you would weigh 130 lb on the moon (except, of course, in the usage of some science teachers, who just ignore the conventional and correct usage). Gene Nygaard 01:14, 15 October 2006 (UTC)
 * Yes, you should, if you're dealing with physics. If you're measuring something in kilograms, it's no longer strictly "weight", but mass. It's common language to treat weight and mass as the same thing, which is fine, but as soon as you get into physics you have to define things more carefully. Neither definition is any more correct, it just depends on context. -- Consumed Crustacean (talk) 04:14, 16 October 2006 (UTC)
 * There's no "treating" about it. That is the proper, legitimate meaning of the word weight, well justified in history, in linguistics, and in the law—the very thing physicists like to call mass in their jargon.  It is improper for you to pretend otherwise; fortunately, nobody ever gave any physics teacher any say-so as to what the word weight means in the net weight of a can of beans or the troy weight of a bar of platinum or the weight recorded on your chart in a hospital. Of course, mass like weight is an ambiguous word with several different meanings; when we are talking about our weight, we wouldn't want it confused with the muscle mass as it is used by bodybuilders, for example.    Gene Nygaard 14:37, 17 October 2006 (UTC)
 * If he's in the centre of the moon, the weight would be 0N.


 * If he is weighing 60kg on Earth (notice the grammar) then obviously his weight on the Moon is 0 because he isn't there. The reason I don't give a proper answer to follow this up is that I don't like being shouted at. DirkvdM 09:55, 11 October 2006 (UTC)
 * Actually, even if the person is at the surface of the Earth, the moon still applies a "weight" on the person. Using universal gravitation, plug in the mass of the Moon, the 60 kg, and the orbital radius of the Moon, and you'll get the lunar weight of the person on Earth. Tito xd (?!?) 04:22, 15 October 2006 (UTC)


 * Just use slugs as your units of mass. Tito xd (?!?) 04:19, 15 October 2006 (UTC)


 * I like slinches better (1 slinch = 1 lbf·s²/in = 386 lb = 175 kg). Gene Nygaard 04:03, 16 October 2006 (UTC) Or, of course, 3 hundredweight 3 stone 8 pounds, for those who think hundred is written in digits as "112".  Gene Nygaard 04:06, 16 October 2006 (UTC)

having laser eye surgery
im thinking about having laser eye surgery so that i don't need to wear glasses or contacts. what is the minimum age for this? what is the average cost? and are there any risks? thanks
 * I had this last year - it was great. In the US, the costs run from $500 to $2500 per eye ($500 is dangerously inexpensive -- doctors who charge this typically do it by cutting corners that SHOULD NOT be cut). It can't be done until after puberty - a candidate in his early-to-mid-20s is ideal. There are a number of risks, but most of them are low (IIRC, the chance of infection is 1 in 500; the chance of the microkeritome malfunctioning is 1 in 500, 'etc). Raul654 17:33, 10 October 2006 (UTC)
 * Also - it's not for everyone. If your vision is too good (say 20/80 or such) they doctors won't risk it; if it's too bad (say, index of -12 or worse) they won't do it and will recommend something else (like an implantable contact lense).


 * It's also important to recognize that as you get older, your lenses lose their ability to accommodate (change focal length). Eventually, you end up stuck at one focal length. So you can get surgery to change that focal length, but you'll still need glasses for any distance outside of your depth of field. Me, I'm happy being stuck at nearsightedness.


 * Atlant 17:54, 10 October 2006 (UTC)


 * Cuba offers free eye surgery, and they're pretty good at it (among the best in the world I believe). Don't know what the requirements are to get it. You'll probably have to live in a country in South or Central America or Africa, where they have a programme. This appears to be called operation miracle, which is now active in 25 countries and aims to do 600,000 operations per year. You probably have to be poor too, to get it for free. But if you pay, it might still be cheaper than elsewhere and the quality of the work seems to be high. Whether they do this specific kind of operation, I don't know. DirkvdM 10:11, 11 October 2006 (UTC)


 * The best person to talk to about this is your eye doctor - your eyes need to be 'stable' (ie not becoming weaker, so your glasses' prescription should not be changing). Also, the doctors need to examine your cornea, some people have a cornea that's too thin for them to remove anything so the operation can't be safely done. &mdash; QuantumEleven 15:18, 11 October 2006 (UTC)

Cell Biology
who is Singer Nicholson and what was his contribution to the cell discovery


 * You're probably thinking of S. J. Singer and Garth L. Nicolson, who proposed "The Fluid Mosaic Model of the Structure of Cell Membranes" in 1972. Melchoir 20:45, 10 October 2006 (UTC)

Why are Magnets RED?
I would like to know why MAGNETS are usually painted RED please ?


 * When i've seen painted magnets, they're usually pained red and blue to distinguish between the poles. Benbread 20:16, 10 October 2006 (UTC)


 * Its a nice color? 8-)--Light current 21:32, 10 October 2006 (UTC)


 * It's probably just something people did to magnets early on that stuck and became universally recognizable as the visual appearance of a magnet. I mean, if you see a piece of wood painted the same as a magnet, you'd probably assume it's a magnet until you pick it up. - O bli (Talk) ? 21:36, 10 October 2006 (UTC)


 * I blame Hollywood! Nearly all magnets I've ever seen were just black, and not shaped like a horse shoe either. - Dammit 21:37, 10 October 2006 (UTC)


 * I'd blame cartoons, actually. Although a horseshoe magnet does have quite a powerful magnetic field near its poles (pity there's no article on them, apparently). Confusing Manifestation 08:34, 11 October 2006 (UTC)


 * Maybe I'm stating the obvious but no one has mentioned it yet, the red wire is always positive, so the positive pole of a magnet is painted red for the same reason. Vespine 03:34, 16 October 2006 (UTC)

magnitude of electrical force
I have been trying to help my son with his homework and we are stumped, please help. How does the magnitude of electrical force between a pair of charged particles change when the particles are moved twice as far apart? Three times as far apart?


 * You want Coulomb's law. Melchoir 20:46, 10 October 2006 (UTC)


 * In fact, the force is given by


 * $$F = \frac{k q_1 q_2}{d^2} $$


 * and what you want to do, is to look at how F (the force) varies with d (the distance). You could say, find the ratio of the forces, something like


 * $$\frac{F_1}{F_2}= \frac{\frac{k q_1 q_2}{d_1^2}}{\frac{k q_1 q_2}{d_2^2}} = \frac{d_2^2}{d_1^2}$$


 * Do you see why finding the ratios of the forces is helpful? Lots of stuff cancels out, and you're only left with what is important. Plugging in some numbers might help. Let's say, you have two charges at d1 apart. And you move them apart 10 times far. That means, the new distance d2 is now d2 = 10 d1...do you see what to do next? Does this help? --HappyCamper 20:52, 10 October 2006 (UTC)


 * For extra points: now determine how electrical force varies with distance for spheres, long cylinders (or wires) and large planar surfaces.Edison 14:25, 11 October 2006 (UTC)

Electricity
Sometimes you hear someone say that a particle appliance " uses up" electricity. What is it that the appliance actually uses up and what becomes of it? Thanks!!!!!!!!!!!


 * What is a particle appliance? --HappyCamper 20:53, 10 October 2006 (UTC)
 * I believe they meant "particular". Dismas|(talk) 21:37, 10 October 2006 (UTC)
 * Energy, in joules, or more commonly known to the average person in Kilowatt hours, because thats what you pay for. And the energy is converted into another type, often kinetic heat or light around the home. Philc  TECI 20:55, 10 October 2006 (UTC)


 * Way back at the power station some form of energy, often heat from burning coal, oil or gas, sometimes heat from atomic fission, sometimes energy from falling water or sunlight, is used to spin a generator. This produces alternating current which is transformed and transmitted to you through the Electric power transmission system. It then provides heat, light, or mechanical motion for your use. What is used up is obviously fossil fuel or nuclear reactor fuel, or water stored behond a hydroelectric dam. In addition, the transformers, substation equipment, and transmission lines, distribution lines, and distribution transformers are "used up" because they are very expensive to install and have a limited life, so even solar, wind, and hydro power cost something to produce and deliver. Incidentally, no one from a utility comany ever said nuclear power would be "Too cheap to meter." That famous quote was from Lewis Strauss of the Atomic Energy Commission, in 1954. What becomes of it is it turn into heat for the most part, excepting for the moment any which is transformed to another form of energy, such as charging a battery or lifting a load to a height. It will be turned to heat later, when that stored energy is used. Edison 22:04, 10 October 2006 (UTC)
 * Sunlight is usually not used to spin a generator, but used to produce electricity via a solar cell. Don't forget about wind either as Zeizmic sais later on.
 * Plus it turned out that nuclear power was vastly more expensive than he had thought, as you have to develop, and refine the technology, build and decommission plants, mine the fuel, and bhide the waste, not just run the plant, which is all he had accounted for. Philc  TECI 22:32, 10 October 2006 (UTC)

Hey, don't diss nuclear! We're tired of the windmills eating up all the birdies, so we're going to warm up some more atoms! --Zeizmic 00:01, 11 October 2006 (UTC)

Nothing uses up electricity because electrons are indestructable!!! 202.168.50.40 00:01, 11 October 2006 (UTC)
 * Are not!!! Clarityfiend 03:30, 11 October 2006 (UTC)


 * Are too!!! —Bromskloss 12:58, 11 October 2006 (UTC)


 * Having said that, electrical appliances uses up (some of) the energy carried by the electrons. What is used up is energy, not electrons.


 * What is "used up" is electrical energy


 * It's too much effort for lazy humans to say "electrical energy" all the time. So they just kept saying "electricity" instead.

dissolving
does iron dissolve in copper sulphate


 * Is this a homework question? And did you check the articles? bibliomaniac15 23:39, 10 October 2006 (UTC)


 * The first article you should check should probably be dissolve. Also, note that copper sulphate decomposes at 650°, and Fe doesn't melt until 1538°.Tuckerekcut 02:54, 11 October 2006 (UTC)


 * The melting point of iron is a bit of a red herring. Salt (NaCl) only melts at 801 °C, but dissolves quite well in water at room temperatures. --Lambiam Talk  06:11, 11 October 2006 (UTC)

This is a (rather badly written) question about metal displacement reactions. And the answer is yes. --G N Frykman 06:47, 11 October 2006 (UTC)
 * Last time I checked nothing could dissolve into something that wasn't a fluid. - 131.211.210.14 10:51, 11 October 2006 (UTC)


 * From the article, "Copper(II) sulfate decomposes before melting." Therefore iron cannot dissolve in copper sulphate, because there is no such thing as a fluid phase of copper sulphate. I also added that iron does not melt 'till 1538° because it would not even be possible to do the inverse: one could not use molten Fe as a solvent for copper sulphate, because the latter would decompose before dissolving. No red herrings, no displacements, it's just not possible by any stretch of the imagination without changing the question. Please remember to be extra cautious when correcting another user because it can lead to even more confusion for the question asker. Tuckerekcut 13:30, 11 October 2006 (UTC)


 * If you're not part of the solution, you're part of the problem. Edison 14:27, 11 October 2006 (UTC)


 * I always thought that was "If you're not part of the solution, you're part of the precipitate."


 * Atlant 15:50, 11 October 2006 (UTC)

The original question probably should have been: does Iron dissolve in a solution of aqueous Copper Sulfate? I think the answer to that would be "no," the iron would not displace the copper... but it has been a while since I did metal displacement chemistry... the place to look would be a table of solubilities. Find out whether Iron Sulfate is more or less soluble than Copper Sulfate... I'm sure this table is in your textbook, since this is probably a homework question. Nimur 18:34, 11 October 2006 (UTC)
 * If we're really talking "just iron" here, a piece of metal instead of ions, you have a redox situation, not a solubility equilibrium. One could use the electromotive series as a rule-of-thumb about what metal will dissolve in a solution of what other metal-ion. DMacks 20:38, 11 October 2006 (UTC)