Wikipedia:Reference desk/Archives/Science/2011 February 23

= February 23 =

An interesting physics problem
Dear Wikipedians:

I am working on this very interesting physics problem:

A spring attached to a ceiling has mass of 500 g suspended from it such that the spring stretches 4.0 cm. Calculate the spring constant.

I have taken two approaches to this question, but they seem to yield different answers!

The first, dynamic approach:

kx = mg

k = mg / x = 0.5*9.8/0.04 = 122.5 N/m

The second, energy approach:

½kx² = mgx

k = 2mg / x = 2*0.5*9.8/0.04 = 245 N/m

I find it very mind-boggling: how can there be two answers???

Thanks,

174.88.35.131 (talk) 02:48, 23 February 2011 (UTC)
 * Somewhat amusing. You might like to consider the spring deflection a little later, in the second case. In mechanical engineering this is why 'suddenly applied loads' are multiplied by two, but not many people know that. Greglocock (talk) 02:54, 23 February 2011 (UTC)


 * The first approach is straightforward Hooke's law - you've calculated the stable applied force mg from the weight, and set the spring to be the right length to be supporting it. But the second approach assumes that the spring is extended solely because the weight was dropped from the point where the spring's length is zero - which isn't necessarily true.  It assumes that the kinetic energy is zero, but this is true when the spring is maximally extended and about to rebound.  In any case, there are two answers because the spring can be extended to that distance for more than one reason - it could be because it is hanging at a constant distance, or because it is bouncing up and down after having been released. Wnt (talk) 03:31, 23 February 2011 (UTC)


 * I agree with Wnt. Considering the wording of your question, your first equation is correct and 122.5 N.m-1 is the answer.  You have called this a dynamic approach but it would be better described as a static approach.  You have called your second equation the energy approach and this is reasonable.  Alternatively it could be described as a dynamic approach.  Dolphin  ( t ) 06:40, 23 February 2011 (UTC)


 * You can solve the equilibrium problem using energy by equating the first derivative of ½kx² − mgx to zero, since an equilibrium configuration is a local minimum of the potential energy. That gives you kx = mg. What you did is equate ½kx² − mgx itself to zero. Since energy is only defined up to an arbitrary additive constant, that will give you different answers depending on, in this case, the point you happened to choose as the origin of x. -- BenRG (talk) 10:08, 23 February 2011 (UTC)

Basically, you have a system in equilibrium. With springs, being in equilibrium and being at rest are two different things. If the body was drooped from the unextended state as above, the spring will reach an extreme position, where energy is zero and it is temporarily at rest. But, the spring force still acts on it. Thus, this is not your equilibrium position. Springs have a tendency to oscillate in simple harmonic motion, so you have to look carefully at the wording. Manish Earth Talk • Stalk 10:31, 23 February 2011 (UTC)

Thanks all for your response! Now I fully understand the difference between the two approaches! 174.88.35.131 (talk) 22:12, 23 February 2011 (UTC)

Pepper seed
I have heard that pepper seed (from bell peppers and hotter varities alike) are mildly poisonous and shouldn't be eaten. This true? 212.68.15.66 (talk) 06:55, 23 February 2011 (UTC)


 * They're not poisonous. However, chefs do tend to remove them, because they're a bit bitter when bitten into.    Sophus Bie  (talk) 10:36, 23 February 2011 (UTC)


 * The seeds of a pepper also tend to be spicier than the flesh. While pepper seeds are not 'poisonous' in any meaningful real-world sense, note that capsaicin is technically toxic at high doses. See e.g Capsaicin. However, there's no real reason to avoid eating pepper seeds; I use them in things like gumbo. SemanticMantis (talk) 15:46, 23 February 2011 (UTC)


 * You may be thinking of apple pips or peach pits, which contain cyanide, though poisoning yourself with those would involve eating the seeds of hundreds of fruits. The article about apple seed oil says it smells of almonds (a giveaway that it contains cyanide) and says vaguely that you shouldn't consume too much of it (however much that is). One food you could feasibly poison yourself with is the bitter almond, but for this reason they don't get much use as an ingredient. 213.122.19.152 (talk) 17:47, 23 February 2011 (UTC)

Keeping soda carbonated longer
I have a 2 liter bottle of regular soda. After opening it, it gets flatter with each day (or rather, with each time) that I open the bottle. Would squeezing the bottle to the point where the liquid was near the top of the spout, and then sealing the bottle closed with the cap, help slow down the rate at which it gets flat? Or would the bottle still expand as the carbonation leaked from the liquid, and this would just be futile? I refrigerate the soda before and after each opening. Also, I did check the archives for similar questions and didn't find anything about this. –  Ker αun oςc op ia◁ galaxies  08:56, 23 February 2011 (UTC)


 * No. By squeezing out the air and hence lowering the pressure above the drink, you encourage the CO2 to come out of the drink. If you wanted it to stay carbonated it would be better to pump CO2 (or just air) into the bottle.--Shantavira|feed me 09:07, 23 February 2011 (UTC)


 * Ah, that makes perfect sense!! Excuse me while I go fix my bottle lol. –  Ker αun oςc op ia◁ galaxies  09:19, 23 February 2011 (UTC)


 * So here I think we landed on a million dollar idea, but there are a half-dozen variations on this invention already (fizz-keeper bottle caps). So much for quitting my day job. –  Ker αun oςc op ia◁ galaxies  09:24, 23 February 2011 (UTC)
 * Fizz keeper caps, appear effective just hours not days. Making your own in a soda siphon would be better. See Sodastream--Aspro (talk) 09:45, 23 February 2011 (UTC)


 * The fizz keeper article consists of an obsessively detailed answer to this vexing question. The best answer is to buy smaller bottles.--Shantavira|feed me 10:09, 23 February 2011 (UTC)


 * These answers are great, thank you! –  Ker αun oςc op ia◁ galaxies  10:42, 23 February 2011 (UTC)


 * Why would squeezing the bottle significantly affect the pressure of air above the drink? You aren't pumping out air while keeping the volume the same, you are reducing the volume of air keeping it at atmospheric pressure. I believe this would actually have a slight improvement although as with fizz keeper it may not be significant. A slightly better solution would be to pour the drink in to smaller bottles, this way the volume of air is reduced and you are only opening one so the others aren't affected every time you open one. Of course buying smaller bottles in the first place is best but that tends to be expensive. Nil Einne (talk) 14:20, 23 February 2011 (UTC)
 * It depends on the bottle. If the plastic resists deformation and exerts a force to rebound to its original shape after it is sealed, then it will generate a (admittedly very weak) partial vacuum above the liquid as the bottle itself counters some of the local atmospheric pressure.  The bottle will tend to expand until it returns to its original volume; the entire headspace in the bottle will then be filled by carbon dioxide lost from the liquid solution.
 * The amount of carbon dioxide that ends up in this headspace is an interesting question. In principle, the final partial pressure of carbon dioxide above the soda should depend solely on the concentration of CO2 in the liquid (Henry's law), so squeezing the air out shouldn't make a difference; in practice, I imagine that there is at least a small departure from ideality which will depend on the partial pressure of air already present.  The best ways to reduce the amount of CO2 lost are (as Nil Einne says) to reduce the amount of headspace in the bottle by transfer to smaller containers, and to keep the temperature of the liquid low when the bottle is open (increasing the solubility of carbon dioxide and decreasing its rate of loss from solution).  TenOfAllTrades(talk) 14:59, 23 February 2011 (UTC)


 * Unfortunately, though, pouring the carbonated drink from the larger bottle into smaller bottles would increase the loss of CO2 during that very action, probably making the action futile. However, being that I currently reside at high altitude (9,000+ feet), I wonder if the expansion of the air in the headspace during shipment would help prevent more loss of CO2 than for a similar bottle at sea level? When I open my bottle, am I starting off with a higher amount of carbonation than a sea-level bottle would have? –  Ker αun oςc op ia◁ galaxies  22:18, 23 February 2011 (UTC)


 * 9000+ feet ? Where's that ?  Do you need an oxygen mask ? StuRat (talk) 23:24, 23 February 2011 (UTC)
 * I'm in the American West, so I guess that makes me some sort of space cowboy? – Ker αun oςc op ia◁ galaxies  00:43, 24 February 2011 (UTC)


 * I'd just stop buying 2 liter bottles. They really are only good for parties.  For individual consumption, you need something smaller. StuRat (talk) 23:25, 23 February 2011 (UTC)


 * But a generic brand of 2 liter soda is 1 cent per ounce, far cheaper than cans (of any brand!). –  Ker αun oςc op ia◁ galaxies  00:43, 24 February 2011 (UTC)


 * This conversation surprises me. I remember that 20-30 years ago, soda pop seemed like it got flat easily, but then it stopped seeming to do so.  I'd assumed that they'd figured out some technical trick to improve the CO2 solubility or something - nowadays, it seems like even open cans of soda seem reasonably fizzy after being left a day, and two liter bottles can be opened many times without trouble.  Or have I just lost the ability to taste the "flatness"? Wnt (talk) 23:34, 23 February 2011 (UTC)


 * I don't know the science behind it, but I do know I can put an open glass of coke in the fridge at night and the next day it's still bubbly. –  Ker αun oςc op ia◁ galaxies  00:43, 24 February 2011 (UTC)


 * Squeezing plastic bottles does make the carbonation last longer. I do this all the time.  My guess is that once you close the cap, because there is a smaller volume of air, less CO2 is drawn out of the pop each time to achieve equilibrium, so there's less gas available to escape into the atmosphere during the time you have it open again (hmmm ... partial pressure doesn't do a good job of explaining the phenomenon). Clarityfiend (talk) 02:04, 24 February 2011 (UTC)


 * No disrespect, but I don't think that's right. I'll go this far though - squeezing the air out before closing will make the bottle have that satisfying PSHT! sound when you crack it open again. But for the opposite reason - now the headspace is at lower than normal pressure rather than greater than normal pressure. When you deform the bottle like that, the plastic will attempt to "right" itself, lowering the pressure of the headroom above the beverage. The effect is similar to boiling water at great elevations - it takes hardly any heat at all to make water boil on Mt Everest, for example. Will that be enough to reduce the amount of CO2 in the soda as some here are claiming? I don't know enough math to say, but your claim that there being less volume of air to bring to equilibrium strikes me as incorrect on the face of it; whether there's 100mL of headspace or 200mL of headspace, only a tiny fraction will be CO2. I don't think equilibrium plays a part (but would be happy to be proven wrong). Matt Deres (talk) 21:04, 24 February 2011 (UTC)


 * I don't think the issue is how much CO2 is in the air at the beginning is the issue but how much is in it at the end? Nil Einne (talk) 18:39, 27 February 2011 (UTC)

Buy one of those computer dusters that shoots CO2 and shoot it into the bottle right befor closing it. this will saturate the volume above the soda with CO2. —Preceding unsigned comment added by 165.212.189.187 (talk) 13:51, 24 February 2011 (UTC)


 * This is a start, but pressure would still need to build up to a higher than atmospheric level. Now if you really want to fix the problem, you need a good supply of dry ice... ;) Wnt (talk) 17:20, 24 February 2011 (UTC)

Stripping all electrons from metal
Has anyone ever taken a very small peice of metal (such as copper) and applied enough electric force to pull all electrons from it? Would this make the metal invisible? Would the metal still stay as a solid? - Anon. —Preceding unsigned comment added by 114.77.191.84 (talk) 09:02, 23 February 2011 (UTC)


 * Any macroscopic piece of metal would explode due to the repulsion between the remaining positively charged nuclei. If you have just a single nucleus, you just get a highly ionized atom which is invisible anyway by virtue of its extremely small size. 157.193.175.207 (talk) 09:30, 23 February 2011 (UTC)


 * Yeah. You first keep in mind that the nature of being a metal requires the presence of electrons (see sea of electrons). Then just think of the forces involved. If you remove just 1% of the electrons from each of two humans standing one meter apart, the force between them would be strong enough propell the planet Earth at many times g. And the force increases asymptotically toward infinity as the distance decreases, and my two human subjects were at 1 meter. A metal would explode with a relatively tiny portion of electrons removed. Someguy1221 (talk) 10:37, 23 February 2011 (UTC)


 * Indeed; the energy cost of stripping all the electrons from an atom (of, say, copper) is much, much, much greater than the binding energy that holds a single copper atom in place in a lump of metal. As you get down to tightly-bound inner-shell electrons, it becomes progressively more difficult to remove them from the correspondingly less-shielded nucleus.  Check out Ionization energies of the elements (data page) for some numbers.  Consider a neutral copper atom with 29 electrons.  Removing the first one costs 7.7 eV; removing the twenty-ninth electron (from Cu28+) costs 11568 eV.   While it is possible to make fully-stripped nuclei, it's something that's done to single nuclei in particle accelerators, not something achievable in bulk materials. TenOfAllTrades(talk) 14:41, 23 February 2011 (UTC)
 * Metal without electrons is in a plasma state, see article. Cuddlyable3 (talk) 14:47, 23 February 2011 (UTC)
 * In a plasma state, a metal is no longer a metal (it loses its sea of electrons). Manish Earth Talk •  Stalk 12:33, 24 February 2011 (UTC)
 * Somehow overcoming the repulsive forces and forcing those metal atoms into an extremely confined space, as for example in core collapse supernovae, produces degenerate matter. ~ A H  1 (TCU) 20:59, 26 February 2011 (UTC)

Effect of atmospheric pressure on reaction?
Our Chemistry class came along a question in a textbook regarding Le Chatelier's Principle, stating roughly: NO2(g) reacts with itself: 2NO2(g) «» N2O4(g) + heat. As the reaction proceeds to the right, 2 moles of gas form 1 mole of gas, so in a fixed volume, pressure would drop as the reaction proceeds to the right. If the total gas pressure is decreased, would the reaction proceed to the right or left?

We cannot decide whether the reaction would shift to the left, favouring more gas, or not shift at all, as a gas fills the container regardless of pressure :/

Thanks! 110.175.208.144 (talk) 10:26, 23 February 2011 (UTC)
 * For background reading, Wikipedia has an article about Nitrogen dioxide NO2 that mentions its existence in equilibrium with Dinitrogen tetroxide N2O4, called a dissociating gas. Le Chatelier's principle says If a chemical system at equilibrium experiences a change in concentration, temperature, volume, or partial pressure, then the equilibrium shifts to counteract the imposed change and a new equilibrium is established. If equilibrium is reached and the container volume is then increased, the equilibrium swings to the left (more gas molecules) to counteract. Cuddlyable3 (talk) 14:14, 23 February 2011 (UTC)
 * I had read the above articles, but it wasn't quite clear exactly what would happen if pressure decreased... thanks for your answer! 110.175.208.144 (talk) 20:14, 23 February 2011 (UTC)
 * Sometimes equilibrium reactions are easier to see one way, and then just recognize that because it's an equiliubrium, it goes the other way in opposite circumstances. For example, it makes sense to me that when you increase the pressure, pairs of molecules combine to make single molecules (in essence, absorbing that added pressure). So "therefore, reducing pressure allows or promotes them to split apart, as the opposite of what happens when the pressure is increased". Not always, but often, one direction is an actual effect, and the other is the absence of the effect (sometimes it sounds weird to say you are "pulling them apart" in this case, because there still is some pressure). Maybe useful, maybe more confusing if you already understand it. DMacks (talk) 22:58, 23 February 2011 (UTC)


 * Since it's an equilibrium reaction, once you progress the reaction one way, there is a counteracting force to move it the other way, so all things being equal the reaction would then return to the left (as the gas is at equilibrium to begin with). If the gas isn't at equilibrium to start, a great help will always be the ideal gas law, but at equilibrium you don't have to say "well, pressure drops so N and/or T must drop, so the reaction goes rightward" because any shift in any of the variables is immediately counteracted: equilibrium means everything stays the same over some small time period (and the equilibrium "reactions" are temporary fluctuations). SamuelRiv (talk) 21:16, 24 February 2011 (UTC)

Electrical Circuit Analysis
Why we use resistor, capacitor, Inductors in the electrical circuits what actually these passive elements doing in the circuits and how to choose those ranges i don't know why when where which what element should be used ? ex-: http://www.circuitstoday.com/wp-content/uploads/2008/03/max2606_single_chip_transmitter.JPG they used 5 capacitors and 8 resistors and one inductor how they know those elements and their values — Preceding unsigned comment added by Kanniyappan (talk • contribs) 13:06, 23 February 2011 (UTC)
 * Most of the circuit complexity is inside the integrated circuit MAX2606 that is connected to external components (resistors, capacitors and an inductor) that are recommended by the manufacturer. You will find a description in the MAX2606 data sheet. Here are simple articles about components: resistor, capacitor and [inductor].Cuddlyable3 (talk) 14:38, 23 February 2011 (UTC)
 * As a young and naive reader of science articles, I learned what the individual components did (caps store and release charge, resistors oppose the flow of electricity, etc) and tried miserably to understand a diagram of a radio, in terms of "The antenna brings voltage and current to the capacitor, which stores then releases charge". That analysis was flawed, because the components act in combination to do certain standard functions (a tuning circuit, an RC filter, a blocking cap, a dropping resistor, a transformer). It helps to understand what larger building blocks do. Edison (talk) 17:03, 23 February 2011 (UTC)
 * May I ask (at the risk of committing the sin of chatting), did you gain this understanding in a way compatible with the first principles (about capacitors, etc.), telling yourself things like "a flip-flop works because the first transistor ..." or did you let go of the need to make this conceptual connection, and simply accept that the standard circuits do what they do? I would find such letting-go incredibly difficult, and the thought occurs that maybe this is why I have never got beyond the stage of trying miserably to understand circuit diagrams in a reductionist way. 213.122.19.152 (talk) 17:58, 23 February 2011 (UTC)
 * I don't know this stuff, but I'll take a stab at it for fun. Let me know if I'm wrong.
 * the 0.47 uF capacitor blocks a net current flow from the input into the chip. It's a fairly big capacitor, and it allows sinusoidal current flows, but (with resistor) it should block very low frequency sound from getting in (high pass filter).
 * the 2200 pF capacitor should cancel out very short-term changes in current going into TUNE and thus (with resistor) act as a low pass filter.
 * the largest 10 uF capacitor should compensate for changes in the power supply - I assume it helps prevent the chip from being damaged when you fool around with the battery.
 * the large variable resistor R1 controls the voltage going into TUNE, except for short-term oscillations coming from the audio input. A volume control, in other words. Doh, it's FM!  Note that its value controls how much power is being wasted whenever the circuit is on.

I think I'd better stop and see if someone corrects me... ;) Wnt (talk) 18:46, 23 February 2011 (UTC)


 * Potentiometer R1 tunes the transmitter frequency by sending a steady voltage into pin 3 of the integrated circuit. It is NOT the volume control.
 * Potentiometer R2 varies the strength of audio (an alternating current) signal into pin 3, causing the frequency to be modulated by the audio. It's an FM transmitter and this is its volume control.
 * The two audio inputs left+right should not make you think this transmits a stereo signal. The two inputs get combined at R2 and transmitted in mono..
 * The 0.47 uF capacitor allows an audio signal in but blocks the voltage from R1 from leaking away
 * The 10 uF capacitor with the 4.7k resistor form a low-pass filter that prevents any ripple on the VCC supply adding noise to the sound. If VCC comes from a battery they can probably be omitted. However the VCC voltage has to be steady otherwise R1 will need continual adjustment to stay on frequency.
 * Note that the 10uF and 1uF capacitors are electrolytic types that have to be connected a particular way round shown by the + symbols. Cuddlyable3 (talk) 21:24, 23 February 2011 (UTC)


 * Hey, thanks - I knew this wasn't my thing, but at least I got someone to answer. ;) Wnt (talk) 23:17, 23 February 2011 (UTC)