Science desk
< November 16	<< Oct \| November \| Dec >>	November 18 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 17[edit]

Just asked for a clarification at Uncoupling_protein[edit]

Please see. Ben-Natan (talk) 08:17, 17 November 2014 (UTC)[reply]

The articles on the individual UCPs listed there each have a brief (or not-so-brief) description of the process. Some are copy-paste wording, others have more/different details. At a minimum, the set of articles should be refactored to centralize the description of this process on its own page. DMacks (talk) 08:28, 17 November 2014 (UTC)[reply]

Soapy water[edit]

What actually happens to the soapy water that gets on the bathroom floor when people are washing their hands. Does the soap component just evaporate with water? Clover345 (talk) 20:14, 17 November 2014 (UTC)[reply]

No. The water evaporates away and the soap is left behind, often leaving a greasy or grimy feel. Same concept as soap scum. Justin15w (talk) 21:21, 17 November 2014 (UTC)[reply]

The residue might eventually get cleaned up by people's socks, and then end up going down the drain from the washing machine. StuRat (talk) 22:47, 17 November 2014 (UTC)[reply]

Gravitation / Relativity / Cosmology[edit]

Since the OP has taken to deleting responses to his/her questions, I am closing this discussion. AndyTheGrump (talk) 17:24, 18 November 2014 (UTC)[reply]

The following discussion has been closed. Please do not modify it.

Hello,

If we compare times between altitudes, we see a gap time. (General Relativity)

Why don't we see differences in the pictures of the skies ?

Thank you for your answers. — Preceding unsigned comment added by 178.194.81.188 (talk • contribs)

The differences are very *very* tiny...a truly microscopic fraction of a second. SteveBaker (talk) 21:13, 17 November 2014 (UTC)[reply]

Should there not be an accumulation of time ?

Please be more specific as to what accumulation of time you expect. As Steve Baker said, even over a very long period of time, the time dilation effect of general relativity can only be measured with extremely precise and sophisticated equipment. I think that even if you do the arithmetic for an accumulation of time, it will be too small to observe except with equipment designed for the purpose of observing the effect. Robert McClenon (talk) 21:35, 17 November 2014 (UTC)[reply]

A stone on a mountain where we can find a telescope stays there since long time. We see the same thing as the stone. So there should be an accumulation of something relative to a billion of years. No ?

A clock on top of a 5 km mountain will move roughly 2 parts in 10¹² faster than at sea level. In a billion years, it would accumulate about 5 hours of extra time. However, your questions suggest that you are imagining that if you climb to the top of such a mountain then you are moving through time, in the sense that the you might suddenly be seeing a world as it was 5 hours earlier or later. That's not how general relativity works. The world is still the same world, but how you would perceive and measure the passage of time would differ (very, very slightly) depending on where you were located. Dragons flight (talk) 21:56, 17 November 2014 (UTC)[reply]

So your glock don't work as you see, or you don't have any time delay.

(Sorry. I am not moving to the mountain. I wait on the mountain and then come back. Be more precise)

Didn't someone raise pretty much this same question, a month or two ago? ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:19, 17 November 2014 (UTC)[reply]

I am talking about gravitation not time travel. Or is the question the same ? If yes, please show me.

Let's frame the answer a little differently. If you are on top of a 5 km mountain you will see the light from distant stars slightly sooner than someone at sea level, solely because you are 5 km closer to the stars. The rough amount is 16 microseconds (5 km / speed of light), and that amount does not change or accumulate depending on how long you stay on the mountain. It only depends on the fact that the mountain is slightly closer to the stars. If you go up to the observatory on top of the mountain, you will also experience time dilation. Relative to someone on the surface, your clocks would appear to move slightly faster than those on the surface and if it could be measured you would seem to age slightly faster. The difference in your clocks would be ~50 nanoseconds per day spent on top of the mountain. That clock difference would accumulate the longer you spend on the mountain, though in human terms it wouldn't amount to much even if you spent a whole lifetime there. Dragons flight (talk) 22:38, 17 November 2014 (UTC)[reply]

""If you are on top of a 5 km mountain you will see the light from distant stars slightly sooner than someone at sea level, solely because you are 5 km closer to the stars. The rough amount is 16 microseconds (5 km / speed of light), and that amount does not change or accumulate depending on how long you stay on the mountain. It only depends on the fact that the mountain is slightly closer to the stars."" I don't think it has something to do with the problem

Think about a human that could have 1 billion years (stone) speaking to another one who just came at the end for 1 second. If I am not wrong there should be an accumulation BETWEEN two beams of stars. 50 nano between the beam at 0.00 and this at 24.00

I'm having trouble understanding what you are trying to suggest here. However, there is no gap or accumulation of events. Any event occurring in the sky will be seen by both the observers at nearly the same time, differing by only a small amount because the light takes a little bit longer to reach the ground. The observers will disagree on how long each event took, because they have clocks moving at slightly different rates, but they won't disagree about how many events occurred. The hypothetical rock on top of the mountain sees the same number of sunrises and sunsets as a rock on the ground, but he thinks each day is very slightly longer. Dragons flight (talk) 23:12, 17 November 2014 (UTC)[reply]

 person A become at 0.00 the Beam 1    (Altitude between A and B: 5km)
 B 0.00 Beam 1

 A 24.00 Beam 2
 B 23.00 (glock A) Beam 2

 A see the stars at 24.00
 B see the stars at 23.00

 Difference between them is not distance but time stayed in gravitation.

 That is what says general relativity (1.00 = 50 nano).
 Then repeat it 1 billion years.
 Gap between stars 5 years.

A and B always see the same stars. However, they perceive those stars as appearing at different times. If A were to call B on a telephone and discuss what they see, they would always agree about what stars are in the sky. They disagree about what time it is when they see those stars. Suppose A calls up B and says: "My current time is 24:00 and I just saw a shooting star." B might reply: "I also just saw the same shooting star, but my current time is 23:00." They agree on what is happening in the sky "right now" but disagree about what time "now" is. There is a gap only in the sense that A's clock / calendar falls out of sync with B's; however, there is no gap in terms of what events they are experiencing. Dragons flight (talk) 00:00, 18 November 2014 (UTC)[reply]

Once again you speak about the distance between them (special relativity). Try if they see a second shooting star and compare the times (with a difference of gravitation between A and B).

Say me if you don't want to answer me anymore.

The gravity has been the same for both beams of light throughout 99.9999... percent of their journey through space. It is only in the last miniscule fraction of a second that the A beam travels through stronger gravity, and this is not enough to make a measurable difference (with current technology). I'm not sure whether this is what your question was about. Dbfirs 01:17, 18 November 2014 (UTC)[reply]

I don't speak about the gravity of the beams but the gravity area of A and B. A and B are people at different altitudes.

 To resume. Someone on a mountain who is 1 billion x 365 days old has 5 x 365 days more than someone (1 billion years old) who is 5km
 below. The difference of nights should be 5 years. The sky from one should be 5 years later than the other one.

Ok I begin to grasp. If time and day are sychronysed, how can someone become older than the other ?

Hey guys, relativity means relativity (it means no time synchronization but referential synchronization). Has anyone read Einstein's work ?

Small exercise

 High gravity on A, low gravity on B.

 :*Synchro0 : 00h00 on the two glocks.
 :*Synchro1 : 01h00 on glock A 00h59 on B.
 :*Synchro2 : sunrise, 02h00 on A 01h58 on B.
 :*Synchro3 : sunset, 06h00 on A 05h54 on B.

 The referential was A, now lets with B:
 :*Synchro0 : 00h00 on the two glocks.
 :*Synchro1 : 01h01 on glock A 01h00 on B.
 :*Synchro2 : sunrise, 02h02 on A 02h00 on B.
 :*Synchro3 : sunset, 06h06 on A 06h00 on B.

 B has to compare data with A at 6h00 for A. It means 5h54 for B

Lets do it for a billion of years.

the value of a 0.022K capacitor?[edit]

i have got non polarized polyester capacitior which has "0.022K400V" printed on it. what is the value of cap in uF? — Preceding unsigned comment added by 223.223.145.15 (talk • contribs) 21:20, 17 November 2014

That's a 0.022 µF capacitor rated for 400V. The letter "K" denotes 10% tolerance. I see that the wiki page on cap markings is pretty slim. See this page for an explanation of component markings. Mihaister (talk) 21:31, 17 November 2014 (UTC)[reply]

Properties of PE-HD (High-density polyethylene)[edit]

I have carrying cases of this material which came with cordless drills. The drills are now obsolete or worn out and the batteries are no longer taking a charge.

But the cases are like new, aside from a bit of dust. It is too bad the tools were not as well-made as the cases.

I would like to reuse the cases to carry other things. Each side of the case has an outer shell and an inner "insert", molded to fit the original contents. Insert and outer shell are strongly joined together.

If it was possible to soften the insert with heat, it could be reshaped to fit other contents.

The question is, can this material be judiciously softened with heat using for example a small propane torch without it catching fire or giving off noxious fumes? Or to put it another way, is the temperature at which it would soften and become malleable far enough below the point at which it would burn, as to make my project possible?

I think "it must be", otherwise how do people manage to make these cases in the first place. But before trying the experiment, I thought I would ask. Thank you, CBHA (talk) 21:39, 17 November 2014 (UTC)[reply]

It's entirely possible they made noxious fumes when formed, polluting the entire area surrounding the factory in China or wherever it was made.

Why not buy another drill, without a case, to match your case ? StuRat (talk) 22:51, 17 November 2014 (UTC)[reply]

I would not use a torch, i'm fairly certain the heat required is far below flame torch. Probably a heat gun would do it. There's loads of articles online about it, here's one that might be useful. Vespine (talk) 23:23, 17 November 2014 (UTC)[reply]

There are quite a few home-made vacuum forming machines out there (Google "home made vacuum former") - one design uses an office waste-bin, a piece of perf-board and a shop-vac and can be put together with duct-tape.

If you use plastic cut from gallon milk containers, you can soften it with a heat-gun. Make a hole in the side of the trash can and duct-tape the shop-vac hose to it. Duct tape a piece of perf-board over the top of the trash can, place the object you want to make a case for over that, perhaps drap a piece of cloth over it to protect it. Then heat the plastic until it starts to 'sag' and fire up the shop-vac.

Obviously, this is a lot of cost/hassle if you only need to do it once - and it does take a little practice. But being able to do vac-forming at home has MANY uses...and you're using recycled plastic...so you can feel good about doing it!

SteveBaker (talk) 23:49, 17 November 2014 (UTC)[reply]

So plastic milk containers (in USA at least) are also HDPE, right? And then it seems you are recommending cutting out the molded plastic inner layer and molding a new plastic insert. That makes sense if we need a close fit, but if OP is willing to cut out the inner molded surface, then the shell could also be fitted with any manner of glued-in foam or plastic inserts that don't have to be custom molded, and could fit a variety of things. Personally, assuming a close fit is desired and the objects are of similar shape to the originals, I'd go at the interior with a heat gun in a well-ventilated area, and then move on the to the other options if that fails. Since the milk container is far thinner than the drill case, I'd worry that softening the inner layer would take a rather long time, and might melt it to the outer shell. SemanticMantis (talk) 00:24, 18 November 2014 (UTC)[reply]

Yes, they should have a recycle-2 symbol on them...you might want to check that before you try this just in case you have some weird milk containers in your part of the world. The nice thing about the gallon milk containers is that you can cut around them with scissors and unwrap a piece of HDPE that's about 18" long and 8" wide and comes out more or less flat and with a fairly uniform thickness - which is great for vacuum forming. SteveBaker (talk) 16:58, 18 November 2014 (UTC)[reply]

This isn't what you asked, but you can usually open up the battery back, and replace the cells with off the shelf rechargeable AA batteries. Ariel. (talk) 23:52, 17 November 2014 (UTC)[reply]

This is interesting, do you have any refs for how/why/in what types of battery packs that might work? (I happen to also have a drill with a similar case and two low-performing battery packs :) SemanticMantis (talk) 00:24, 18 November 2014 (UTC)[reply]

The battery packs (except Li-ion) are usually made of ordinary AA (and sometimes AAA) batteries with no name on them and sometimes a solder tab. Open up the pack and take a look. Just trace the connections - most often it's simply all the batteries in series. Then get new batteries and solder them together in the same pattern and put them back in the pack. Soldering a battery is hard (it's a large heat sink, and you have to avoid heating the battery much), but it's possible. Google for instructions on how, a soldering gun is easier than a soldering iron. Ariel. (talk) 00:50, 18 November 2014 (UTC)[reply]

In regards to safety concerns, here's a MSDS for HDPE: [1] -- looks pretty safe to my (inexpert) estimation. SemanticMantis (talk) 00:28, 18 November 2014 (UTC)[reply]

voltage divider looking to the load like two resistors in parallel[edit]

hello, on this page here it reads (among other things): "As you can see, this puts 11 mV of DC across the load. If the load is 32 Ω at DC (...), 0.34 mA is forced through the load. This current can only come from the rail splitter, which looks like two parallel resistors to the load."
I just don't get how the two resistors can look as though they were in parallel, to anything. Do they because the two rails are the same point electrically (are they?) due to the (ideal) battery's internal resistance of 0?
Generally, what is the "algorithm", if there is one, for how to arrive from the original circuit to this equivalent circuit (3rd box from the top here)? Thank you very much Asmrulz (talk) 23:19, 17 November 2014 (UTC)[reply]

The "algorithm" for converting a circuit is Thévenin's theorem or Norton's theorem, depending on your favorite representation. An even more generalized representation is to treat the circuit as a two-port network and model the impedance as viewed at any two points on the circuit net. In practice, the process of formalizing these representations means application of network analysis techniques. If you're very practical, you can concoct a lot of "rules of thumb"; if you're very theoretical - or if your hobby is to write electrical CAD software - you can apply techniques of linear algebra (like row reduction, Gaussian elimination, or the Gram–Schmidt process) to re-represent a linear network.

Two resistors in parallel "look like" one single resistor; by the commutative property, one resistor also can also be said to "look like" two resistors in parallel. Nimur (talk) 23:47, 17 November 2014 (UTC)[reply]

There's a voltage divider that biases the opamp (it's a single-supply design) in the schematic. The two 4k7 resistors are in series, but in the paragraph discussing the DC equivalent circuit the author says that to the load, they look as being parallel, thus Rsplit=2.35k. That's the bit I don't get....and many others, but this one intrigues me most Asmrulz (talk) 00:37, 18 November 2014 (UTC)[reply]

It's superposition. There is a 9V battery across two 4k7 resistors in series, so a certain current flows through them. The virtual ground of the opamp is connected to the midpoint, so it should be at 4.5V. However, as the article discusses, the opamp adjusts its output voltage until 0.34mA flows through the load and into the opamp. That current can be thought of as due to a superposition of the current normally in the 4k7 resistors combined with a current change in each resistor. The top resistor has 0.17mA extra current flowing down, and the bottom resistor has 0.17mA extra current flowing up (that is, 0.17mA less total current in the bottom resistor). That gives a 4k7 × 0.17mA = 0.76V voltage drop at the midpoint, which is the same conclusion as the article (effective power rails of +3.7V and −5.3V). Johnuniq (talk) 02:35, 18 November 2014 (UTC)[reply]

What the hell is "4k7?" It is a nomenclature that may be unfamiliar to some readers. Is it a British or European standard? Is there a different US standard (ANSI or MIL) or is it now official in the US? Does the Wikipedia manual of style prefer that as opposed to "4.7 k ohm"? Edison (talk) 13:51, 18 November 2014 (UTC)[reply]

It's a pretty common convention for component values - I've seen it on both sides of the Atlantic for at least 30 years. When you're writing something like 4.7k quickly, it's easy for the '.' to get lost - and because (especially) resistor values are chosen to be available in decades (4.7k, 47k, 470k). Putting the 'k' where the decimal point would be makes things much less error-prone. SteveBaker (talk) 16:49, 18 November 2014 (UTC)[reply]

...and the reason that the op-amp adjusts its output in this way is because of a non-ideal behavior, the input offset voltage. Nimur (talk) 03:24, 18 November 2014 (UTC)[reply]

Oops, looking again shows I got the current directions reversed. Each direction I mentioned should be inverted to give a 0.76V rise. Johnuniq (talk) 06:19, 18 November 2014 (UTC)[reply]