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February 12

sodium chlorate

what are the specifics of making sodium chlorate? or sodium perchlorate? —Preceding unsigned comment added by Steweydewey (talk • contribs) 01:20, 12 February 2008 (UTC)[reply]

Our articles on sodium chlorate and sodium perchlorate describe the synthesis of each. Someguy1221 (talk) 01:22, 12 February 2008 (UTC)[reply]

chemistry

magnesium carbonate is added to an added to an iron binding capacity determination in order to a. allow color to develope b. precipiate protein c.bind with hemoglobiniron d. remove excess unbound iron —Preceding unsigned comment added by 72.155.150.73 (talk) 01:38, 12 February 2008 (UTC)[reply]

Yes, quite probably one of a, b, c, d is the reason (or maybe a combination of those?). DMacks (talk) 01:40, 12 February 2008 (UTC)[reply]

A search eg http://www.google.co.uk/search?q=iron+binding+capacity+determination+magnesium&hl=en&start=10&sa=N suggests it may be 'd'. Who that helps if anyone?87.102.81.140 (talk) 12:55, 12 February 2008 (UTC)[reply]

Decrease of Earth’s orbital angular momentum (around the Sun)?

The motion impulse is sustained in case of zero gravity only as a straight line (?) If a turn or curve takes place this will lead to a reduction of the motion impulse. This applied to the motion impulse of the Earth would imply that the velocity of the Earth around the Sun is decreased. This because the gravitational impact of the Sun is such that the straight trajectory of Earth is curved. That is my question respectively my statement. This implies for Earth’s orbit around the Sun, that the radius is decreased. On the other hand leads the Sun’s burning to a decrease in mass and thus also to a decrease in gravitation. And that leads to an increase of Earth’s orbit. The total effect of these two partial effects remains open. Is this sound reasoning? Or can somebody explain me why the orbital angular momentum of the Earth should remain the same during all these millions of years (despite the gravitational impact of the Sun)? 77.57.61.59 (talk) 01:49, 12 February 2008 (UTC)[reply]

The Earth is in freefall. It doesn't require any energy input and can go on forever. --Sean 02:42, 12 February 2008 (UTC)[reply]

Additionally, the Sun's loss of mass, while existent, is quite negligible. For instance, if the Sun were to completely exhausts its hydrogen supply, it would have lost about 0.7% of its mass. In practice, the Sun won't come anywhere near fully consuming its mass of hydrogen (most of it remains well outside the core where fusion takes place), so something like a tenth of a percent is a better estimate for the mass lost in such fashion. I would expect that solar wind loss is similarly small. — Lomn 14:11, 12 February 2008 (UTC)[reply]

Using a 200V machine in a 220-240V country.

A machine has these power specs:
3 phase 200V
50/60Hz
1.6kw

Australia is rated at 220-240V.

If I acquire this machine, is it okay to use?
If not, is there a feasible solution?

Phrased properly: What difference in voltage ratings between a machine and a country is acceptable?
Rfwoolf (talk) 03:02, 12 February 2008 (UTC)[reply]

Volunteers on an online encyclopedia are not a suitable source of information for such a question. Consult a licensed professional electrical engineer in your country for advice. That said, the supplied voltage will be connsiderably greater than the rated voltage. This could over stress the insulation and contacts, causing catastrophic failure of the equipment. You might contact the manufacturer to see their recommendations. One remedy would be using a transformer to adapt the machine to the excessive voltage. Another would be to have qualified personnel rewind or otherwise modify it to operate on your voltage. It is surprising that the machine has one rated voltage rather than a range of allowed voltages, since power equipment may see voltage somewhat above or below the "official" supply voltage during actual operation. Edison (talk) 03:20, 12 February 2008 (UTC)[reply]

She'll be fine mate. Give it a whirl.Shniken1 (talk) 03:54, 12 February 2008 (UTC)[reply]

Does it have a motor in it?Tuckerekcut (talk) 04:50, 12 February 2008 (UTC)[reply]

Yes. Rfwoolf (talk) 05:00, 12 February 2008 (UTC)[reply]

Standard residential power is single-phase, not 3 phase. I suspect that the motor you are looking at is what we call here (Canada) 208 volt 3 phase. Such motors will not work using 230V single-phase power (which is what will be available in residential buildings in Australia). This motor will be equipped with three conductors (plus ground) instead of two and there will be no way to wire it. As stated above, we are not a reliable source of information, but I can warn you that connecting 3-phase equipment to a single-phase power supply is by no means safe. —BradV 07:02, 12 February 2008 (UTC)[reply]

Thanks for all the help. I've only just found out about the 3 phase thing, which I did look up, and now I realise that only certain factories are connected to 3-phase, and we wouldn't have that available, so that puts this issue to bed.
However, for my own education, the question still remains, what difference in voltage ratings between a machine and a country is acceptable? What would happen if the above machine was a 1-phase 200V. Could I use it in Australia which has 220-240V?
Rfwoolf (talk) 07:40, 12 February 2008 (UTC)[reply]

This is however an extremely theoretical question since no country has a 200V supply (see Mains power systems) so no one bothers to manufacture equipment for a 200V supply Nil Einne (talk) 09:01, 12 February 2008 (UTC)[reply]

I think you'll find that Japan has areas of the country where industrial power is 200V (and residential power is 100V). This diversity (between the low of 200 V and the high of 240 V) is a recurring challenge for the equipment that my company designs.

All houses in Japan receive 200V. Most machines are 100v so only two wires are used. But there are many machines that are 200V used in houses in Japan. For example, larger air conditioners, cookers, dishwashers, and so on. The government had a plan to convert everything to 200V because of the more efficient energy costs, but this has not happened. All the same, 200V machines are not unusual and easy to install.

Atlant (talk) 13:11, 12 February 2008 (UTC)[reply]

If that device can tolerate such overvoltage, it will work (although it will heat up more)(it also might reduce it's lifespan). If that device cannot tolerate such overvoltage, it will either blow fuse or it will get damaged. -Yyy (talk) 09:57, 12 February 2008 (UTC)[reply]

I didn't answer before because I couldn't get my head around what was meant by a voltage in a three-phase system. I was hoping an Australian EE would answer. Our article "Mains electricity#Voltage levels" says that very often loads for 230V supplies will be marked "200V". --Milkbreath (talk) 11:56, 12 February 2008 (UTC)[reply]

Have you seen Three-phase electric power, Polyphase system and Three-phase? Nil Einne (talk) 15:39, 12 February 2008 (UTC)[reply]

Many large industrial scale electrical devices are built with tapped windings which allow the same machine to be connected to several different voltage levels. The full winding is used when it is connected to the highest rated voltage, and there are taps along the winding to allow it to be connected to somewhat lower voltages. The same motor might be capable of connection to a delta or wye three-phase supply as well. A supplier who has a motor with various taps and connection options needs stock only one motor to handle demands for, say 208 volt or 240 volt supply. Likewise transformers or autotransformers can be used to supply the desired voltage, but this is not a cheap option. If a factory had to install their own transformer because their local utility was supplying them with a higher voltage such as 4,000, 12,000 or 34,000 volts, then it might be quite economical to step the supply voltage down to the quirky voltage required by the electrical device. If a motor is operated on a higher than normal voltage, it may actually draw less current (depending on the type of motor), while it would draw more current if operated at lower than normal voltage. For these motors, low voltage would cause overheating due to high current, but high voltage would stress the insulation rather than overheating the conductors. When only single phase power is available, a phase convertor can sometimes be used to allow the use of a three phase motor. Or a second phase can be brought to the location from the utility's nearest three phase supply point, and the motor could be operated open delta. The most expensive option, and the one with the best results, is to bring in the other two phases. Again, an EE qualified to analyze and design power applications is the best person to advise on this. Edison (talk) 15:24, 12 February 2008 (UTC)[reply]

But you'd still have major problems getting the device to work in Australia though wouldn't you? Since three phase power in Australia is of course 415 V not 208 V... Nil Einne (talk) 15:33, 12 February 2008 (UTC)[reply]

The voltage tolerance will depend on the specific machine. I worked with one that had transformer taps for every common voltage from 100 to 415, plus additional taps around those voltages for better matching. Proper installation required matching voltages to +/- 2.5 volts. (Incidentally, it's also the only machine I've ever used to have water-cooled wires.) --Carnildo (talk) 21:45, 13 February 2008 (UTC)[reply]

Strange light

Hello. A few nights ago, I saw a golden sphere in the sky. At first I thought it was a star, but I realized it was moving. Then the light started to dim, and then it disappeared. I don't know what this was. Does anyone have any ideas? Thanks. JetLover (talk) (Report a mistake) 03:32, 12 February 2008 (UTC)[reply]

How did you know it was a sphere?Shniken1 (talk) 03:52, 12 February 2008 (UTC)[reply]

From my viewpoint, it looked like a sphere. JetLover (talk) (Report a mistake) 03:54, 12 February 2008 (UTC)[reply]

See Ball lightning --S.dedalus (talk) 04:50, 12 February 2008 (UTC)[reply]

Since this appears to be an UFO I would suggest you look at the page and associated external links. Avoid anything which talks about conspiracy, aliens etc seriously. Look for stuff which discuss which try to explain or suggest logicial possibilities for UFO sightings Nil Einne (talk) 16:02, 12 February 2008 (UTC)[reply]

You don't say where you are but a few nights ago, Saturday to be precise, there was an extremely bright sky transit of the International Space Station from west to east at 18.14 hours over southern England. It took about 4 minutes to go from the western horizon to the eastern and it certainly rises as a golden/red colour and sets the same colour but is a bright white overhead. Richard Avery (talk) 09:16, 12 February 2008 (UTC)[reply]

Similar to a question lower down, could it be a satellite flare? They're bright, they move, they dim. -- Consumed Crustacean (talk) 21:47, 12 February 2008 (UTC)[reply]

Hmm, sounds like an iridium flare to me. You can also use Heavens-Above to search for Iridium flares and other satellites, in the future, as well as in the past. Be sure to use your exact city, otherwise the predictions for both the past and future will be off. Hope this helps. Thanks. ~A H 1^(TCU) 22:31, 12 February 2008 (UTC)[reply]

Oh, by the way, sometimes the ISS will also hit a point in the sky, start to dim, then dissapear, as it climbs over the shadow of the Earth. You can use Heavens-above for ISS too. I even wrote a Wikihow article on finding an iridium flare/ISS. Hope this helps. Thanks. ~A H 1^(TCU) 22:34, 12 February 2008 (UTC)[reply]

It probably was a satellite flare. Thanks for the help. JetLover (talk) (Report a mistake) 00:58, 13 February 2008 (UTC)[reply]

I doubt it was an iridium flare as these tend to be fleeting - a second or three, whereas the ISS flare, if it can be called a flare can last for 3 - 4 minutes as it tracks across the sky. Iridium flares are, in my experience, quite difficult to observe due to their unpredictability and dimness. Richard Avery (talk) 07:55, 13 February 2008 (UTC)[reply]

time of twilight

is there a mathematical equation to calculate the amount of monutes from sunset to dusk in different parts of the world. for example in NYC how many minutes after sunset will it be completly dark.Yribowsky (talk) 03:49, 12 February 2008 (UTC)yribowsky[reply]

I doubt it. I think it depends a lot on landscape. AlmostReadytoFly (talk) 09:51, 12 February 2008 (UTC)[reply]

The length of time from sunset until it gets dark - twilight - depends on the latitude and on the declination of the sun. The further north or south one goes from the equator, the longer will be the period of twilight. The declination of the sun depends on the season of the year. Therefore in a particular place, the length of twilight will vary throughout the year. To do a mathematical calculation on the length of twilight requires a knowledge of spherical trigonometry.Simonschaim (talk) 13:26, 12 February 2008 (UTC)[reply]

The Twilight page has several external links to twilight time calculator websites, including Length of Day and Twilight (Formulas), which begins by saying: This article describes, how the length of day can be calculated for any given Northern latitude and any day of year. It also includes calculation of the twilight duration. Pfly (talk) 06:07, 13 February 2008 (UTC)[reply]

Ok, so much for the time bizzo. Now, how do we calculate the number of “monutes” Myles325a (talk) 01:26, 15 February 2008 (UTC)[reply]

Nearest star of spectral class O

What Class O star is nearest to Earth? --Cam (talk) 04:30, 12 February 2008 (UTC)[reply]

Zeta Ophiuchi maybe? 458 light years away. Clarityfiend (talk) 06:11, 12 February 2008 (UTC)[reply]

Thanks! Looking around on the web, it does seem like it's the closest, but I can't find a definitive statement anywhere. --Cam (talk) 02:22, 13 February 2008 (UTC)[reply]

Space?

Watching my usual satellite passing over at 0645 this morning, (in Scotland), I was surprised not to be able to any stars. Since I believe that satellites circle 2-300 miles above the earth, what could have been between the satellite and the stars to obscure them. Too high for clouds, or am I wrong? Any ideas please.--Johnluckie (talk) 08:29, 12 February 2008 (UTC)[reply]

Are you sure it wasn't simply too bright at the time to see anything else (too much light pollution or perhaps it was dawn?)? You didn't mention what satellite this was but I assume you are referring to a satellite flare rather then simply observing the satellite. Satellite flares can sometimes be fairly bright, for example Iridium flares can sometimes be -8 magnitude which is a lot brighter even then Venus (~-4 i.e. about 40 times dimmer), and Sirius (-1.5) doesn't even come remotely close Nil Einne (talk) 08:52, 12 February 2008 (UTC)[reply]

Thanks for this, how do I differentiate between the satellite and it's flare? —Preceding unsigned comment added by Johnluckie (talk • contribs) 09:02, 12 February 2008 (UTC)[reply]

Well, the satellite itself would just look like a star moving across the sky in a straight line. A flare is just a flash. —Preceding unsigned comment added by Mattbuck (talk • contribs) 12:12, 12 February 2008 (UTC)[reply]

To clarify, a flare lasts a few seconds and you can see it moving during that time. You can look up Iridium flares at Heavens-Above. Icek (talk) 15:34, 12 February 2008 (UTC)[reply]

Thanks for these answers. I did see the satellite traversing for at least five minutes as I sat in my hot tub in the garden. It was dawn, I did see one faint star, but I still ask where were the other stars!--Johnluckie (talk) 16:44, 12 February 2008 (UTC)[reply]

According to apparent magnitude a non-flaring ISS can still be ten times brighter than any star, under the right conditions. Algebraist 17:30, 12 February 2008 (UTC)[reply]

You can look up ordinary satellite passes at Heavens-Above as well (but just now the website seems to be down). Icek (talk) 18:14, 12 February 2008 (UTC)[reply]

Huh? Heavens-above seems to be working fine for me. Anyway, if you were able to see one "star" at dawn, then most likely it was Venus or Jupiter. Maybe the dawn twilight washed out all the stars, or maybe there were thin clouds that were thin enough to let the ISS and the brightest stars be visible, or maybe there was too much light pollution in the area. Anyways, you can use http://fourmilab.ch/yoursky for creating star maps of your area (try finding out yoor coordinates and entering them into the yoursky. Hope this helps. Thanks. ~A H 1^(TCU) 22:27, 12 February 2008 (UTC)[reply]

Thanks for all these splendid answers and links; much appreciated.--Johnluckie (talk) 07:44, 13 February 2008 (UTC)[reply]

Golf

At home, I have a golf club that is much lighter than normal. But why does it hit things with the same force as a normal golf club? Interactive Fiction Expert/Talk to me 11:31, 12 February 2008 (UTC)[reply]

The lighter club may allow you to swing faster, which can compensate for the loss of mass. Additionally, many modern woods and drivers are engineered with a degree of "springiness" to better hit long drives. Our golf club and wood articles discuss this to some extent. The howstuffworks article on clubs may also be useful. — Lomn 14:05, 12 February 2008 (UTC)[reply]

Calculating power use

I am trying to figure out how much power my computer uses in standby mode. I found rates of 1-6 watts on several reference sites. What I can't figure out is how to convert that to kiloWatt hours, the units used on my electricity bill. Watt explains that the time is included in the unit, specifically one joule per second, so does 1 watt mean I have to multiply by 3600 to get the amount of power used in an hour? That gives me an unrealistically high number. It's been so long since I've done this kind of simple physics (ten years!) that I can't remember how to do it, and I am consternated. --Ginkgo100^talk 16:01, 12 February 2008 (UTC)[reply]

1 watt is one joule per second. However, no-one uses joules for electricity bills, they use kilowatt-hours. A kilowatt hour is by definition the energy expended in maintaining a power of 1 kilowatt (=1000 watts) for one hour (so it =3,600,000J). Thus if your computer is running at 1 watt (say) it will take a thousand hours (approx. six weeks) to use a kilowatt-hour of electricity. At 6W, it'll take about a week. Algebraist 16:07, 12 February 2008 (UTC)[reply]

or to put it another way 1W = 1/1000 (0.001) kWHr if used for an Hour. So 1kWHr of electricity will run a 1W computer for a thousand hours. and a x watt computer will run for 1000/x hours (divide) for the price of 1 kilowatthour.87.102.81.140 (talk) 16:40, 12 February 2008 (UTC)[reply]

So in other words I divide by the time rather than multiply. Thanks, I knew I was doing something horribly wrong! --Ginkgo100^talk 16:45, 12 February 2008 (UTC)[reply]

Wait... that's not it either... but I will use your formula. --Ginkgo100^talk 16:46, 12 February 2008 (UTC)[reply]

Multiply by the number of hours you use it, to get Watt-hours, then divide by 1000 to get kiloWatt-hours. 81.174.226.229 (talk) 16:55, 12 February 2008 (UTC)[reply]

Actually it's (watts of computer )x 3.6 x (hours used) to get the number of kilowatthours required —Preceding unsigned comment added by 87.102.81.140 (talk) 17:03, 12 February 2008 (UTC)[reply]

Huh? to get from watts to kilowatts you multiply by 3.6? that would cost me over $10000 a month to run one light bulb. 81.174 has it right —Preceding unsigned comment added by Mad031683 (talk • contribs) 17:19, 12 February 2008 (UTC)[reply]

Damn SineBot wastes no time. Mad031683 (talk) 17:20, 12 February 2008 (UTC)[reply]

No, it's not. You're thinking of Joules. Watts to Watt-hours is a multiplication, which is why they're called Watt-hours.

If you take Watts and multiply by the number of seconds (i.e. 3600 times the number of hours) you get W.s=Joules.
If you take Watts and multiply by the number of hours you get W.h=Watt-hours
If you take Watt-hours and divide by 1000 you get kWh

(I'm IP 81.etc.) AlmostReadytoFly (talk) 17:56, 12 February 2008 (UTC)[reply]

There's an article Kilowatt hour "One watt hour is the amount of energy expended by a one-watt load (e.g., light bulb) drawing power for one hour."

which is of course (3600 x 1 watts){answer is in joules} since there are 3600 seconds in an hour, and time (seconds) x power (joules per second = watts) = energy (in joules).87.102.9.73 (talk) 19:20, 12 February 2008 (UTC)[reply]

I understand now thanks to the various comments here... I know how 87.102 got 3.6 even though it's wrong, and I know how to figure that it's not a big deal to leave my computer on stand-by at night. Thanks to everyone who responded! --Ginkgo100^talk 03:43, 15 February 2008 (UTC)[reply]

Killing a cold virus by Exercising?

Good morning, At work, we’ve had some sort of cold virus that has infected many of my team members. One colleague claims that he was “not feeling well” and running a slight fever on Saturday. By Sunday, his temperature was just over 100 degrees Fahrenheit (38 Celsius). Despite the high body temperature, he went to the gym Sunday afternoon to exercise and workout. By today, he claims that he is much better and now has a normal body temperature. He said he “cooked” the virus (killed it) by exercising. He said when he exercises, his body temperature rises, and that temperature is too high for the virus to survive. Thus he is cured by exercising at the onset of his cold. Is this plausible and make any sense? Thanks! Rangermike (talk) 16:50, 12 February 2008 (UTC)[reply]

Hmm, I'm pretty sure that (at least heavy) exercise while sick can cause myocarditis. Not good! -- Aeluwas (talk) 17:02, 12 February 2008 (UTC)[reply]

See fever. An elevated body temperature can inhibit the reproduction of some pathogens while increasing the production of white blood cells. So, it is valid to claim that a fever helps fight an illness. As far as exercising, increased blood flow and raised blood pressure will move blood-borne pathogens throughout the body at a faster rate. It is valid to claim that this could lead to further infection. All in all, unless you do a controlled experiment all you will have is opinions. -- kainaw ™ 17:11, 12 February 2008 (UTC)[reply]

(Question undeleted. If you want it deleted, you must discuss it on the talk page. -- kainaw ™ 18:22, 12 February 2008 (UTC))[reply]

Wouldn’t it be similar and less dangerous just to raise the thermostat, take a hot bath, bundle up, and drink hot liquid? --S.dedalus (talk) 02:36, 13 February 2008 (UTC)[reply]

I have heard of "sweating out" a cold in a sauna or steam room. It did feel like it was working, but you know... AlmostReadytoFly (talk) 07:09, 13 February 2008 (UTC)[reply]

When I was a little kid and had a high fever my mother would put me naked in the bathtub and pour rubbing alcohol on me with the fan blowing on me (to evaporate the cooling alcohol) because it would make me really cold. She thought this was lowering my temperature but it was sheer torture which I remember as akin to child abuse. As an adult, perhaps in traumatic response to this, whenever I get a fever and feel sick and chilled, I immediately take the longest and hottest bath I can manage and then get back in bed with lots of covers to sweat some more. I can say that I have taken antibiotics just once since 1984 and have not had any fever/sickness that lasted over a day. That said, this is a personal anecdote from my life and it not intended as information to be used to diagnose, treat or cure any disease experienced by another human being. Saudade7 08:52, 13 February 2008 (UTC)[reply]

Hmm, I'm pretty sure that (at least heavy) exercise while sick isn't a good idea since you are wasting energy on working out, when you want to conserve that energy for fighting off the virus. 64.236.121.129 (talk) 15:00, 14 February 2008 (UTC)[reply]

Questions about EMP

Will EMP affect electronics that are turned off? There are electrical signals in the human brain. Could an extremely powerful EMP affect humans at all? 64.236.121.129 (talk) 17:15, 12 February 2008 (UTC)[reply]

It is possible for turned off electronics to get a shock from nearby turned on electronics. An overly-simplified way to think of it is to consider dunking the electronics in water. If you put two radios in water, one turned on and one turned off, it is possible that the one turned off could still be damaged. As for humans, I do not know of any EMP tests on humans. I have seen studies on rats - including a good one that had two control groups and tested the rats for three days on maze-learning tests. In general, they find that there is a reduction in brain function following extremely dangerous EMP shocks, but those are repaired over time. It is theoretically possible to do enough brain damage to kill a person (or animal) with EMP. -- kainaw ™ 17:39, 12 February 2008 (UTC)[reply]

There is a technique called Transcranial magnetic stimulation that does exactly that: sends an EMP through the skull into the brain, affecting the brain areas closest to the EMP source. The EMP source is a particular configuration of coils placed over the scalp. --Dr Dima (talk) 19:45, 12 February 2008 (UTC)[reply]

I'm pretty sure that magnetic resonance imaging uses intense magnetic fields, but doesn't scramble brains. I might have my forces confused, though. --Mdwyer (talk) 23:50, 12 February 2008 (UTC)[reply]

Let's make a distinction between these three electromagnetic phenomena.

MRI uses a strong static magnetic field, plus a very small radio-frequency electromagnetic wave. A static magnetic field doesn't induce currents.
Transcranial magnetic stimulation essentially uses the brain as the secondary winding of a transformer. A changing magnetic field induces an electric field that drives currents in one region. There is no electromagnetic wave propagating in free space. Getting a desirable outcome, instead of damage, is due to the practitioner and the manufacturer controlling the induced electric field.
EMP is an electromagnetic wave generated in a medium of charged particles. It propagates through free space.

When you want to know the "damage" caused by any of these things, you primarily need to ask "how much current is made to flow?" Bigger electric fields push more current; more current causes more heating, burning, and other damage. JohnAspinall (talk) 16:08, 13 February 2008 (UTC)[reply]

I suggest an experiment: Construct an aerial system connected via a (working) semiconductor diode to ground. Then let off a family size nuclear device fairly nearby (but shielded from the heat and blast etc). When the explosion has died down, examine the diode for functionality. I think you'll find its dead (like a lot of people in the vicinity) —Preceding unsigned comment added by 79.76.189.102 (talk) 00:50, 13 February 2008 (UTC)[reply]

Your use of the term "family size" made me imagine them being sold at Sam's Club. 206.252.74.48 (talk) 13:46, 13 February 2008 (UTC)[reply]

They are used as a sterilizing agent; you'll find them in the cleaning supplies section. -SandyJax (talk) 19:08, 13 February 2008 (UTC)[reply]

Reminds me of Back to the Future where 1955 Doc believes plutonium to be readily available in 1985 drugstores. 206.252.74.48 (talk) 20:34, 15 February 2008 (UTC)[reply]

Are all indoor heaters 100% efficient?

If an electric space heater is heating up a room, and is not completely efficient in its method of operation, presumably all inefficiency is lost as waste heat. But if you consider that waste heat to be useful in itself (it also heats up the room), can the heater be said to be 100% efficient (thereby breaking the law)? If not, where did the lost energy go? Thanks. --Sean 17:18, 12 February 2008 (UTC)[reply]

Yes indeed. It is 100% efficient. But it doesn't break the 2nd law of thermodynamics. If the goal is to get heat, then the so called "waste heat" is not waste. 64.236.121.129 (talk) 17:22, 12 February 2008 (UTC)[reply]

Not quite, if it produces light and sound (like a vibration say), then not all of the energy is heat. Wisdom89 _{(_T / ^C)} 17:25, 12 February 2008 (UTC)[reply]

What if it wastes energy producing ozone or even X-rays? There is a lot of room for energy "loss". -- kainaw ™ 17:30, 12 February 2008 (UTC)[reply]

If it produced x-rays or ozone, the person using the space heater would be dead. 64.236.121.129 (talk) 17:34, 12 February 2008 (UTC)[reply]

Wouldn't you also have to take the generator into account? Mad031683 (talk) 17:37, 12 February 2008 (UTC)[reply]

Why would the user be dead? While X-rays are not extremely common in household appliances, ozone is common in those ionizing air cleaners. I haven't heard of anyone getting killed by using an ionizing air cleaner. -- kainaw ™ 17:40, 12 February 2008 (UTC)[reply]

X-rays are actually common in CRT monitors, but they are stopped by the glass. Space heaters are not 100% effient because they do not release all their energy as heat, some of it is lost to other spectrums, including visible light if you can see it glowing. 206.252.74.48 (talk) 18:44, 12 February 2008 (UTC)[reply]

Well, assuming there are no windows, any light produced is still 100% efficiently heating the room, right? The electricity gets turned into heat, which makes the element do some black-body radiating, and those photons then plink me in the eyeball and warm me up a little, so there's no loss of efficiency. Same goes for sound, but perhaps not for the ozone production mentioned above, since that stores energy in chemical bonds. --Sean 19:13, 12 February 2008 (UTC)[reply]

You were talking about electric heaters I assume - in which case the answer is (excluding doubters and niggles) YES!87.102.9.73 (talk) 19:30, 12 February 2008 (UTC)[reply]

Guys. Learn some thermodynamics. please. Not only is it possible to make a heater exactly %100 efficient, it's possible to make a heat pump provide more heat output than the total free energy input. See Coefficient of performance. It doesn't break the first law, because the energy is coming from an external heat source, and it doesn't break the second law, because the total entropy increases, making it an irreversible process. So, if you want to heat up a room, a %100 efficient heater is actually pretty wasteful compared to a heat pump. —Keenan Pepper 19:10, 12 February 2008 (UTC)[reply]

The question wasn't whether or not it was possible to construct a 100% efficient heater, it was whether or not all indoor heaters are 100% efficient. The answer to this question is indubitably "no". Wisdom89 _{(_T / ^C)} 19:43, 12 February 2008 (UTC)[reply]

except they (fan, convection, oil storage) practically are 100% effecient eg (99%) making the answer an indubutable 'yes' ?87.102.9.73 (talk) 21:20, 12 February 2008 (UTC)[reply]

Um except 99% is not 100%... This is important, not nitpicking, if something is 100% efficient then it is 100% efficient, not nearly 100% efficient. Nothing can be 100% efficient which I presume was the point Nil Einne (talk) 09:07, 13 February 2008 (UTC)[reply]

Don't I need something (a resovoir) that is already hot (to put it explicitly) to get a heat pump to pump more than 100% in heat of the energy it expends - ?? Just checking, not really sure......87.102.9.73 (talk) 19:28, 12 February 2008 (UTC)[reply]

If sound is produced most of it will be absorbed within the room and heat it up, light (even a lot of xrays) will absorbed within the room and heat it up. Ozone will decay to oxygen, releasing heat. So yes a heater is 100% efficient if you define what you want as heat (in any form). If you define what you want as 'heating the air' or 'increasing the temperature of the air in the room' then no it will not be 100%.Shniken1 (talk) 23:03, 12 February 2008 (UTC)[reply]

Most but not all Nil Einne (talk) 09:08, 13 February 2008 (UTC)[reply]

By the definitions here (no windows, sound stays in the room, etc) many appliances (electric fans, vibrators, etc) would be 100% efficient space heaters. Edison (talk) 19:09, 13 February 2008 (UTC)[reply]

Well, if it emits even a single neutrino, it's no longer 100% efficient :)) --V. Szabolcs (talk) 19:53, 18 February 2008 (UTC)[reply]

Eyeglass prescription relation to 20/20 scale

I've been trying to figure out what my numbers on the 20/20 scale are, and the Snellen Chart doesn't help because I can't see any letters on it at all. My eyeglass prescription is roughly -8 dioptres for each eye. Is there some formula or trick to find out what my eyesight is using this number? 206.252.74.48 (talk) 17:31, 12 February 2008 (UTC)[reply]

You can't see anything on the Snellen chart? Are you considered legally blind? -- MacAddct  1984 ^{(talk  contribs)} 18:10, 12 February 2008 (UTC)[reply]

At -8, quite possibly if they want to be. Without my glasses, I'm not convinced I could see anything on it from the proper distance. Skittle (talk) 18:27, 12 February 2008 (UTC)[reply]

Ah, from the article "The biggest letter on an eye chart often represents an acuity of 20/200, the value that is considered "legally blind." Many people with refractive errors have the misconception that they have "bad vision" because they "can't even read the E at the top of the chart without my glasses." But in most situations where acuity ratios are mentioned, they refer to best corrected acuity. Many people with moderate myopia "cannot read the E" without glasses, but have no problem reading the 20/20 line or 20/15 line with glasses. A legally blind person is one who cannot read the E even with the best possible glasses." Skittle (talk) 18:33, 12 February 2008 (UTC)[reply]

So, 20/20 just refers to how far you can see compared to normal people. You know that your vision is worse than 20/200 without glasses, but with glasses it's probably rather better. I've had a look for any easy correlations, but can't find any. Skittle (talk) 18:37, 12 February 2008 (UTC)[reply]

To clarify, when I say "see" I mean "see clearly". I can see the letters on the chart without my glasses, but they are just black and gray blobs. My eyesight with glasses is a little better than 20/20, but I want to know what it is without them. 206.252.74.48 (talk) 18:41, 12 February 2008 (UTC)[reply]

Yeah, I've been there... The Snellen chart simply isn't really that useful to extreme myopes. The 20/200 number just says that you can see at 20ft what someone else can see at 200ft. So to find the right number, you need to find a gigantic 'E', or do some extrapolation. I think the math is in the linked article. --Mdwyer (talk) 19:54, 12 February 2008 (UTC)[reply]

I did some math a while ago, it came out to 20/14400 - which for some reason I don't think is quite right. That means that I have to be within 20 feet to read an enormous 'E' that someone with good vision can read from nearly 3 miles away. The article never states what prescription would be given to someone with 20/200 vision. 206.252.74.48 (talk) 20:06, 12 February 2008 (UTC)[reply]

The article does point out that the 20/40 line is twice the size of the 20/20 line. So, if there was a 20/14400 line, it would be 720 times as big as the 20/20 line. It doesn't give the exact size of the 20/20 line. Assume it is 1 inch. You are talking about a 20/14400 line that is 720 inches (60 feet) tall. It is perfectly reasonable to assume that a person with 20/20 vision could see a 60 foot letter E that is 2.7 miles away. Just looking out my window right now, I can see a few 5-7-story buildings, which is roughly 60 feet, that are at least 3 miles away. I can clearly make out the windows. So, if they pasted a big letter on the side of the building, I'm sure I could tell you what letter it was. Now, could your vision be so bad that you couldn't make it out while standing on the opposite side of the street? It is possible. I don't know how bad your vision is. -- kainaw ™ 20:17, 12 February 2008 (UTC)[reply]

Ah, thanks for clarifying that. If I stand across the street and look at a 60 ft tall letter, I can only tell what it is because of the general shape (since at these scales seeing a letter clearly is not needed), but I won't be able to see it clearly at all, it would just be a blur in the general shape of an 'E', which probably means that my calculations are correct. I am suddenly reminded of an ad at my local mall for eyeglasses that is painted in huge letters on the wall. I can't read any of the 10 ft tall letters from 20 ft away without my specs. Like Mdwyer said, the Snellen chart becomes quite useless after 20/200, but I'm sticking with 20/14400 until greater logic or mathematics prevails. 206.252.74.48 (talk) 20:31, 12 February 2008 (UTC)[reply]

Keep in mind that the Snellen chart isn't intended to measure diopters. It is only a subjective measure of acuity. So, I don't think there really is a snellen::diopter mapping. For that matter, diopters change depending on the position of the lens. That's why my glasses were -8, my contacts with -9.5, and my implants are something like -11. I really like the 20/14400 number, though. I always just told people I was off the chart. --Mdwyer (talk) 23:48, 12 February 2008 (UTC)[reply]

Photosynthesis vs photovoltaics

Which one is more energy efficient? Which method derives more energy from a given amount of sunlight? 64.236.121.129 (talk) 17:33, 12 February 2008 (UTC)[reply]

The photosynthesis and high efficiency solar cells both give values for efficiency. Have to be careful with your terms: does "given amount" mean overall, or of the usable part of the spectrum for each method, and do you mean useable energy (after processing into some externally-accessible form) or just the initial light→energy conversion? DMacks (talk) 20:46, 12 February 2008 (UTC)[reply]

Tandem Repeats

Are tandem repeats the same as regulatory genes that sit above a certain gene and the lengths of the repeats are when to turn on or off? —Preceding unsigned comment added by 67.121.105.145 (talk) 20:33, 12 February 2008 (UTC)[reply]

No, tandem repeats are not the same thing as regulatory sequences. --mglg_(talk) 21:31, 12 February 2008 (UTC)[reply]

But don't the tandem repeats control the turning on or off of a gene? —Preceding unsigned comment added by 67.121.105.145 (talk) 02:32, 13 February 2008 (UTC)[reply]

In a (genetically) healthy individual, no; tandem repeats are not part of a normal cell's mechanism of controlling gene expression. However, there are some diseases, most notably Fragile X syndrome, which are due to trinucleotide repeats. You may be interested in reading about trinucleotide repeat disorders. (EhJJ)^TALK 03:37, 13 February 2008 (UTC)[reply]

Minor nit: There's no strong reason to consider fragile X more notable than other trinucleotide repeate disorders. --David Iberri (talk) 03:43, 13 February 2008 (UTC)[reply]

Good point. It was just the example that kept coming up at my med school, and was the first that came to my mind. (EhJJ)^TALK 16:39, 13 February 2008 (UTC)[reply]

Well it depends, the number of repeats in this case do cause the transcription to abort i.e. a little like a regulatory defect, whereas in Huntington's disease the repeats lead to many extra amino acids in the protein. In this latter case the protein does not function but there is no change in the production of transcript. Given the original question was from a regulatory perspective Fragile X is a more appropriate choice. David D. (Talk) 03:44, 14 February 2008 (UTC)[reply]

empty space?

I read some where that on a moecular level the human body is 99% free space. When atoms bond there is space between them, much like filling a box full of balls. Is this true and if it is what is in the empty space? cris —Preceding unsigned comment added by 24.183.237.19 (talk) 22:21, 12 February 2008 (UTC)[reply]

There's some but not much space between atoms, and only sometimes any at all: a covalent bond is literally when the electron clouds of adjacent atoms overlap. However, atoms themselves are mostly empty space: a very small nucleus with a very large electron cloud extending far beyond it. DMacks (talk) 22:30, 12 February 2008 (UTC)[reply]

In the Standard Model of particle physics, elementary particles are assumed to be point-like objects with no size - so every object, including the human body, is actually 100% empty space. In various alternative (and currently speculative) theories such as string theory, elementary particles have extension and are not dimensionless points - but they are still unimaginably small compared to the distances between them. However, "empty space" is, in fact, a sea of transient virtual particles, so it is not exactly "empty" anyway. If all this seems weird, that is just because our usual working models of reality which we have built by interacting with everyday objects just don't apply at sub-atomic scales. Gandalf61 (talk) 22:37, 12 February 2008 (UTC)[reply]

I'm not sure I'm happy with either of the answers above. The Standard Model is a field theory; space is occupied to the extent that fields can be said to occupy space. Field strength is never exactly zero anywhere, so you can't draw a distinction between empty and nonempty space without inventing an arbitrary cutoff where you consider the field to be weak enough that it doesn't matter. You can choose the cutoff such that the whole universe is occupied, or such that interstellar space is mostly empty but human beings are solid, or such that human beings are mostly empty except for widely-spaced nuclei, or such that the whole universe is empty. As for the zero-point energy, I'm not sure I believe in it at all. It doesn't have any physical consequences in the Standard Model (not even the Casimir effect, as discussed in this paper), and the one physical consequence you might expect it to have in quantum gravity is famously absent (the cosmological constant problem). Describing the zero-point energy as a sea of virtual particles is even more dubious, since the virtual particles only appear in individual Feynman diagrams, and only a sum over many diagrams (with different virtual particle lines) has any physical significance. -- BenRG (talk) 17:26, 13 February 2008 (UTC)[reply]

BenRG - yes, the Standard Model is a field theory, but nevertheless it contains point-like particles and uses quantum fields to describe these point-like particles - our quantum field theory article says "Quantum field theory thus provides a unified framework for describing "field-like" objects (such as the electromagnetic field, whose excitations are photons) and "particle-like" objects (such as electrons, which are treated as excitations of an underlying electron field)".

And when you say "the virtual particles only appear in individual Feynman diagrams" are you saying that virtual particles have no actual existence, and are only introduced into Feynman diagrams as a calculational fiction ? The first sentence of our virtual particle article says "In physics, a virtual particle is a particle that exists for a limited time and space ...", which certainly implies that they are not solely aids to calculation. Gandalf61 (talk) 09:33, 14 February 2008 (UTC)[reply]

It may help to imagine looking closely at something - like a TV, magazine picture, or pencil line. When you look closely, what looked like a solid clear-cut thing is actually made up of lots of dots, grey fuzz and is actually pretty fuzzy or patchy. Solid matter, like our bodies, is a bit like that. While at a distance it looks solid, if you could make a powqerful enough microscope, you'd see that it's made up of a cells, which are made up of molecules, made up of atoms... and at this level "matter" is tiny dots in a lot of space, held apart by forces such as the electromagnetic force and strong force etc (there are others), which prevent these points from coming too close and keep them separated, and stop them collapsing into almost nothing. It is in that sense that people are "mostly made up of empty space".

If you looked even closer you'd see the dots and the space itself are a sort of fuzzy haze of potentials and quantum theory entities that may-or-may-not exist and can-and-cannot be seen, which is a lot harder to describe... but by then you're into quantum theory.... FT2 ^{(Talk | email)} 17:42, 13 February 2008 (UTC)[reply]