The Thing

Colourful demonstration, you say? One cool demonstration would be mixing colourless potassium iodide and coluorless lead (II) nitrate and watch the yellow lead iodide precipitate form "almost from nowhere". Very freaky and colourful.

A couple of methods I came up with all require solving of 4th degree polynomials, which I can't do by hand at all. But anyways, here are my methods: Method 1: Let x be the length of the smaller (bottom) triangle of the two similar triangles. [math] \frac{20-x}{6}=\frac{x}{\sqrt{x^2-6^2}} [/math] Simplifying this, including squaring both sides and removing the x from the denominator turns it into a 4th degree polynomial. I can't solve it. Method 2: Let x be the same as in Method 1. [math] sin^{-1}(\frac{6}{x})=cos^{-1}(\frac{6}{20-x}) [/math] Turns out to be...something along the lines of a 4th degree polynomial. Method 3: Let X be the bottom left angle of both similar triangles. Let x be the same as in Method 1 and 2. [math] sinX=\frac{6}{x} [/math] [math] cosX=\frac{6}{20-x} [/math] [math] \frac{6}{sinX}+\frac{6}{cosX}=x+20-x=20 [/math] I dunno what this turns out to be. Haven't tried simplifying it yet but I'm not very optimistic about it turning out t be something other than a 4th degree polynomial again.

Compute the ratio of the area of the triangle ABC to that of the area of triangle made by drawing lines CF,AD, and BE. LINES CF, AD, and BE. Complicates the problem very dramatically. I did almost exactly what you did but just as I was about to post it I saw that. Lines CF, AD, and BE, NOT lines EF, FD and ED. EF, FD and ED just creates another equilaterial triangle and the Triangle FDB on your picture would be a 30-60-90 triangle (side of 1/3, side of 2/3, 60 triangle in between) and hence its area wouldn't be hard to figure out. Drawing lines CF, AD, and BE would create 3 lines that intersect to make some kind of triangle in the middle of the larger triangle.

Many computers, especially older ones, still use Machin's formula or some form of arctan infinite series.

Oh AHAHAHAHAHAHA I get it!!11!1!eleventyone! No really, I don't think there are right answers. My answer is that you can fit nine cars like this: [hide] N I N E C A R S [/hide]

Okay, sorry for reviving this ancient thread again. I have a new question on the subject. Could we create some kind of...thing (antibody, cell, whatever) that can deliver specific chemicals to a tumor site by targetting the high telomerase activity in the cancer cells? If so, what kind of...thing can do this and how can it target the telomerase? Or is it better just to stick with creating antibodies that could order cytotoxic T cells to attack the telomerase-expressing cells (such as pulsing dendritic cells with telomerase proteins)? EDIT: Also, is it possible , instead of targetting telomerase, to target angiogenin during tumor angiogenesis with the same methods (monoclonal antibodies or creating antibodies that orders cytotoxic T cells to attack cells expressing angiogenin)? However, is angiogenin highly expressed (as much as during tumor angiogenesis) during normal blood vessel growth? If so then I'm sure we don't want to target it. Thanks.

Beauty! Nice find. Thanks a lot! This is going to be especially helpful since I don't have any books concerning Quantum Physics.

Okay, you're probably thinking this is another "can I use quantum entanglement to achieve superluminal-speed communication?". Well, it's not exactly that issue. I understand that an entangled pair of photons behave like there is no space between them and that once one has been observed it has an instantaneous effect on the other. But then, WHY? How is the information of photon A being observed travel to photon B? Does THIS information (that says, "oh, photon A's been observed as something") travel faster than light? Does it perhaps travel in another dimension? Or is this information not needed at all to affect the other photon and I'm just thinking in classical mechanic terms? One more thing: a similar thing I think has been asked by bascule before, but he couldn't get any answers: A photon (or photons) is shot through a double slit, and is then split into two entangled photons by two down converters (1 at each slit). The two photons are then sent to the two ends of the apparatus, which are a light hour apart. Detectors at the end show interference patterns. But, if a path from one of the down converters to one of the detectors is blocked by a screen, then the other detector which is a light hour away, will immediately show a solid line instead of interference. Isn't this a set up for FTL communication? Thanks.

Well, coincidentally, while I am making my mousetrap catapult, my drafting class is assigning us a mousetrap CAR fun assignment, wanting us to design a car powered on a mousetrap that would go the furthest. So, what kind of things do I need to note for a long distance mousetrap car? It can be slow to a crawl, but really long distance traveller.

Oh frick... I know what I did wrong... Didn't convert from km to meters... I'm pretty sure my equation's right: [math] 25x+x^2-\frac{350}{9}x=\frac{350}{9}-25 [/math]

What are the 13.88 and the 25 and the 38.88?

Hmm...why am I off by 10x?

If I'm not incorrect, I think I got an answer of 1.588 seconds. Could someone else check on that? Someone that actually KNOWS Physics? (lol not me, I used all math)

I'm trying to build a mousetrap catapult. It would be launching acorns. I was wondering how I can get the acorn to fly as far as possible? Is a shorter arm or longer arm going to throw the acorn further, and what other features, such as rubber bands, can I add onto the catapult to make it more powerful? Thanks.

I agree, this is among the most ridiculously stupid posts ever, but my curiosity is getting the better of me. As it is the sickness season, I inevitably fell victim to a cold & fever. Funny thing was, I got it Friday, but my fever disappeared yesterday. I also had a dentist's appointment yesterday for plaque removal (those giant machines that the dentists use to remove all sorts of junk from your mouth, including a small drill, a scraping hook and a suction tube). As usual, lots of bleeding occured and everything. But I didn't mind it. However, something weird happened. That night, my fever for some inexplicable reason reappeared. Why is that? I've heard that you shouldn't have a tooth pulled while you're sick, is it true for this as well? I just don't get why this seemingly frivolous event would bring back my fever - and an annoying headache. So, speculations and suggestions? Incidentally, can HIV be transmitted during any sort of dentist operations or plaque removal, or do the dentists ALWAYS sanitize their equipments?

That would be what you get using those three vertices. Assuming that those three vertices ARE correct. And after that absolute fiasco regarding the contest, someone should check whether I have the right vertices.

Aaarg I died horribly at today's Cayley contest. On one question, it wasn't until now, about 3 hours after the contest, did I remember that 0.56cm does NOT equal 56 millimeters. By far the worst reason to lose a whole chunk of 8 marks. So for the question above, the three vertices are (0,0), (5,0), and (5/2,5/2)? Then just use those vertices to find the area.

Err... No, coterminal angles are not that. They are angles which share the same arm when put in standard position. So imagine a Cartesian plane and an arm extending a certain angle from the x axis, rotating counterclockwise. That's an angle in standard position. So, say you have an angle at 30 degrees at standard position. Its coterminal angles would include 390, because you have a 360 rotation about the origin more then add the 30 degrees upon that 360. In that sense, 750 degrees is a coterminal angle of 30 as well, but with two rotations (720 degrees more), and so are 1110 (3 rotations), 1470 (4) and so on. So, this way, 870 and 150 are coterminal angles. Why? Well, 870 when put in standard position is definitely more than 2 full rotations because it is larger than 720 degrees. So subtracting the 2 rotations from the angle gives an angle with the same arm, or a coterminal angle. 870-720=150, thus 870 and 150 are coterminal angles. Also, negative angles can be coterminal angles of positive angles too, and vice versa. A negative angle in standard position is starting from the x axis, just like a positive angle, but instead of rotating counterclockwise, it goes clockwise. So, a 30 degree angle in standard position is coterminal with a -330 degree angle. So, I think your textbook's wrong. EDIT: I've drawn a picture for you (yes yes, you can marvel at my artistic sk1llZ - one of the many talents I was born with jk). The 45 denotes, well, the 45 angle in standard position. The 405 is coterminal with the 45 degree angle, because it goes from the x axis all around 360 degrees, then end up at THE SAME ARM.

It's been invented and I think commercialized (Tempra & Crown, the name of the brand, methinks). Meh, I even considered it for a project myself before, and 3 seconds in google delivered a sledge hammer blow to the death to my idea. Google self cooling beer can or something. And I used to have a textbook that devotes an infoclip to it. It's really cool. I've read the infoclip so many times I memorized it. Basically it (the self cooling can) uses a phase change of CO2. NO chemical reaction occurs. It is able to cool the liquid inside to about 1 degree Celsius from a starting temperature of about 40 degrees in about 90 seconds. It contains a cone containing liquid CO2 under high pressure. When you open the can, you open that cone, and the CO2 immediately changes into a gas due to the much lesser pressure, and thus it absorbs heat. The cone can chill to -50 degrees Celsius in 90 seconds, and thus it cools the liquid (coke, beer, whatever) around it. Okay, time to stop. I sound like a salesman. I think the "self cooling beer cans" you find on the web will be different from this one I just mentioned. I'm not too sure. EDIT: Oh, Tempra & Crown uses evaporation, not compressed gas. Google Self cooling cans for more info.

Meh, I've seen it n times so it fails to surprise me now. Virtually every forum has some people using sigs consisting of that. I don't think SFN allows pictures in the signatures though...that's probably why no one has seen it before here.

Diamond is not even the hardest substance known to man. Dun dun dun dun... Ultrahard fullerite is harder than diamond. NO it's not a giant chunk of material that you can use to scratch glass. Small particles of it left nanoscale scratches on diamonds. Boron nitride is not harder than diamond. Rather, its hardness is just below that of the diamond, and of course, ultrahard fullerite. Two wikipedia links for you: Ultrahard Fullerite http://en.wikipedia.org/wiki/Ultrahard_fullerite And an EVEN HARDER MATERIAL: http://en.wikipedia.org/wiki/Aggregated_diamond_nanorods As for the bunker, 5 kilometers of concrete (stop neutron radiation), with another 5 kilometers of steel coating (iunno), then another 5 kilometers of lead (basically more protection) beyond that, and then another 5 kilometers of tungsten (heat resistant), then all coated with Teflon (dunno why), which has another 5 kilometers of tungsten beyond that. The entire bunker should have walls that are about 25 kilometers thick. Haha, the nukes got sp0on3d by t3h ultra 1337 bunker designed by t3h ub3r k3wl 3ng1n33r1ng g3n1us p0on0rz sk1llz0rz0rz0rz! C4mp3d by t3h h4x0r! EDIT: Oh, shoot. Thinnest wall wins.

Err, hokay. Let's try this. So let x be the time during which Bill works. So x-1 will be the time during which both Bill and Jane work. In 1 hour, bill completes [math]\frac{1}{B}[/math] of the job, and same with Jane, completing [math]\frac{1}{J}[/math]. So, multiplying them by the time x and x-1 (for Jane). [math]\frac{x}{B}+\frac{x-1}{J}=1[/math], 1 being the entire job, the entire wall being painted. Err I think all to do now is to simplify and solve for x: Multiply the equation by BJ to get rid of the denominators, ending up with [math]xJ+xB-B=BJ[/math]. Then factor out the x, and isolate it: [math]x(J+B)-B=BJ[/math], equals to [math]x=\frac{BJ+B}{J+B}[/math]. Factor out the B on the top and we're left with [math]\frac{B(J+1)}{J+B}[/math] I think that's A. Assuming I'm correct, then EvoN1022v is correct as well! But he gets all the credit for answering first. How very unjust.

Gravity's the weakest of the fundamental forces by far. Heck, the nuclear "weak" force is not as weak as gravity, in fact, much stronger. But gravity is a long range force.

For H3O+, it appears that its correct name, according to IUPAC nomenclature, is oxonium. Hmm...confusing. I searched for hydroxonium in wikipedia, it got redirected to hydronium. The article is here: http://en.wikipedia.org/wiki/Hydronium

It is a soft c. As in a sound that sounds like an s. Who do you know that pronounced it with a hard c? I've never heard of anyone who has done that.