mooeypoo

Excuse me, but neither the sun NOR THE EARTH expand over time. The Earth doesn't expand at all. The Sun expands *AND* contracts. It does NOT expand over time. As for particles/subparticles in space - naming them means someone found them and recognized them. Who was that someone? you? Show us how you found these? what are they? what do you mean by "subparticles"? really really small dustlike-particles or muons/gluons/whatever-else ? You can't continue to a conclusion until you explain this (and the rather ludicrous statement about the earth and sun expanding). Also, it seems some new visitors to the forum, lately, are insisting on looking only one ONE set of forces in the universe, and insist they are the ONLY forces in existence. Be it "F=ma" driving all (preposterous. Explain sun spots with F=ma?), or Electromagnetism being everything (preposterous. Explain the fall of the twin towers by electromagnetism alone). Physics has a COLLECTION of laws and forces we know and recognize that explain multiple phenomena. They sometimes work quite well together (Gravity is, many times, very similar in behavior to electrostatics) and sometimes they don't (which is why physicists still look for the mysterious "unified theory"). If you ignore a set of rules and decide to ONLY look at one set and one set alone, you will have serious problems explaining the phenomena in the universe that do not conform, necessarily, to this set of rules. Unless you are prepared to explain *EVERYTHING* in the universe with only a single set of force equations, I suggest you open your collection just a tiny bit, and see what other laws and forces exist that may affect and explain phenomena you may think otherwise have no explanation. ~moo

Yes, tht's what I claimed too. What do youthink of my hypothesis? (it's continuing in the link i supplied)

Read more here. Feel free to criticize, discuss, and make your own hypotheses (based on evidence, of course)

Well, they seem to claim they have.. I just can't find anything about it online :\ I want to see exactly HOW they did this experiment.

uh.. okay.. I'd love to see the actual research that led to this, but I was wondering what you guys thought on this, based on the many many failed experiments thus far.. http://dsc.discovery.com/news/2008/10/21/magnet-depression.html did they prove it right? Anyone has any more info on this? The actual research made, or the results? ~moo

I think the problem is that the *subject* of this thread says "perpetual motion" while the question in the poll asks about a perpetual motion machine. Different concepts. Perpetual motion is a motion that never stops, right? Well, technically, theoretically, you already have that in space; theoretically, a tiny marble will go on moving forever if nothing obstructs it. That's not against the laws of nature, because that movement doesn't create/spend energy. The problem starts when we start talking about a MACHINE. Perpetual motion machine means, usually, that you have more energy exherted than what is put in. That is against the laws of nature, and is just impossible, at least unless we find something extremely mindbogglingly different than what we know so far, which is highly unlikely. ~moo

Wouldn't this affect "The fabric of space" in a way that would be detectable, though?

Anyone read "Sphere" by Michael Crichton? The book rocks (the movie much much much less.. lacking the good message of the book, I must say). Great philosophical conclusion there.. if anything, I'd believe in these type of life forms, the ones we won't even know are alive because they don't quite fit our definition of life.

Yes, sorry, the (3\pi\epsilon)-1 is supposed to be (4\pi\epsilon)^-1 --- sorry, that was a typo, indeed. The kr'/r' comes from the calculation of E with a gaussian surface r>R [math] 4\pi r'^2 E = \frac{2\pi kR^4}{\epsilon} [/math] but I didn't use it in my final answer (as I didn't need, I think, an "Outside" v®. My final calculation was [math] W = \frac{1}{2} \int \rho ® V® d\tau = \frac{1}{2} \int (kr) \frac{kr^3}{3\epsilon} r^2 dr sin\theta d\theta d\phi = \frac{4\pi k^2}{2\epsilon} \int r^6 dr = \frac{4\pi k^2 r^6}{6\epsilon} [/math] Using [math] V® = \frac{kr^3}{3\epsilon} [/math] Right.. so just to make sure -- my limits are "inside" the sphere only, from R to 0, right? That would make the final answer: [math] W = \frac{4\pi k^2 R^6}{6\epsilon} [/math] ((OOPS! wrong.. )) [math] W = \frac{1}{2} \int \rho ® V® d\tau = \frac{1}{2} \int (kr) \frac{kr^3}{3\epsilon} r^2 dr sin\theta d\theta d\phi = \frac{4\pi k^2}{6\epsilon} \int r^5 dr = \frac{\pi k^2 r^6}{9\epsilon} |_{0}^{R} = \frac{\pi k^2 R^6}{9\epsilon} [/math]

Yeah, I just thought of this again. My professor did this question in class using the other method (integral with E^2) , and calculated both mediums (inside and out). But... I think I overthought this. In the case of using E, I *do* have a medium outside and inside... in this case I don't.. so.. there shouldn't be a need to calculate the outside. I think... Does that suond reasonable? I'll try to solve it and post the steps and answers, see if I got this right (at least in method). Thanks Okay, so here are my computations, thinking that there is only one medium to calculate from: [math] V® = \frac{kr^3}{3\epsilon} [/math] [math] W = \frac{1}{2} \int \rho ® V® d\tau = \frac{1}{2} \int (kr) \frac{kr^3}{3\epsilon} r^2 dr sin\theta d\theta d\phi = \frac{4\pi k^2}{2\epsilon} \int r^6 dr = \frac{4\pi k^2 r^6}{6\epsilon} [/math] Does that look reasonable?

I am so confused, specially (AGAIN!!) about my vectors in this equations.. The question: (V is potential). The way I understood it, the energy stored in this configuration is the one involving the potential inside + the one involving potential outside of this sphere. They're both, obviously, different. So, I tried to see if I can first define the two Vs (in and out) and then integrate the W equation. For the potential inside the sphere, I created a "gaussian surface" with r smaller than R, here's what I did: [math] V® = \int_{0}^{r} 4\pi \left( \frac{1}{3\pi \epsilon} \right) \frac{kr'}{r'} r'^2 dr' = \frac{1}{\epsilon} \int_{0}^{r} kr'^2 dr = \frac{kr^3}{3\epsilon} [/math] That's the potential inside the sphere. Now, outside the sphere, I'm getting stuck, primarily because the rho outside the sphere is zero (??). but I figured I might have a chance calculating it from the other way, by figuring out the Electric Field E: [math] \int E \dot da = \frac{Qenc}{\epsilon} [/math] [math] Qenc = \int_{0}^{R} \rho(r') d\tau ' = 4\pi k \int_{0}^{R} r' r'^2 dr = 4\pi k\frac{r'^4}{4} |_{0}^{R} = \pi k R^4 [/math] [math] da = 4\pi r'^2 [/math] The electric field E is in the same direction as r, so the dot product is settled, and we have: [math] 4\pi r'^2 E = \frac{2\pi kR^4}{\epsilon} [/math] [math] E = \frac{kR^4}{2\epsilon r'^2} [/math] And now I find the potential: [math] V(r')=- \int E dr = - \int^{\infty}_{R} \frac{kR^4}{2\epsilon r'^2} dr' = -\frac{kR^4}{2\epsilon} \int^{\infty}_{R} \frac{1}{r'^2} = \frac{kR^4}{2\epsilon r'} [/math] So, I guess i have both my potentials, but now I am having trouble figuring out how I deal with \rho in my W integral.. [math]W = \frac{1}{2} \int d^3 r \rho \text{(inside)} V \text{(inside)} + \frac{1}{2} \int d^3 r \rho \text{(outside)} V \text{(outside)}[/math] Help! (1) Did I do it correctly up until now? Is that the way to "attack" the question or should I use another method? (I know of "poisson's equation" but.. err can't figure out if or how to use it here?) (2) I don't quite have a \rho outside... what do I do??? Help me make sense of these type of questions, pleaaase! ~moo

Metric system rules, DrP

This is a very interesting debate, but please, just a side note (and important one, at that) - First, please please please use the "Quote" button when you're quoting another member. It's very hard to see their quotes with no reference to where they said it and what the context was. Second, please supply references to your sources, *ESPECIALLY* if you're quoting them.. and particularly here, can you please give the complete URL this paragraph is in? I want to read the entire section

I was thinking that maybe this concept will help you make sense of it, Booxx: I always had a "trick" that I used to remember these (I am used to working with dots moving around, since I was educated on the metric system, and that's how you convert units there, mainly): The significant number will always be the first digit, dot, and however many digits left for the SF, rounded properly, of course (1.4 --> 2 SF, 1.37 --> 3 SF, etc) The dot, though, is not there originally. It "jumped" places from the beginning of the number to the digit before last (The original number can be written as 136,500,000.0 - hence the dot is at the end...). How many places did it jump? 8 places, which is the number that is the power of the 10. It might sound silly, but it helped me remember how to convert to SF back and forth without mixing up my meters, centimeters and milimeters (all related to one another by powers of 10!) Hope that helped ~moo

Nerd boredom can be dangerous... And just for the record, *I* rule. You two are nerds.

You need to review triangle properties for this question. Specifically: What is the sum of all the angles in a triangle? What does it mean if a triangle has 2 equal vertices? What does it mean if a triangle has a 90 degree angle? Those questions are very easy once you go over the properties of triangles. Here's a few resources: http://mathworld.wolfram.com/Triangle.html http://www.mathwarehouse.com/geometry/triangles/ http://euclid.barry.edu/~mat476/Apang/apangppt2/sld003.htm Good luck!

I had an epiphany. I think... i used a sort of double substitution for this integral, and I hope it's "legal". I don't see anythign wrong with it, since I merely called my constants in a different name, and whatever isn't a constant I did a proper U substition (with differetiating for du, etc). But... just want to make sure. So, I start with: [math] \frac{1}{4\pi \epsilon} \int_{-L}^{0} \frac{\lambda}{\sqrt{a^2+b^2+(c-z)^2}} dz = [/math] and then do a double substitution; first: [math]u= c-z[/math] [math]du=dz[/math] second, just for 'convinience' to see the formula, I renamed my constants: [math]m^2 = a^2+b^2[/math] And so I have: [math] \frac{\lambda}{4\pi \epsilon} \int_{-L}^{0} \frac{1}{\sqrt{m^2+u^2}} dz = [/math] Which is a formula (given in our hw, too), and solved like so: [math] - \int \frac{1}{\sqrt{m^2+u^2}} dz = - ln \left(u + \sqrt{m^2+u^2} \right) [/math] So, replacing back the constant and u-substitution, I have: [math] = \frac{\lambda}{4\pi \epsilon} \left( -ln(c-z+\sqrt{a^2+b^2+(c-z)^2}) \right) |_{-L}^{0} = [/math] [math] = \left( \frac{\lambda}{4\pi\epsilon} \right) \ln \left( \frac{c-z+\sqrt{a^2+b^2+(c+L)^2}}{c-z+\sqrt{a^2+b^2+c^2}} \right) [/math] Ha! Well, it seems like my substitutions are okay, but.. there's no actual problem in substituting m^2=a^2+b^2 , specifically since I'm not sqrting it or doing anything to it (and hence risking losing signs), right? it's just "calling it" a different name..? Anyhoo, I think it's solved fine.. if you spot an error, do tell. Thanks again!! ~moo Just an update -- the professor solved this question in class today, and my solution is, indeed, true (including the 'trick' I've used, yay! ). Thanks a lot! ~moo

Woah, I took statistics a long time ago, but here's my input based on some internet research.. hope it helps, at least for a starter: To achieve the "mean value", you add up the ages of all the students and divide by the number of students (source: http://www.mathsisfun.com/mean.html also this: http://www.sosmath.com/calculus/diff/der11/der11.html with a bit more advanced level, I am not sure which level you are in. The first should suffice for a basic level statistics question). The standard deviation is, basically, the amount of values who "stray" off the mean. If you have your mean as 5'8", with most people something close to that but a few 7'1 and some 4'2" (...kindergarden? ) that's your 'deviation'. You use the mean to calculate the deviation. (this is a good example of the steps you need to do to calculate that: http://en.wikipedia.org/wiki/Standard_deviation#Example ) So, basically, you have the value of the standard deviation (look at the example from wikipedia on what to do with it) and you have the equation for the mean, that you can work with: [math] \frac{\text{sum of the ages of all students}}{20,000 \text{ students}} = 5'8" [/math] Working with both values, you can figure out how many potential leprechauns there are (hence, how many men that are 5'8" high are in the 20,000 total) and how many are your deviation (taller than 6'5")... Hope that helped!! Good luck ~moo

Uhh.. didn't you just quote it in your previous post, as adding credibility to your claim? So.. you treated it as credible when it suited your claim, but now it's "making stupid claims" when you disagree with the suggestions in it? It's either credible or it isn't.. which is it?

There hasn't been conclusive results, and no real experiments using real antimatter as of yet. There are, however, some speculations based on mathematical and theoretical concepts. Here's a useful explanation from the NASA website: http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/000531a.html CERN's ATHENA is a good promise for some experiments... soon, hopefully.

Hm, wow, cool! I'm just doing a small research into some of the shuttle missions, specifically about shuttle design and where crew quarters are found in it, etc. This was a cool update, thanks for posting it !

Yes, well, proofs are required for this statement, was my point.

UFOs are "Unidentified Flying Objects". If you see a plane but you can't identify the fact it's a plane because it's dark outside, or whatever other circumstances, that's also a UFO. In that sense, UFOs exist. As to concluding that they're visitations from aliens or an elaborate government conspiracy, well.. that requires much much more proof, to say the least.

Was this close to sundown? most UFO-related sightings that are next the horizon at these times are related to sightings of Venus. The red color can be the light breaking at an angle in the atmosphere... I am not sure, but it's hard to gage distances at night -- very big but very far airplane might look the same as a very small but rather close aircraft. I did hear about such incident, where they found out that lights from a nearby road reflected in the damp atmosphere and looked like they're weird UFOs. Could that be it? I am guessing here without much info..

oh, shoot! sorry [math] f(\theta) = \Bigl{\lbrace} \aligned \sin^2 \theta &\quad \text{when $\theta \in [ 0,\pi ]$} \\ 0 \phantom{\sin^2} &\quad \text{when $\theta \in ( \pi,2\pi ]$} \endaligned [/math] (ha! fixed alignment thanks D H for the sample Sorry 'bout that, the function is 0 in the second condition, and then periodical over other domains.