The Sandbox

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science forums science forums science forums Posted light switch on and off. New user issue?

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point 1 point 2 point 2.A point 2.Bindent a paragraph of lines on a single bullet point 2.C point 2.C.a point 2.C.b point 2.C.c point 2.D point 2.E point 3 point 4 point 4.i point 4.ii point 4.ii.1 point 4.ii.2 point 4.iii point 4.iv point 4.v point 4.vi to retrieve the code use the quote button

Any clues. I can just about login to Science Forum and have a reasonable knowledge of Word processing and Excel ( enough knowledge to make a Table in Word and change a cell value in Excel). So, my knowledge of using code is very limited. I would really like to present something like this in a post as an animation: http://highered.mheducation.com/sites/0072495855/student_view0/chapter14/animation__transmission_across_a_synapse.html

the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog the quick brown fox jumps over the lazy dog

Very Special Relativity SRN=Special (..) Relatively New We (can) use the (kinetic energy) KE=m*((v1^2)-(v0^2))/2 for mass m with final_speed v1 (gained) from initial_speed v0 ((is) set to the) ether's (light) speed c to find the (ether's (mass's)) kinetic energy KEo=m*((v1^2)-(c^2))/2 i.e. with respect to (wrt) the ether. The final speed v1 is the speed (that was) lost (wrt the ether, although gained by us.) That is now, all in absolute coordinates. The ether (coordinates). Commonly wrongly known as the rest mass energy (from Einstein & others). It is simply the (kinetic) energy the ether has to offer. An energy lost when…

Well, I didn't get any support from this forum, as my comments suggest.

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<a title="By MissMJ [CC-BY-3.0 (http://creativecommons.org/licenses/by/3.0)], via Wikimedia Commons" href="http://commons.wikimedia.org/wiki/File%3AStandard_Model_of_Elementary_Particles.svg"><img width="512" alt="Standard Model of Elementary Particles" src="//upload.wikimedia.org/wikipedia/commons/thumb/0/00/Standard_Model_of_Elementary_Particles.svg/512px-Standard_Model_of_Elementary_Particles.svg.png"/></a> Standard Model of Elementary Particles By MissMJ [CC-BY-3.0 (http://creativecommons.org/licenses/by/3.0)], via Wikimedia Commons

Question: Is surface tension holding the photosphere together more than the Gravitational fields ??? Judging by the way Solar storms rip thru the Photosphere then contracting back to reform, it seems that the Photosphere is more in surface tension not gravitational compression. http://solarscience.msfc.nasa.gov/feature1.shtml Storm watching: http://www.youtube.com/watch?v=91H4qgFCauQ zorro

http://www.youtube.c...h?v=n9KaI5T0zRw Hooray! Now to see if I can get the two in the one post. http://www.youtube.com/watch?v=n9KaI5T0zRw Seems that two in the one post broke it.

[math] \frac{a}{b} [/math]

My link

How can I start a poll on this site? I thought it might be useful to get the opinions of the community. Are there any special rules for polls? Am I allowed to ask about religion, political standing, and other sort of touchy subjects? Thanks.

The quick brown fox jumps over the lazy dog.

The forum FAQ is here: http://www.scienceforums.net/forum/faq.php The list of BB code we use is here: http://www.scienceforums.net/forum/misc.php?do=bbcode

[math] m\left(\sqrt{(x - j)^2 + (y - k)^2}\right) = n\left(\sqrt{(x - r)^2 + (y - s)^2}\right) [/math] get rid of roots [math] m^2\left((x - j)^2 + (y - k)^2\right) = n^2\left((x - r)^2 + (y - s)^2\right) [/math] multiply out brackets [math] m^2\left((x^2-2jx+j^2) + (y^2-2ky+k^2)\right) = n^2\left((x^2-2rx+r^2) + (y^2-2sy+s^2)\right) [/math] multiply through [math] m^2x^2-2m^2jx+m^2j^2 + m^2y^2-2m^2ky+m^2k^2 = n^2x^2-2n^2rx+n^2r^2 + n^2y^2-2n^2sy+n^2s^2 [/math] gather like terms [math] m^2x^2 - n^2x^2 + m^2y^2 - n^2y^2 -2m^2jx +2n^2rx -2m^2ky +2n^2sy = +n^2r^2 +n^2s^2 - m^2k^2 - m^2j^2[/math] factorise a bit [math] (m^2 - n^2)x^2 + (m^2 - n^2)y^2 -x(2m^2j -2n^2r…

hello the second post [math]x^2+4/2[/math] worked