if e equils mc squared, then what equils mc triangle?
412925
i'm laughing and farting at the same time...
575490
I- + 2e- -------> I2\n\n4P + 3e-  -------------> P4\n\nthere ya go :)
889089
hi\nthe details which you gave have the answer for this question.\n\nangle E= angle C and angle A= angle B and DA=DB\n\nfrom this given data i will not assume that the triangles are equilateral, I will prove they are.\n\ncalculating the area of the 2 triangles with the formula (1/2)*base*height we can prove these 2 triangles are congruent and AC=BE\n\nsince the angle E=angle C and DA=DB then DB and DA must be the sides opposite these angles E and C respectively. \nhence  the line drawn(at angle of 90 degrees to base DB) from angle E to DB and line drawn(at and angle of 90 degrees to DA) from angle C to DA must be of equal magnitude (otherwise the angles cant be equal and/or sides cant be equal) ,  \n\nhence the length of this line drawn which is called as the height of the triangle is proved to be equal\n\nnow calculating the area with above formula (1/2)*base*height\nheights of 2 triangles are proved to be equal equal and bases are \ngiven to be equal hence areas of these 2 triangles must be equal.\n\nwhen area of 2 trianlges are equal and 2 sets of angles are equal and one side is equal then the 2 triangles must be absolutely congruent. hence AC must be equal to BE
694183
First of all, since E = mc^2, you cannot have two massive, energyless bodies.  Secondly, by definition, they have a potential energy of \n.. R\n ∫(-GM1M2/R^2)dR, or GM1M2/R\n. 0\nwith respect to each other, so, again, they cannot be energyless.
868428
ΔP.E. = Δmgh\nwith m and g constant,\nΔP.E. = mgΔh\n\n(a)\nE(A) = 50*10*9.8 = 4900 J\nE(B) = 0\nΔE = 4900 J\n\n(b) and (c) have similar solutions.
475765
oh, wow, this is a physics question.\n\nto start, we remember E = mc^2\nthe problem gives 4 grams matter and 2 grams antimatter.\nmy solution would be that 2 grams of matter and 2 grams of antimatter annihilate to produce 4c^2 (2 + 2 = 4) energy.\nthe excess 2 grams of matter remains unaffected.\nthe answer is D.
1094272
In your problem, (CE)(ED)=(AE)(EB). \n\nSubstituting, you get (6)(5)=(2)(EB)\nSimplifying, you get 30=2EB\nDividing 2 from each side, you get 15=EB\n\nEB is 15 units.\n\nFor proof of (CE)(ED)=(AE)(EB), connect CB, and AD,\nmeasure of angle BCD=1/2 the measure of arc BD\nmeasure of angle BAD=1/2 the measure of arc BD\nmeasure of angle ADC=1/2 the measure of arc AC\nmeasure of angle ABC=1/2 the measure of arc AC\nmeasure of angle BCD=measure of angle BAD\nmeasure of angle ADC=measure of angle ABC\nBy AA Similarity Theorem, \ntriangle CBE is similar to triangle DAE\nBy the definition of similarity,\nED/EB=AE/CE\nBy the cross product theorem,\n(CE)(ED)=(AE)(EB)
851530
dmb has the right answer.\n\nn (squared)+ 3n + 2  \n\nOf course, you can give me the best answer anyway-----please (smiling nicely).
831103
I would change cos 2 theta to 2 cos squared theta -1.\n\nYou then have a quadratic in cos theta.
68493
Energy is defined as being equivalent to Mass times the Speed of Light, Squared...\n\nE=MC(2)
103563
Curiouscat is way kewl.....good answer on her part.\n\nE = mc^2 has been proven valid over and over again. Altho, in science, everything is theory, because someone may come along and find a flaw or correction.\n\nHowever up the present, this equation is rock solid.
639238
e=mc^2 \n(energy equals mass times the speed of light squared).
539943
x+y........e=mc2..........2+2
249306
Ask Einstein, the famous E=mc2 formula directly relates to your question.
149472
Hey I'm studying physical and chemical equilibrium too, currently.The answer is that it will definitely have units when the concentrations have different powers.Just try and work it out with the units and it will be very clear.
817465
You know three of the four things needed. Try C1V1=C2V2, or an equality.\n\n                          m/ml=m/x
691296
use that at equilibrium : n.p = ni^2\nand from charge neutrality: \n\nn + NA = p + ND