at what angle do the equipotentials intersect the edge of the conduction paper? why?
360483
If it is a perfect conductor, it will have equal potential everywhere along its edge as charge will spread itself out evenly.
497010
DUE TO SURFACE TENSION POSITIVE ANGLE OF CONTACT AND INTERNAL PRESSURE.
920208
If the current is flowing into the paper from up to down, then using the right-hand rule, we determine that the magnetic field lines are in concentric cirles rotating in a clockwise direction.
815817
You must have the CELL CONSTANT to do this properly.\nResistance and Conductance are reciprocals.   But to go from\nConductance to ConductIVITY, you need the Cell Constant.\nIt's different for each probe, and should be easy to find on the probe you're using.   Find that and get back to me.
975601
consider this eqn,\n\ndsin(theta) = n (wavelength)\nd = slit separation\nn = no. of order\n\nthe max angle is 90, and sin90 = 1\n\ntherefore, d = n (wavelength)\nwhen d is reduced n is reduced too.\nwhen less fringes appear on the screen, then distance between them increases
569885
This is a good time to learn dimensional analysis.  It will come in handy for all of your physics courses.\n\nFirst, µ x M cancels one another leaving ohms x Farads\n\nOhm = volt/amp\n\ncharge = capacitance x voltage  so capacitance = coluomb/voltage\n\nOhm x Farad = volt/amp X coulomb/volt\n\nso the volt cancels leaving \n\nOhm x Farad = coulomb/amp\n\nAmp = coulomb/sec\n\nOhm x Farad = coulomb/(coulomb/sec)\n\n= (coulomb/1) x (sec/coulomb) = sec\n\nBy the way, Tau = RC is the timeconstant of an RC type circuit.. you remember for a cap charging it is I = V/R[exp (-t/RC)  RC is in units of time = sec\n\nI used to have to do this stuff all the time in my undergrad physics courses.
826187
I am guessing you want the vector projection onto one of the 6 sides rather than one of the 12 edges.\n\nThe diagonal vector is (1, 1, 1).  Projecting it onto any side simply removes the coordinate in the direction of projection.  Projecting it onto the side in the xy plane gives (1, 1, 0). \n\nIf you did mean an edge (the line where 2 sides meet), then the projection is always two 0s and a 1.  The one is in the direction of the edge.
418785
Two ways of doing this:\n\nThe way to do these questions is as follows:\n\nIf you put your pen on each side and follow it around teh polygon - then it will make a full turn - ie 360º.\n\nSo all of the external angles (posh name for the angle you get if you extended the side beyond the corner and measured the angle between that sticky out line and teh next side) add up to 360º.\n\nHence each external angle is 360º ÷ n\n(where n is the number of sides).\n\nNow since the internal angle and the external angle are on the same straight line - they add up to 180º (since a straight line is half a full turn).\n\nSo you can get the internal angle by subtracting the external angle from 180º.\n\nSo for the dodecagon.\n\nExternal angle = 360 ÷ 12 = 30º\nInternal angle = 180 - 30 = 150º\n\nAnother way of doing this whole thing is to note the pattern in the total angles inside.\n\nTriangle 3 sides = 180º\nQuadrilateral 4 sides = 360º\nPentagon 5 sides = 540º\nHexagon 6 sides = 720º\nSeptagon (or Heptagon) 7 sides = 900º\nOctagon 8 sides = 1080º\nNonagon 9 sides = 1260º\nDecagon 10 sides = 1440º\nUndecagon 11 sides = 1620º\nDodecagon 12 sides = 1800º\n\nThe total of the internal angles goes up by 180º each time.\n\nA short cut rule would be:\n\ntotal internal angles = 180 × (n-2)\n\nSo for our dodecagon:\n\n180 × (12 - 2) = 180 × 10 = 1800º\n\nIf all the inside angles are the same, then each one would be:\n\n1800 ÷ 12 = 150º\n\nHope that helps!
653555
Not considering the spheric harmonic of high orders I would have to tell you that my physics spirit tell the surface area  is what counts and there you could have k0.2Pi.R=2Pi <=>k0=1/R\nk1=1/(2R), k2=1/(distance of the 3 points more away from each other)........................
577456
basically all things tend to have minimum potential energy. this can be achieved when the charges stay as far away as possible from each other. \n\nhence when charges are given to a conductor the are distributed uniformly,as in the case of a symmetrical conductor, on the outer surface of the conductor to minimise potential energy. \n\nif the conductor is asymmetrical, then the charge distribution is inversely proportional to the radius of curvature.
48873
well i would like to pretend its a regular hexagon, in this case you would have to draw a line from one vertex to an opposite one, the line should meet at the center of gravity (the point where all 3 lines intersect)
262981
All sides and angles are equal to a side and angle on another figure
703405
If you connect the opposing corners of a parallelogram you make two triangles, thus the area of the triangle is always half that of the parallelogram. Conversely, the area of a parallelogram is always double that of the triangle.
712373
The rule can be simplified thus: If you have to rotate the speed vector clockwise to get fastest parallel with  the field vector then the resulting force enters the paper. (positive charge.)\n1) Inwards (marked with an x on your paper)\n2)(negative charge) magnetic field points to the ceiling. opposite to the result for a positive charge.
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
654251
No, because the charges will have to distribute to opposite sides of the strip (as if it was a loop rather than a mobius strip), however since we are dealing with a one dimensional object (essentially), the charges cannot do this.
221019
Have a look at the reference for an explanation of Bragg's Law\n\nIf the surface plane is plane 0, the next one plane 1 etc.\n\nThe path difference (PD) between plane 0 and plane 1 is 2d sin theta = 1.01 lambda\n\nThe PD between plane 0 and plane a is 2ad sin theta = 1.01*a*lambda\n\nCancellation occurs when the waves are out of phase by half a wavelength, i.e. the path difference is lambda times [an integer (whole number) plus 0.5]\n\nSo we’re looking for a vale of a where 1.01*a = integer + 0.5\n\nSuitable values are plane 50 (PD = 50.5), plane 150 (PD = 151.5), plane 250, 350 etc.\n\nTip: Don't get into X-ray crystallography, it'll drive you nuts!