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According to Gaussess law. Total outward flux across any closed volume surrounding the charge q is q $$ E •dS = q/eo $$ E • dS = $$$ ddV/eo $$$div E dV = $$$ d dV/eo $$$$(∆E-d/eo) dV = 0 As the result of integration is equal to 0 then ,integrant should be equal to 0 So , ∆E= d/eo If no charge then ∆E=0 Again as E = - grad V So , ∆^2 V=0 This is refer as LAPALCES EQUATION. For Cartesian Coordinate system d^2V/dx^2 +d^2V/dy^2 + d^2V/dz^2=0 For cylindrical coordinate systen (r, @,z) The Laplaces equation become 1/r^2 d^2 V/dr^2 +1/r dV/dr +1/rsin@ d^2 V/d@^2 + d^2 V/dz^2=0
Spark should have only two identical charges in contact but while the top head connected then as contact took place so as whole it is only one kind of charge and at the same time in the same body like wire .So based on theoretical conception wise no spark should be observed sir while the top may joined .
The best mathcamp for econ I have ever learned, detailed and clear. Easy to catch on. Helped me figure out many questions I couldn't deal with before. Thanks to UCI and Professor Kronewetter!
But why the density if the olate even matters for subduction? I mean yes the oceanic olate is more dense but it was still floating before colliding with the continental plate so it is less dense than the liquid it is floating on and that's what matter for floating what actually happened at the collision that the denser plate gets subducted??....i mean what are the forces involved...at the time of collision?
Seems there's been a mistake.even100/3<x<500/9，you still can't guarantee that you can get a positive profit.see，we use x=50 for exam(like professor)，then，you loss 50$ and win 62.5 so finally there 62.5 in you pocket（case1）or you win 100$ and loss 50 so finally there 100 in your pocket. That can't let you 100% win.because when case1 happens(I don't use one team name because it buzzles me),you finally have 62.5$ in your pocket,but remember, you have 100$ first, that's your own original money,so you really lose 37.5$ when case 1 happened. The promble is (-9/4x+125)or(3x-100)is not your profit,it just your finally money you get in you pocket,the profit should minus 100,and let the displayed formula>0:(and I find the money in you pocket displayed formula also is a mistake,It should be:5/4(100-x) and 2x) P=[5/4(100-x)]-100=[2x]-100,then you get x≈38.5，then P≈-23,yeah, you still lose money（If you don't believe, you can do this,x= 38.5:case1 happens：you loss 38.5 in one guy but you win1.25*(100-38.5)=77(76.785) in another guy,so finally you get 77 in your pocket ,and77-100=-23;case 2 happens: you win 38.5*2=77 in one guy but loss(100-38.5)=61.5 in another guy, so finally you get 77 in your pocket,and77-100=-23.perfect!x=38.5 is the extremum） SO，if you want use your 100$ win some change，the odd in this example should be changed, in this example you can't win, you gonna lose
Maxell Thermodynamics based equations depends on four variables S P T & V dS/dP| T =dT/dV| P dV/dT | S=dP/dS| V dP/dS| T = dV/dT| S dT/dV| P =dS/dP | T All are partial derivatives refer as Maxwell Thermodynamics Equations
It is analogous to TRANSFORMER,the emf induced in the secondary coil due to induction e= - dN/dt where N is the magnetic flux which is define as number of lines of force crossing unit area NORMALLY .
Unit of charge is define by considering test charge qi with parent charge Q so untill the ELECTRIC FIELD E= 1/4πe0 Q/r^ 2 as F =qiE untill this value not given constant value then for the specific of charge for that F= qi E = Constant then such test charge is define as unit charge.
Last five years the ocean having raised temprature 1° C. So this is supporting statement that in each day about 4500 cm^2 area received heat energy in 1 second 675 JOULE then estimation suggest this statement.
l2=l1 [1+a(t2-t1)] where a is the coefficient of linear expansion having in relation a= b/2 = y/3 where b is surface expansion & y is the volume coefficient of expansion respective ly
Nice problems. However, in the last problem, why was the volume of the lake immediately rounded off? Shouldn't the rounding off be done at the last part of the calculation? This would lead to 6*10^5 kg (631508.1681 kg) instead of 7*10^5 if my calculations are correct.
