I need help solving these electrical formulas?
818849
looks like straightforward algebra, and isn't really relevant whether the formulas are electrical
321670
Using the sheet from this website http://www.bced.gov.bc.ca/exams/search/grade12/english/release/additional-material/0508ch_databooklet.pdf \n1/2 rxns:\n1) 2 Ag+1 + 2 electrons --> 2 Ag(s)+.80 V\n2) Mg(s) --> 2 electrons + Mg2+ +2.37 V\nBalanced) 2  Ag+ + Mg(s) --> 2 Ag(s) + Mg2+ Exp. voltage = difference between 0.8 and 2.37 = 1.57 V
1102009
q = I t\nq = 2.5 C/s x 900 s = 2250 C = 2.3 kC\n\nP = 120 V x 2.5 A = 300 watt\nE = P x t\nE = 300 J/s x 900 s = 270 kJ
733256
Single phase circuits have two terminals one "hot" and the other ground.  v(t)=120*cos(2*pi*60*t)\n\nThree phase have three hot terminals (each 120 degrees out of phase with one another) and one ground:\nv1(t)=120*cos(2*pi*60*t+0 deg)\nv2(t)=120*cos(2*pi*60*t+120 deg)\nv3(t)=120*cos(2*pi*60*t+240 deg)
368244
V (Voltage) = I (Current) x R (Resistance)\n\nOr (although you probably don't need it):\n\nJ (current density) = sigma (conductivity) x E (electric field)\n\nI always use wikipedia to look up stuff like this, it's very helpful for all kinds of info.
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
315859
you got Ohm law : Ur=R*i(where i stands for current) and what's more, for the coil we have Ul=-L*di/dt . And for the condenser we have C*Uc=q and i = dq/dt.Now you have a differential equation with only one variable i.\n note: if the circuit is parallel then U should be a constant , in this case you shuold use U instead of i.Goodluck
864021
[a b]\n[c d]\n\n1. E=[2 -2]\n ____[-1 2]= 2*2-(-1)(-2)=4-2=2\nE^-1=1/2 [2 2]  = [1   1]\n________[1 2] = [1/2 1]   \nYou multiply 1 divided by ad-cb and switch a and d and make b and c negative.\n\n2.  E= [1 -1]= 1+1=2\n______[1 1]\nE^-1=1/2 [1 1]   =   [1/2 1/2]\n________[-1 1] =  [-1/2 1/2]\n\nHope this helped!
238179
Resistance=Voltage/Amperes
160821
1. Dy = 1/2 A*t^2 + Vy*t, Vy = 0 (speed is horizontal),\nso t = sqrt(66m/(1/2*9.8m/s^2)) = 3.67 s.\n   Dx = Vx*t = 6 m/s * 3.67 s = 22 m.\n\n2.(a)  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso t = sqrt(1.150m/(1/2*g)) = 0.484 s\n  (b)  Dx = Vx*t, so Vx = Dx/t = 0.385m/0.484s = 0.795 m/s\n\n3.  Dy = 1/2 A*t^2 + Vy*t, Vy = 0\nso Dy = 1/2*9.8m/s*(2.9s)^2 = 41.2 m\n\n   Dx = Vx*t = 2.4 m/s * 2.9 s = 6.96 m\n\n4. Dy = 0, Dx = 97 m, a = 50 degrees\n   Dx = Vx*t = V*cos(a)*t, \nso t = Dx/(V*cos(a))\n   -1/2*g*t^2 + Vyt = Dy = 0 ( Vy = V*sin(a))\ndivide by t, so -1/2*g*t + V*sin(a) = 0, or\nt = V*sin(a)/(1/2*g)).  Equate the two expressions for t:\nV*sin(a)/(1/2*g)=Dx/(V*cos(a)), multiply both sides by V\nV^2*sin(a)/(1/2*g) = Dx/cos(a)\nV= sqrt(Dx/cos(a)/sin(a)*(1/2*g))\nV = 31.1 m/s
972428
C=80uF\nV=Voc1.5 volts\nI=Isc=5.0 amps (assuming short circuit current)\n\nSo thevnin's theorem implies the \nRth = Voc/Isc = 1.5/5 = 0.3ohms\n\nSo the circuit is an RL circuit,\n\nTime constant T = RC = 0.3*80uF = 24us\n\nNow you can use that formula for charging
741999
whats the equation or formula u need help with?
653976
You need to use Ohm's law.\n  E=I x R  \n  E=Voltage. I=Current/Amps  R=Resistance/Ohms\n  E= 9voltsDC\n  I= 3amperes.\n  R= ?\n  So:\n  R = 9volt/3amp\n  R=  3ohms
663777
SOrry, I can't help you. If I had the formulas I could do it, but there is no way I remember that off the top of my head! Good luck
97533
+12VDC, -12VDC, +5VDC, -5VDC
788402
You can find the current flowing through a Wheatstone bridge by using Kirchoff's Loop Rules and Junction Rules.      \n\nThe sum of the Voltage Sources - Voltage Drops = 0 in a closed circuit.  \n\nThis setup can be solved using a matrix or a CAS. \n\nGood Luck
1143501
you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp