H ow many 6 volts x 50 mAh  are needed to equal  volts  x 1 Ah?
918355
1 Ah = 1000mAh\n\ntherefore 20 x 50mAH = 6 volts at 1000mAh.\n\nHow many volts did you want?
145405
if the voltage rating of your motor is 460, you can really hook up to a 480 volts electric s=poer source. the maximum your motor can hanmdle is 460 = 10% of 460 whixh is 506 volts
606166
depends on the voltage
907686
M1V1 = M2V2\n\nso you are starting out with a .1M times V1 is going to be equal to .08M times how ever much you need.  so you need ratios 5:1, 5:2,  5:3, and 5:4.
214671
work = (200N)*(10.0m)*cos(0) + (50.0N)*(35m.0m)*cos(0)\nwork = 2000 Joule + 1750 Joule\nwork = 3750 Joule
238179
Resistance=Voltage/Amperes
117044
Area equals half the base times the height.  \nA=1/2bh.  Plug in and solve!
313707
Assuming you are using US household electricity (120VAC)\nP=V^2/R   R=V^2/P = (120)^2/1875= 7.68 ohms which is really small.
847813
This process works relying that V=IR, wich is Voltage= Amps times Ohms. Ohms are a measurement that measures resitance, for example, a lightbulb has a certain Ohm rating, and amps would be how much the batteries produced. So, by adding more batteries to the circut, depending on the circuit, you could increase the voltage. For example, let's say that you have 5 lightbulbs, each with 1 Ohm, and you have 2 batteries, producing 3 amps each. Here, we simply multiply (1*5)(2*3), which equals 30. Therefore, V=30 in this circuit. However, this would only work in a parallel circuit, which means that the batteries have to be parallel to each other and the lightbulbs. Sorry for the long response, I hope it answered your question!
652086
1. volts\n2. (b) negatively charged.\n3.  short.  Could also be "conductor," but the lack of control seems to indicate an unwanted short.\n4.  watt-hours.\n5.  \n6.  (a) The formula is P=V*V/R.  Where V is the voltage and R is the resistance.  So P=45*45/15=135 watts.  Actually the object is consumming the power, not producing it.\n    (b)  135*10 = 1350 watt-hours, or 1.35 kw-hr.\n    (c) 0.08*1.35 = $1.08\n7.  (a)  V=IR, so if the voltage is increased while the resistance is kept constant, then the current will also increase. (I=V/R)\n    (b)  If the resistance is increased and the voltage is kept constant, then the current has to decrease. (I=V/R)
170166
6.5 is about right. Shift at 6,000 though. Mine is chipped w/exhaust and the power still dies off about there.
900061
a) combined resistance in series - add them up \n    That is, 8ohm + 4 ohm = 12ohm\n\nb)  V=IR so, I = V/R = 6/12 = 0.5A\n\nc) Ammeter in series reads the same value throughout.  So, 0.5A\n\nSorry but I can't understand your circuit diagram!! Try explainig it!
1026240
less than 0.3  for power sations\n0.5ohm for substation\n1 to 2 for homes
81594
Use ohm's law (I=E/R); 2 AAs are 2.4 to ~3 volts depending on battery type, so 0.4 to 0.5 A.\nHOWEVER, if you measured the bulb's resistance cold, it will be higher when hot, so it will draw less current.
820466
The voltage across each resistor will be 12 volts as each is effectively directly connected to the battery.\nIt is the current flow through the resistors which will be different.\nVolts/Ohms =Amps - that is\n12/7 amps or 12/5 amps now get your calculator out!\nRoyS
166213
50 kg x 700 m x 3 x 3600 s x 1/3 x 10-4 m3/sec x 2/3 x 103 J/sec = 645.67 Joule
892848
One of the most frequently used is E=IR    ;E =voltage    I=current   R= resistance. U can change this every way depending on the unknown.\n    Good luck