what is 50 kilo volts when written in volts?
924875
50,000V\n\nOtherwise, if you use Kilo, you just use the K - eg: 50 KV
1130014
V=IR\n\nWire 1:\n50 = 2 * R1 => R1 is 25 ohms\nCurrent = 2 amps\npower = V*I = 2*50 = 100 watts\n\nWire 2: \n100= 2* R2 => R2 = 50 ohms\nCurrent=2amps\npower = V*I = 100*2 = 200 watts\n\nare you looking for ratios when you say "compare" ?\nif yes: ratio of resistances is  1:2\nratio of currents is 1:1\nand ratio of power is 1:2
514134
i think this is the equation\nE=VxI\nE is energy. V is volts . I is current which is measured in Amps\nE= 120 x 12.5\nE=1500Watts
768635
Well we are given voltage, and power.  We are asked to find the current. Well the formula we have to use here is P=VI.\nRearranging this formula we get; I=P/V\n\nHowever; there is only one thing wrong with this problem.\nWe do not know what the voltage is correctly.\nIf this diode is a ideal diode. An ideal diode acts like a short circuit with no voltage drop across it. Then we would use .8V as the Voltage Source across the diode.\nIdeal Diode:\nI = .05 W/.8V\nI = 62.5mA\n\n\nIf it is not an ideal diode and there is a .75V drop across the diode then we would use .05V as the Voltage Source across the diode.\nI = .05W / (.8V - .75Vdiode drop)\nI = .05W / (.05V)\nI = 1A\n\nHere this might give your more information!\nhttp://en.wikipedia.org/wiki/Diode_modelling
786456
a) 2900 * (54 + 0.29 v)\nb) 9 * 2900/v\nc) 2900 * (54 + 0.29 v + 9/v)\nv = 5.57
713107
range/amount\n\nrange = 120 mV (90mv - (-30mV))\namount = 2^16 = 64*1024\n\nplug-n-chug
188896
K = mv^2 = (10kg)(4.5 m/s)^2 = 202.5 J\nI would suggest you try doing your homework, asking for all the answers will only hurt you in the future.
761484
II think the letters you are looking for are RMS: root mean square. The value you see on a voltmeter is the RMS value of the voltage, which is about 70% of the zero-to-peak voltage of the waveform.\n\nRMS can be calculated by taking the input voltage waveform, say a sine wave, squaring it (which makes all the voltages positive), taking the mean (average) voltage, and then taking the square root (to undo the value change from the squaring operation while leaving the value positive).
697895
E = mc^2 = 1kg * (3*10^8)^2 = 9^10^16 Joules
648766
The other 2 answers are basically right, but Kes has no clue.\n\n3 phase voltages are normally specified by the phase to phase voltage (say 480V).  If a neutral is present from a wye transformer, the phase to neutral may be specified second, such as 480Y/277.  But just saying 480Y means 480 phase to phase in the wye configuration.  The phase to neutral voltage is the phase to phase voltage divided by the square root of 3 (about 1.73).\n\n480V 3 phase is not something to play with.  If you don't know what you are doing, you shouldn't even be in a position to be able to measure the voltage.\n\nTo Kes, normal home wiring is single phase 240/120.  2 phase is a system that existed before 3 phase became standard.  For what you call "440 V" for motors, it would be 440 V between phases.
283818
nihilistic is incorrect, kVA is actually a measure of power, not voltage.  high voltage or low voltage, the units are still volts.  \n\nhowever for high voltages, rather than saying 125,000 volts, people often say 125 kilovolts or 125 kV.
906536
These are a published value.  The actual voltages vary as you can see from this quote from a power plant site \n\n"POWER PLANT CAPABILITY: The two hydroelectric power generators boast a generating capacity of 92,000 kilowatts. The estimated annual energy output is 205 million kilowatt-hours."\n\nAnd none of the values quoted are a multiple of 11.  It appears that the source your using just happened to have values that were multiples of 11.\n\nA quick look on Google listed quite a number of plants that produced values that had 13800kv to 205 million kv.\n\nChance that the ones you looked at were multiples.\n\nBest of luck in your project.
246402
Latent heat of vaporization of water is 2260 kJ.kg-1.\nConverting 50g to kg = 50/1000= 0.05kg\n\nTherefore 2260 KJ/KG x 0.05KG = 113 KJ or Kilojoules or 113,000 J or Joules is your answer
871859
Nominal voltage required for excitation of a circuit.\n\n- - - - - - -s-
499258
.5kg =  500.0  g\n50mg =    0.05 g\n          2.5  g \n       =======\n        502.55 g
657790
They are not proportional.  Watts are units of power, volts are not.  Similarly, you can't ask "how many kilometres per hour are in a kilogram?"\n\nIn direct current conditions, the power expended in a device (in Watts) is equal to the voltage across the device (in Volts) times the current passing through the device (in Amperes).
918355
1 Ah = 1000mAh\n\ntherefore 20 x 50mAH = 6 volts at 1000mAh.\n\nHow many volts did you want?