A kilo watt hour question?
953813
You are right, perhaps your teacher calculated for 10 hours of operation.\n\nand technically you are not allowed to multiply by .h unless you want to calculate the cost for a definite time, but it works here because you want to find the cost for one hour (i.e 1.95 kW for One hour gives you the 1.95 kWh)\n\nRemember: kW represents power\n                     kWh represents energy
327062
kWh is NOT Kilowatts per hour. It is Kilowatt Hours and is the measure of the amount of energy drawn in one hour hour if it is drawn at a rate of one Kilowatt.\n\nKilowatts are a measure of Energy per unit time ie. Joules/s so kWh are J/s x 3600 s ie. measured in Joules.\n\nThe peak is just the highest energy usage rate in a given time period
1031597
That 1kg of hydrogen would convert to about 0.954 kg of helium, the other 0.046 kg is changed to energy.\n\nE=MC^2\nM=0.046 kg\nC=300,000,000 m/s\nE=4.14X10^15 joules\n  = 1,150,000,000 Kilowatt-hours
393670
‘watt’ is the unit of power. It means joule per second.\n\n12 watt means it can impart 12 joule per second.\n\nTherefore, the answer to your question is it will impart 12 watt to the battery.\n\nIn one hour it will impart  12x60x60 joule; 12 watt-hour (energy-not power).\n\nIn electrical unit one watt = 1 Volt-Ampere. \n\nIf the battery of 12 V is charged for an hour using 12 watt power, then it can give a current of one ampere for an hour. If the current is reduced to 0.5 ampere then it will work for 2hours. If the current is reduced to o.1 ampere, it will work for 12 hours.
2752
A. Potential energy = Force * Distance\n     = (50 * 9.8) * 10\n     = 4900 J\nB. Power rating = Energy / time = 4900 / 5 = 980 Watts
1187482
Kilowatt\nCost depends on method of generation and location
1186419
That would depend on the voltage. Current = Watts / Volts\nA couple of examples:\n110V - 90.9 Amps\n220V - 45.45 Amps\n240V - 41.67 Amps\n440V - 22.7 Amps\n10,000V - 1 Amp
129261
Kilowatts  Equations\nDirect Current  =        I x E/1000\nAC / 1phase 115v or 120v =I x E x PF/1000\nAC / 1phase 208,230, or 240v= I x E x PF/1000\nAC 3 phase All Voltages     =    I x E x 1.73/1000\n\nE = Voltage \nI = Amps \nPF = Power Factor
1015451
Power = work done per unit of time.\n\nP=W/T
254996
If PC power supply is 300 watts,during 12 hours,it consumes 3600 watts ( 3.6 kilowatts) or less
38597
P= work/time and work is = to force x distance so its 500x4 =2000joules\nand 2000/4 =500 watts
655289
The units of:\n[Power] = Watts\n[Current] = Amps\n[Time] = Seconds\n[Energy] = Joules\n     \n      also\n\n[Energy] = kWh\n\n1kWh = 3.6 X10^6 Joules\n\nPower = Current X Voltage\n\nEnergy = Power X Time \n\nAfter performing the previous calculation, the units of energy must be converted from Joules to kWh, since electric companies charge in dollar per kWh.\n\n\n$ = Energy (measured in kWh) * Cost of energy in Dollars per kWh\n\nUsually, the cost of electric energy is in the neighborhood of $0.05 and $0.10 per kWh.
824682
The energy absorbed is that which is not outputted.  In this case, 75% of the energy is absorbed, so the energy absorbed per cycle is .75*8500 J.  One watt = 1J/sec.  So seconds = J/watt.  The power is constant, so the power in each cycle is 6,00 kW; this times the duration of the cycle will be the energy in each cycle.
89877
joules are an amount of power with an equivalent in calories btus and kilowatt-hours
1030224
The answer in SI units is 1200 joules or 1.2 kilojoules. the definition of a joule is one watt second. in Calories, or English units, which shouldn't really be used with watts as it mixes systems of measurements, is 286.806883 calories or 0.286806883 kilo calories.
121569
Mmm. 9.2 hours of sleep sounds nice.\n\n4900 kcal
833466
Volts times amps = watts.