this unit of H/P power , what does it means
133629
Pferdestärken (German: horse power)
339136
Power is the rate at which work is done. It is the work/time ratio. \nWork = force * distance\n\nWhat force was used to stop the car?\nWhat distance did it travel?
482697
h^2 = p^2+b^2\n\nwhere h--- hypotenuse\n          p--- perpendicular\n          b--- base
156010
The unit ma (correctly written as mA) is a unit of current, not voltage.\n\nTo be perfectly technical, mA refers to miliAmps, where as MA refers to MegaAmps.  1 MegaAmp = 1 billion milliAmps.I can only assume your meant mA since you would have to have one huge solar panel to be outputting MegaAmps of current.\n\nAlthough there is a releationship betweent electrical current and the voltage in a circit as given by ohms Law,\nV = R * I,\nyou cannot use it in this case to determine voltage.\n\nYour best bet to figure out the voltage output in your solar panel would either be to look at the documentation (if any), or just get out a voltage meter and measure it directly.
270066
Hmm, I knew that one horsepower was equal to 746 watts, but I didn't know the original definition.  From Howstuffworks, 1 horsepower is equal to 33,000 ft-lbs of work in one minute (or as the example in the link says, lifting 330lb of coal 100ft in a minute).  Interesting.\n\nAnyway, a motor's power output rating on the nameplate should be equivalent regardless of the voltage it uses.  The only difference will be that the 110 volt motor will draw more current than the 550 volt motor given the same loading (edit: and motor efficiency).
92914
Starting with Ohm's law:   P = I * V\n\nSo in your example: \n P1 = 120V x 25A = 3000W\n P2 = 120V x 30A = 3600W\n\n Ptotal= P1 + P2 = 7600W or 7.6kW\n\nNow, energy consumed in kW-hr is just the power multiplied by the duration in hours:\n\n  Say that you held this load for 2 hours, then:\n  \n  E = P x t = 7.6kW x 2 hours = 15.2kW-hr
516679
If a voltage spec (v volts) is given on the resistor, then the power\nrating, p = v^2/r  ( r = resistance in ohm, you already know).\n\nIf v is not given then the only way is to connect the resistor to the poles of a battery cell via an Ammeter, and see the electric current measurement in ampere (say, i amperes).\n\nThen p = i^2 x r  (p, in watts).
634649
Look on the AC/DC power adaptor that came with your computer.  It should say how much power it supplies.  It's on the brick that is between the AC outlet and the DC power plug on your computer.  For mine, it says OUTPUT: 19V - 6.3A(6.3A).  Multiply those two numbers together, and that's how many watts your computer requires.  I have a P35, and I require almost 120 watts.
530920
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
661732
I assume DH is delta H ? If so - it's the change in heat. Final minus initial.
51178
Volts is not a measure of power. It is a measure of potential energy. To measure power you should be thinking watts. So a 100W lightbulb is twice as powerful as a 50W lightbulb (W stands for Watts).\n\nNow suppose you have a 100W bulb at 100 volts (100V). It must be consuming 1 Ampere (1A) of current because Watts = Volts x Amperes. But a 100W bulb at 10V would draw 10A (10x10). It would be wrong to say that the 10V bulb is less powerful than the 100V bulb if they were both 100W.\n\nThat being said, real light bulbs are rated in watts at their operating voltage. So a 100W bulb at 110V draws about 1A, but if you put 12V through it, it would not draw enough current to be a 100W bulb anymore. In other words, don't think that because a 100W bulb is marked 100W that it always draws 100W under any circumstances -- it is just a convinient marking to guage relative strength.
68024
It stands for Raw horsepower and Raw torque. Did you need to know what horsepower and torque were too? Wasn't sure.......maybe I should add that too eh??\n\nHorsepower: Awhile back, a gentleman by the name of Watt (the same gent who did all that neat stuff with steam engines) made some observations, and concluded that the average horse of the time could lift a 550 pound weight one foot in one second, thereby performing work at the rate of 550 foot pounds per second, or 33,000 foot pounds per minute, for an eight hour shift, more or less. He then published those observations, and stated that 33,000 foot pounds per minute of work was equivalent to the power of one horse, or, one horsepower. \n\nTorque: A foot pound of torque is the twisting force necessary to support a one pound weight on a weightless horizontal bar, one foot from the fulcrum. \n\n\nOK. Remember Watt? He said that 33,000 foot pounds of work per minute was equivalent to one horsepower. If we divide the 6.2832 foot pounds of work we've done per revolution of that weight into 33,000 foot pounds, we come up with the fact that one foot pound of torque at 5252 rpm is equal to 33,000 foot pounds per minute of work, and is the equivalent of one horsepower. If we only move that weight at the rate of 2626 rpm, it's the equivalent of 1/2 horsepower (16,500 foot pounds per minute), and so on. Therefore, the following formula applies for calculating horsepower from a torque measurement: \n\n\n                                Torque * RPM\n\n        Horsepower      =       ------------\n\n                                    5252\n\nLol...besides the math....here is an article that might help you understand what you want to look for in your car. Good luck :) (but pay off the house first!!)
391074
HP is horsepower 746 watts is one HP
585297
I agree with 'keulenae' but the only exceptions are a cubic meter is approximately 35.3 cubic foot and a power plant has an efficiency less than 1 (at the best case reaches 64%). Therefore:\n1GWH=10^9WH=3.6x10^12 joules\nassuming the average efficiency of power plant be 50%:\nRequired input energy = 7.2x10^12 joules\nHaving each 5.2x10^7joules equal to thermal energy of one Kg gas:\nRequired Gas = (7.2x10^12)/(5.2x10^7) ~ 138461.5 Kg\nAssuming 0.7Kg = 1 cubic meter = 35.3 cubic foot we get:\nRequired Gas = 138461.5 x 35.3 / 0.7 = 6982415.6 cubic foot\n\nSo it burns about 7 million cubic feet to get 1GWH electricity!\n(Don't forget that the above gas volume is almost the normal DAILY production of an standard treating train in gas refineries).
596607
go to http://www.hp-lexicon.org/about/sources/jkr.com/jkr-com-tips.html#ps-man and u will find out
240135
Lower voltage means less electrical pressure, so likely less power consumption.\n\nProbably what you are seeing is line voltage fluctation due to induction devices coming on-line (motors, fluorescent lights, etc.)  Induction devices cause a power phase shift ("power factor") which definitely cause a negative power impact (higher kWh).\nWhen it gets bad enough, the power company will make you install capacitors to counteract the inductive load.  \n\nWhen the phase shift reaches 90°, all power goes into oscillating the current (and making heat) and none into the load.
360713
I kept reading anyway.\n\nThe heater is already telling you the power consumption.\n900W = 0.9 kW\nTherefore running the heater for one hour uses 0.9 kWH.\n\nYour formula is wrong. Watt/1000  * cost/kWH = cost per hour\n\nFor heaters, it is true that Volts * Amps = Watts.\nTherefore if you know the supply is 120V and the power is 900W, you can calculate that the current load is 7.5 A.