1 mega watt is equal to 1 million watt or 1000000 watt.
1 kilowatt = 1,000 watts 1 megawatt = 1,000,000 watts or 1 thousand kilowatts.
Depends on which kind of waste is used for the production.
5 HP MOTOR WOULD CONSUME ENERGY OF 1342800 JOULES IN AN HOUR.EXPLANATION :-1 HP = 746 WATTSTHEREFORE 5 HP = 3730 WATTS.= 373O Joules/sec (since 1WATT= 1 Joules/sec)=3730 x 3600= 13428000 Joules/ hour.
Depends on which kind of waste is used for the production.
1Watt = 1 J/sec Therefore: 100 J/sec * 10 sec = 1000J
You can't directly convert one to the other. A joule is a unit of energy, and a watt is a unit of power. 1 watt = 1 joule per second.
The SI unit for a quantity of energy is the Joule.The SI unit for the rate of moving energy is the watt = 1 joule per second.
we know that power=work/time therefore, p=1 joule/1 second =1js-1 =1watt (W)
A micro-amp is a measure of current. Power in Watts is a function of the current and voltage. Impossible to answer your question without knowing the voltage. For example, 1 uA (micro-amp) x 10 Volts = 10 micro-Watts, but 1uA x 1,000.000 Volts = 1Watt.
The conversion is not direct, since watts are units of power, while BTU/Hr represents energy per hour of equipment operation. So, in this case, watts will represent the power dissipation of a product. 1 watt = 3.412141633 BTU/Hr It works both ways, of course: 1 BTU/Hr = 0.29307107 watts The above is mathematically correct, but I think how it represents the time value is confusing. I've left it for reference purposes. A Watt is an instantaneous measure of power. It assumes no unit of time. A Watt-Second on the other hand, is 1 Watt of power for 1 Second, which *is* a unit of energy. 1 Watt-Second is equivalent to 1 Joule. A 60 watt lightbulb turned on for exactly 1 second uses 60 Watt-seconds or 60 joules of energy. BTU on the other hand, is a measure of energy and time is implicit in the unit. 1 BTU is equivalent to approx 1055 Joules. 1 Watt-Second = ~0.00095 BTU. A 1000 Watt heater, running for 1 hour(3600 seconds) would consume 1KiloWatt-Hour(KWH) of energy. Or 3,600,000 Joules. 1KiloWatt-Hour = 3412 BTU. OR 1Watt-Hour = 3.412 BTU While you could divide both sides by Hours and get a comparison of power wherein 1W = 3.412 BTU/hr. I think comparing energy in the forms of KWHs and BTUs is more useful than the other way around.
In a.c. circuits, the watt is used to measure the true power of a load, and is determined by multiplying the supply voltage by the load current by the power-factor of the load. The volt ampere is used to measure the apparent power of a load, and is determined by multiplying the supply voltage by the load current. So the relationship between the watt and the volt ampere depends on the power factor of the load. For example a 100 VA load with a power factor of 0.8 (leading or lagging) will have a true power of 80 W.
Impossible question, like asking "One Apple is equal to how many oranges" Simply set : Decibels it the unit of volume, how loud things go hertz is how fast things change, the frequency (number of changes per second). When you talk about music, different frequencies can have a different volume. You can even measure the volume in frequency ranges - look up : paragraphic equalizer for the principle. Geert Fieuw Beyond The Labyrinth ----------------------------------------------------------------------- +db= 10*log(p2/p1) +db= 10*log(2watts/1watt) +db= 10*log(2) +db= 10*.30103 +3.0103= 10*log(2watts/1watt) so double the power you gain 3db's so double the sound you gain 30dbs and p2 is 1000 times p1 +db= 10*log(p2/p1) +db= 10*log(1000watts/1watt) +db= 10*log(1000) +30= 10*3 test signal say 60 Hz
No, four zener diodes in parallel will not increase the power rating of the set. This is because one zener will always conduct first, so it will be the only one that has current through it. If, however, you put four zener diodes in series, each with a voltage rating one quarter your desired voltage, you can achieve a higher power rating for the set.
the first convert the power in dBm to MW, the define of dBm=10 log (P MW) -10 log ( 1mw). example: let P=-2 dBm convert this to dB? answer: Pmw= inv log(-2/10)=0.630mw*1000 micw/mw=630 microw 10log(630)=28dB
Depends on the current. Put a resistor in-line with the current, then measure the voltage across the resistor. V=RI. So, divide the measured voltage by resistor value. Be careful with the size of the resistor, as Power dissipated in a resistor is R*I^2 or V^2/2. So, a 1-Amp current into a 1 Ohm resistor will result in a 1Watt power dissipated in the resistor. If it's too small, it'll burn. Also, notice that if you do that, you haven't measured the current in the original circuit. You've measured the current when an extra resistor is installed in the original circuit, and that's different.