Power input = 700 WEfficiency = 75% so power output = 75% of 700 = 525 W
Time = 30 seconds => Work done = 525W * 30s = 15750 W.s = 15750 Joules = 15.75 kiloJoules.
The result is 600 seconds or a decrease of 150 seconds.
40/120 = 1/3 = 331/3 %
75%
45/60 x100=75%
First, you need to determine how many seconds there are total in the month of June: 30 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute = 2,592,000 seconds. Now, multiply that result by 15 percent, or 0.15: 2,592,000 seconds x 0.15 = 388,800 seconds.
A Watt is a joule per second. So in a whole hour there are 360 seconds. 500 x 360 = 180. kilojoules in a whole day there are 8640 seconds. 500x 8640 = 4320 kilojoules
Let's assume that this is an incandescent bulb with a light efficiency of 35 percent ... i.e. 35 percent of the electric power it consumes is converted to light, and the other 65 percent becomes heat, which the bulb then dissipates into the room. So the bulb is dissipating heat at the rate of (0.65 x 40) = 26 watts = 26 Joules per second. 2 hours = (2 x 3,600) = 7,200 seconds 26 joules per second x 7,200 seconds = 187,200 Joules of heat
The duration of Electric Ink is 1800.0 seconds.
The duration of The Electric House is 1320.0 seconds.
The duration of The Electric Grandmother is 3600.0 seconds.
The duration of Electric Earthquake is 480.0 seconds.
30 seconds is 50% of a minute.
what makes a electric motor get very hot and then shutoff after 20 seconds
50.4 seconds.
A kilowatt hour is a metric unit, although it is a slightly confused one. A watt is one joule per second. This means that 1 kWh is 1000 joules multiplied by 360 seconds (1 hour) per second. So 1 kWh is 360 kilojoules.661 kWh is 661 times 360 kilojoules, which is 237960 kilojoules, or 237.96 megajoules.
150%
An electric current is the amount of electrons per unit time or seconds.