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What is the total kilowatt hours used if a microwave oven operates on 5 amperes of current on a 110-volts circuit for one hour?
Power = (voltage) x (current) = 110 x 5 = 550 wattsIf operated at this level for 1 hour, thenEnergy = (550 watts) x (1 hour) = 550 watt-hours = 0.55 kilowatt-hour.Another AnswerAs microwave ovens don't operate continuously, but in short bursts, it's not really possible to determine an answer to your question without knowing the frequency and duration of these bursts of energy.
Convert Bs381 224 to RAL equivalent?
bs 381 c 110
How do you you write know in binary code?
k n o w ? First convert it to ASCII code ... 107 110 111 119 (all decimal numbers) Then convert to binary : 1101011 1101110 1101111 1110111
The moving rod is 13.2 cm long and generates an emf of 110 mV while moving in a 0.98 T magnetic field that is facing outwards. What is its speed?
Since E=BLV (Equation for a moving conductor) Solve the equation for V to get V = (E/BL) E = EMF given, which in this case is 110. Convert it to Volts. E = .11 volts V = .11volts / (.98 T * .132 M) = .850 m/sec
If a heater delivers 1500 watts at 110 volts how much the electricity bill will be a month if it is on all the time I leave in Chicago?
Electricity is nearly always sold in kilowatt-hours (kW h). It is important to note that this is not a measure of kilowatts per hour (kW/h) but the product of kilowatts and hours. 1500 watts is 1.5 kilowatts. Assuming a month containing 31 days, with 24 hours per day the month would contain 744 hours. The heater uses 1.5 kW when operating, so multiplying the two gives the result that your heater is using 1116 kW h per month if it is left running all the time. Without knowing the price of a kW h in Chigaco I cannot answer further. However in Maine I pay on average about $0.16 per kW h in my market. So for me to run your heater non stop for a month it would cost me $178.56. To find this result for yourself, determine how much you pay per kW h and multiply that number by 1116 kW h.
