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What does code J stand for on the electric motor identification tag?

It is a locked rotor kva code. See related link for an excellent description.


How long would it take a 500 W electric motor to do 1.50x105 J of work?

500 miles long


How much thermal energy is produced in an electric motor j of electrical energy are converted into 92j of mechanical energy?

If 92 J of electrical energy are converted into 92 J of mechanical energy in an electric motor, then all of the electrical energy is being converted into mechanical energy. No thermal energy is being produced in this scenario. The efficiency of the motor in this case is 100%, meaning all the input energy is being converted into useful work without any energy loss in the form of heat.


10. How long would it take a 500 W electric motor to do 1.50 x 105 J of work?

not sure


What is the j code for bupivicaine?

The J code for bupivacaine is J3490.


A motor with an efficiency of 75 percent must supply 240 J of useful work. What amount of work must be supplied to the motor?

At least 320 J


What has the author J Richard Johnson written?

J. Richard Johnson has written: 'Electric circuits' -- subject(s): Electric circuits


What has the author J A Harrison written?

J. A. Harrison has written: 'The essence of electric power systems' -- subject(s): Electric power systems


What has the author John J O'Neill written?

John J. O'Neill has written: 'Prodigal genius' -- subject(s): Electric engineers, Biography, Electric currents, Alternating, Electric engineering, History, Alternating Electric currents


What does the 210 mean?

It's a code for oral sex, since b is letter 2 and j is letter 10.


What is the difference between electrical potential energy and potential difference?

Electric potential is the electric potential energy per unit coulomb. So unit for electric potential is J/C and that of electric potential energy is simply J


How long would it take a 500 W electric motor to do 1.50 x 105 J of work?

To find the time it takes for a 500 W electric motor to do (1.50 \times 10^5) J of work, we can use the formula: [ \text{Power} = \frac{\text{Work}}{\text{Time}}. ] Rearranging gives us: [ \text{Time} = \frac{\text{Work}}{\text{Power}} = \frac{1.50 \times 10^5 \text{ J}}{500 \text{ W}} = 300 \text{ seconds}. ] Thus, it would take 300 seconds, or 5 minutes, for the motor to perform that amount of work.