how height does a 2.5 Kg object rises when 255J of work done against gravity A 637.5 B 20.5 C 15m D 10.2 m
2 answers:
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Answer:
which corresponds to the second option shown: "voltage times amperage"
Explanation:
The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.
Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:
Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.
This corresponds to the second option shown in the question: "Voltage times amperage".
