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Answer :
Final answer:
The power dissipated in a wire with 65 ohms of resistance and 2.5 A of current is 406.25 W. None of the given choices are correct.
Explanation:
The power dissipated in a wire can be calculated using the equation P = IV, where P is the power, I is the current, and V is the voltage. In this case, since the resistance is given, we can find the voltage using Ohm's law: V = IR. Substituting the values, we have V = (65 ohms)(2.5 A) = 162.5 V.
Therefore, the power dissipated in the wire is P = (2.5 A)(162.5 V) = 406.25 W which is not included in any of the options.
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