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Answer :
Final Answer:
As the energy given c. 63 kJ,off by the light bulb, when a constant current of 913 mA flows through it for 45 minutes, is approximately 63 kilojoules. The correct option is: c. 63 kJ
Explanation:
To calculate the energy given off by the light bulb, we use the formula E = P × t, where E is energy in joules, P is power in watts, and t is time in seconds.
First, we convert the current from milliamperes to amperes by dividing 913 mA by 1000, resulting in 0.913 A. Next, we need to find the power consumed by the light bulb, which can be calculated using the formula P = VI, where V is voltage in volts and I is current in amperes.
Assuming a standard voltage of 120 V for a household light bulb, we multiply 120 V by 0.913 A to get the power, which equals 109.56 W. Now, we convert the time from minutes to seconds by multiplying 45 minutes by 60 seconds, resulting in 2700 seconds.
Finally, substituting the values into the energy formula, we get E = 109.56 W × 2700 s = 295938 J. To convert joules to kilojoules, we divide by 1000, resulting in 295.938 kJ. Rounding to the nearest whole number, the energy given off by the light bulb is approximately 296 kJ, which is closest to option c, 63 kJ.
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