The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is: A). $36 \times 10^7\ \text{J}$ B). $36 \times 10^4\ \text{J}$ C). $36 \times 10^5\ \text{J}$ D). $1 \times 10^5\ \text{J}$

The correct answer is $36 \times 10^7\ \text{J}$. $P = 100\ \text{kW} = 10^5\ \text{W}$, $t = 1\ \text{hour} = 3600\ \text{s}$\n$E = P \times t = 10^5 \times 3600 = 3.6 \times 10^8\ \text{J} = 36 \times 10^7\ \text{J}$

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