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A 0.200-kW computer runs 6.00 hours per day for 30.0 days, with electricity priced at $0.120 per kilowatt-hour. What is the total cost for the month?
A$0.72
B$21.60
C$7.20
D$4.32
Answer & Solution
Correct answer: D. $4.32
1. Energy consumed is E = Pt, so first find the total hours of use: 6.00 h/day times 30.0 days equals 180 hours.
2. The energy used is E = (0.200 kW)(180 h) = 36.0 kW*h.
3. The cost is the energy used multiplied by the price per kilowatt-hour, cost = (36.0 kW*h)($0.120 per kW*h).
4. That multiplication gives a cost of $4.32 for the month.
5. $21.60 comes from using the daily hours alone without multiplying by the number of days, and $0.72 and $7.20 come from misplacing a decimal point in the final multiplication.
_Source: OpenStax College Physics (CC BY 4.0), Ch 7 "Work, Energy, and Energy Resources", section 7.7 Power_
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