#### wakefreak90

A transistor radio operates by means of a 10.0 V battery that supplies it with a 48 mA current.

(a) If the cost of the battery is \$0.55 and it lasts for 300 h, what is the cost per kWh to operate the radio in this manner?
3.819 dollars/kWh

(b) The same radio, by means of a converter, is plugged into a household circuit by a homeowner who pays 6¢ per kWh. What does it now cost to operate the radio for 300 h?
??? cents

I got Question A but I have tried everything on Question B and I do not understand what I am doing wrong. For some reason, I think it is easier than what I'm trying to do.

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Work out the number of kilowatt hours and multiply by 6.Note that the question gives the current in mA and not A and note also the the power must be expressed in kW and not W.

#### wakefreak90

So if one kWh = 3.6E6 J, and the total energy is 518,400 J..

(3.6E6)/(518400) = 6.9444

--> (6.9444)(6) = 41.666

Is 41.666 cents the answer? I only have two more tries left and I don't want to get it wrong. I'm not sure if those are the correct unit conversions