Sagot :
[tex]750W=0.75 kW\\ \\ 0.75*\frac{13}{60}=0.1625 \ kWh\\ \\ \boxed{0.1625*0.10=\$ \ 0.1625}[/tex]
The '120-V' is what we in the engineering profession call a "red herring".
You don't need it to solve the problem. The answer would be the same
if the blow drier were operating from a 440-V outlet, or any other kind
(just as long as the blow drier is designed to operate on whatever voltage
it was plugged into.)
Power = 750-W = 0.75 kW
Time = 13 minutes = 13/60 of an hour
Energy = (power) x (time) = (0.75 kW) x (13/60 of an hour) = 0.1625 kWh
Cost per unit of energy = $ 0.1 per kWh
Total cost = (energy) x (cost per unit of energy)
Total cost = (0.1625 kWh) x ($ 0.1 per kWh) = $ 0.01625 = 1.625 cents
===============================
The other answer has his significant figures correct, but shows what happens
when you don't carry your units all the way through.
You don't need it to solve the problem. The answer would be the same
if the blow drier were operating from a 440-V outlet, or any other kind
(just as long as the blow drier is designed to operate on whatever voltage
it was plugged into.)
Power = 750-W = 0.75 kW
Time = 13 minutes = 13/60 of an hour
Energy = (power) x (time) = (0.75 kW) x (13/60 of an hour) = 0.1625 kWh
Cost per unit of energy = $ 0.1 per kWh
Total cost = (energy) x (cost per unit of energy)
Total cost = (0.1625 kWh) x ($ 0.1 per kWh) = $ 0.01625 = 1.625 cents
===============================
The other answer has his significant figures correct, but shows what happens
when you don't carry your units all the way through.