Data;
- Power = 1200W
- Time/day = 10 mins
Power
The is the rate of energy used to time. It is the measure of expedience of energy.
a). How many minutes is the dryer used in a month.
Assuming there 30 days in a month;
\(1 day = 10 mins\\30 days = x mins\\x = 30 * 10 \\x = 300 mins\)
This shows that the dryer is used 300 minutes in a month.
b). How many hours is the dryer used in a month.
To solve this problem, we simply need to convert the value of (a) from minutes to hours.
\(1hr = 60 mins\\y hr = 300 mins\\y = \frac{300}{60} \\y = 5 hours\)
The dryer was used for a total of 5 hours in a month.
c). The power of the dryer in kilowatt.
To convert the power in watt to kilowatt, we divide by 1000.
\(P = 1200W\\P = \frac{1200}{1000} \\P = 1.2kw\)
The power in kilowatt is 1.2kw
d). How many kilowatt/hr of electricity used in a month.
\(1.2 * 5 = 6kw/h\)
The dryer consumes 6kw/h in a month.
e). If the town charges $0.15/kwh, the cost in a month
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The cost of the dryer in a month is $0.9.
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