Physics - hair dryer problem - multiple partsKilowatts-hours, power, etc…

Explanation
Step 1
a)Power
this is an electric power measure of the rate of electrical energy transfer by an electric circuit per unit time,it is given by the expression
[tex]\begin{gathered} P=VI \\ (\text{unit; watt ot }\frac{joule}{s}) \\ \text{where v is the voltage and I is the current} \end{gathered}[/tex]then
Let
V= 120 v
I=15 A
now,replace in the formula
[tex]\begin{gathered} P=VI \\ P=120v\cdot15A=1800\text{ J} \\ P=1800\text{ Joules} \end{gathered}[/tex]therefore, teh answer is
a) 1800 Joules
Step 2
if runs for 3 minutes
how many joules of energy
Power can be expressed as P = Work/Change in Time}
so
[tex]\begin{gathered} \text{work = Power }\cdot\text{ time} \\ \text{work = 1800 Joules}\cdot180\text{ s=}324000\text{ joules} \end{gathered}[/tex]so
b)324000 Joules
Step 3
c)Kilowatt -hour
to do this conversion of units, we need to know the equivalence:
[tex]\begin{gathered} 1\text{ Kwh= }3600000\text{ Joules} \\ so \\ 324000\text{ Joules(}\frac{1\text{ kw hour}}{3600000}\text{)=}0.09\text{ Kwh} \end{gathered}[/tex]so
c) 0.09 Kw-h
I hope this helps you