Question

1) A light bulb takes in 30 of energy per second. It transfers 3j as use

Answer 1

Answer:

$\boxed{\text{10 \%}}$

Explanation:

The formula for efficiency is  

$\begin{array}{rcl}\text{Efficiency} & = & \dfrac{\text{useful energy out}}{\text{energy in}} \times 100 \,\% \\\\\eta & = & \dfrac{w_{\text{out}}}{w_{\text{in}} } \times 100 \,\%\\\end{array}$

Data:

Useful energy =  3 J

Energy input  = 30 J

Calculation:

$\begin{array}{rcl}\eta & = & \dfrac{\text{3 J}}{\text{30 J}} \times 100 \,\%\\\\\eta & = & 10 \, \%\\\end{array}\\\text{ The efficiency is }\boxed{\textbf{10 \%}}$