Answer:
c. 80%
Explanation:
Given;
load raised by the machine, L = 360 N
distance through which the load was raised, d = 0.2 m
effort applied, E = 50 N
distance moved by the effort, e = 1.8 m
The efficiency of the machine is calculated as follows;
[tex]Efficiency = \frac{ 0utput \ work }{1nput \ work} \times 100\% \\\\Efficiency = \frac{Load \ \times \ distance \ moved \ by \ load }{Efort \ \times \ distance \ moved \ by \ effort} \times 100\%\\\\Efficiency = \frac{360 \times 0.2}{50 \times 1.8} \times 100\%\\\\Efficiency =80\%[/tex]
Therefore, the efficiency of the machine is 80%
