Question

1. In a hydraulic machine, a force of 12 N is applied on area of cross section 48 cm2. What force is obtained on its piston of area of cross section 120 cm2?​

Answer 1

Answer:

In first condition, force= 12N and area= 48 cm2

• The pressure obtained in first condition-

$\frac{Force}{Area}$= 12/48= $\frac{1}{4}$ pascal or 0.25 pa.

• In 2nd condition if the pressure is same, then force-

pressure*area=120*0.25=30 N

Explanation:

Answer is 30 N. Make my answer Brainliest plz