Answer:
1400.38N
Explanation:
Step one
Given data
P1= 250N
D1= 0.02m
A1= πD1^2/4
substitute
[tex]A1= 3.142*0.02^2/4\\\\A1=3.142*10^-4[/tex]
D2= 0.15m
A1= πD2^2/4
[tex]A2= 3.142*0.15^2/4\\\\A2=1.76*10^-3[/tex]
Required
The load P2
Step two:
Applying the hydraulic expression for a non-compressible fluid
we know that
Pressure= force/are
P1/A1=P2/A2
[tex]250/3.142*10^-4= P2/1.76*10^-3[/tex]
cross multiply we have
P2= 1.76*10^-3*250/3.142*10^-4
P2=0.44/3.142*10^-4
P2=1400.38N
