Physics: Pressure Gauge
Physics

Tutor: None Selected  Time limit: 1 Day 
The spring of the pressure gauge shown in the figure below has a force constant of 1,080 N/m, and the piston has a radius of 1.14 cm. As the gauge is lowered into water, what change in depth causes the piston to move in by 0.750 cm?
Thank you for the opportunity to help you with your question! the solution is as such:
area of piston = pi x r^2 = pi x (1.1/100)^2 = 3.80132 x 10^4 m^2To move 0.750 cm force required = ( 0.75 /100) x 1071 = 8.0325 N
Pressure required = Force / area = 8.0325 / 3.80132 x 10^4 = 21130.817 N/ m^2
1000 x 9.81 x h = 21130.817
h = 2.154 m ans
oh sorry just a second
there is little calculation mistake. i will just correct it.
area of piston = pi x r^2 = pi x (1.14/100)^2 = 4.028 x 10^4 m^2
To move 0.750 cm force required = ( 0.75 /100) x 1080 = 8.1 N
Pressure required = Force / area = 8.1 / 4.028 x 10^4 = 20109 N/ m^2
1000 x 9.81 x h = 20109
h = 2.051 m ans
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