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^2

To 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

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Jul 8th, 2015

oh sorry just a second

Jul 8th, 2015

there is little calculation mistake. i will just correct it.

Jul 8th, 2015

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

Jul 8th, 2015

thank you. please contact for more help on any question.

Jul 8th, 2015

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