##### 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?

Jul 8th, 2015

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

Please let me know if you need any clarification. I'm always happy to answer your questions.awaiting a positive feedback. It was lovely helping you. Please do contact or text me directly for more work assistance. Looking forward to help you again.
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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Jul 8th, 2015
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Jul 8th, 2015
May 26th, 2017
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