##### Physics: Pressure Gauge

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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

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

Jul 8th, 2015

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Jul 8th, 2015
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Jul 8th, 2015
Sep 23rd, 2017
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