Posted by dale on .
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 1.43 cm. As the gauge is lowered into water, what change in depth causes the piston to move in by 0.750 cm?
k = 1080 N/m, r = 0.0143 m,
A = πr²= π•0.0143²=6.42•10⁻⁴ m²
The pressure P = F/A = ρ•g•h,
where water density is ρ = 1000 kg/m³
g = 9.8 m/s² , x = 0.0075 m
For the elastic force of the spring
F = kx.
h = kx / A•ρ•g = …