Post a New Question


posted by 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?

  • physics - ,

    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 = …

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question