The spring of the pressure gauge shown in the figure below has a force constant of 1,160 N/m, and the piston has a radius of 1.22 cm. As the gauge is lowered into water, what change in depth causes the piston to move in by 0.750 cm?

Answer 1

k = 1160 N/m
r = 1.22 cm = 0.0122 m
x = 0.750 cm =0.0075 m
Change in the pressure, when the gauge is taken into water
ΔP = F/ A= k•x /π•r² =
=1160•0.0075/ π•(0.0122)²= .....
ρ(water) =100 kg/m³
ΔP = ρwater• g •h,
h = ΔP/g• ρ(water)=…

Answer 2

ρ(water) =1000 kg/m³ !!!!!!!!!!!!

Answer 3

(k•x)-(ρ•g•v) = zero
v=h•A

Answer 4

(k•x)-(ρ•g•v) = zero
we have: k= 1160 N/m
x= .0075 m
ρ(water)= 1000 kg/m³
g= 9.8 m/s²
r= .0122 m
and v=h•A A= π•r² = π(.0122)²
h=??? (Change in the pressure)
so: h=(k•x)/(ρ•g•A)
=(1160*.0075)/(1000*9.8*(π(.0122)²)
h= 1.898557011 m :)