I keep getting the first box wrong but I'm not sure why?

The spring of the pressure gauge shown in the figure below has a force constant of 1 155 N/m, and the piston has a diameter of 1.50 cm. As the gauge is lowered into water in a lake, what change in depth causes the piston to move in by 0.650 cm? Vacuum kPart 3 of 3 - Analyze Assuming the spring obeys Hooke's law, the increase in force on the piston required to compress the spring an additional amount Ax, is given by AF =F - Fo = (P - P) A = k(4x), where A is the surface area of the piston. The gauge pressure at depth h beneath the surface of a fluid is given by P - Po = pgh, so we have pghA = k(Ax). Solving this equation for h, gives us the following. h = k (4x ) pgA We know that the density of water is 1 000 kg/mu, and the area of the piston is given by A = (d/2)2, where d = 1.50 cm, so for Ax = 0.650 cm, we have k(AX ) h = pgA 1 155 x N/m 0.0065 m (1 000 kg/m3) (9.80 m/s2 )71 0.0075 m = 4.34 m . Submit Skip (you cannot come back)