The rectangular tank in Fig. 2 (next page) has width Ly in the dimension out of the paper. Unlike most tanks, there is a partial partition that is always vertical (that is, constrained to lie in the y-z plane) but can slide in the x direction. It forms perfect seals with the sides of the tank in the x-z planes. The bottom of the partition is at z = z0. The top of the partition is above the fluid surfaces. The left-hand side of the tank has a cross-sectional area of Ly (in the x-y plane), while the right- hand side has area Ly(Lx ). Note that the value of depends on the fluid pressures, and that its value must be determined (by you) such that in the x direction, the partition has no net forces acting on it. The volume of fluid 1, with density 1 and height h1 is V1, so V1 = h1Ly. (2) The volume of fluid 2, with density 2 and height h2 is V2, so V2 = h2Ly(Lx ). (3) If the volume of fluid 0, with density 0 is V0, find an equation for V0 in terms of Ly, h0