(7%) Problem 7: A cube of mass m = 680 kg is totally immersed in a liquid densit
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(7%) Problem 7: A cube of mass m = 680 kg is totally immersed in a liquid density e = 0.92 g/cm3. The cube has an edge length of 1.1 m and is held at a depth of d= 1.9 m from the top of the cube to the surface of the liquid. of Ctheexpertta.com the top surface, in terms of the defined quantities and the acceleration due to gravity.g the top surface. and the magnitude of the force it exerts on the top surface, in terms of the defined quantities and the acceleration due to gravity, g. This is the 20% Part (a) Enter an expression for the difference between the fluid pressure acting on the bottom surface of the cube and that acting on 20% Part (b) Calculate the difference, in pascals, between the fluid pressure acting on the bottom surface of the cube and that acting on 20% Part (c) Enter an expression for the difference between the magnitude of the force the liquid exerts on the bottom surface of the cube . magnitude of the net vertical force the liquid exerts on the cube. That force points up and is called the buoyant force, denoted F 20% Part (d) Calculate the magnitude of the buoyant force, in newtons, that the liquid exerts on the cube. Grade Summary Deductions 4%Explanation / Answer
(a) delta(P) = rho g delta(H)
= rho g L
(b) delta(P) = (0.92 x 10^-3 kg / 10^-6 m^3) (9.81)(1.1)
= 9927.7 Pa
(C) F = delta(P)A
= rho g L^3
(d) F = 920 x 9.81 x 1.1^3
F = 12012.5 N
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