(11%) Problem 6: A cube with edge length 0. 19 m and density .-0.37x 103 kg/m3 f
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(11%) Problem 6: A cube with edge length 0. 19 m and density .-0.37x 103 kg/m3 floats in equilibriurn in a liquid of density P1= 1.3× 103 kg/m3, with the top of the cube a distance d above the liquid's surface, as shown in the figure. Pi Otheexpertta.com 33% Part (a) Enter an expression for the distance d, in terms of the defined quantities. 33% Part b) If the density of the cube is equal to the density of the liquid, how high, in meters, will the top of the cube be above the surface of the liquid? Grade Summary Deductions 25% Potential d 0.63 7596 Submissions Attempts remaining Z (59% per attempt) cotanasinOacosC atan0 acotan) sinh) coshO tanhO cotanho "Degrees Radians detailed view : 596 5% END NO BACKSPACEELCLEAR Submit Hint I give up 596. Hints: deduction per hint. Hints emaining:- Feedback: deduction per feedback. M se 33% Part (c) The given values ofedge length and dens tes are L 0 19 m, p.-0.87×103 kgm3, and ,-1.3×103 kg m3 values, how high, in meters, is the top of the cube above the liquid's surface. With theseExplanation / Answer
(A) Fnet = Fb - m g = 0 ( for eq.)
=> ((L-d) L^2) (rhol) g - (L^3 rhoc) g = 0
L - d = L (rho_c / rho_l)
d = L (rho_l - rho_c) / rho_l
(B) when rh0_l = rho_C then rho_l - rho_c = 0
so d = 0
(C) d = (0.19 / 1300) (1300 - 870)
d = 0.063 m
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