open with Google Docs ant Name: A solid cylinder length 2 m and cross-section ar

Question

open with Google Docs ant Name: A solid cylinder length 2 m and cross-section area of A, is made from Aluminum with density of rv 2.7 grCm and wood with density of r, 0.50 grCm The cylinder is then floating vertically in the water and 1.75 m of the cylinder is submerged into the water (r.n- 1.0 grCm a) Determined the height h, of Aluminum part of the cylinder b) lf the radius of the cylinder is 0.10 m, then, how much foroe would the water applied on the bottom area of cylinder? c) We put the cylinder in the new fluid with density of 1.17 grCm How much of the height of the cylinder is out of the fluid.

Explanation / Answer

volume of water displaced = (1.75A)

buyont force = 1*(1.75A)*g

Weight of cylinder = [2.7x(Ah)+0.5(A(2-h))]g

solving the equation we get:

h = 0.340 meter

b)

A = 0.0314 m2

Buyont force = 0.538 N

c)

height out of the fulid with different density = 1.027 m

let x' be the length inside liquid then:

1.17A*x' =2.7*A*0.34+ 0.5A*1.66

x' = 0.972

hence part outside liquid is = [ 2 - 0.972 ]

= 1.028 meters Answer

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