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Ship draft (draught) word problem buoyancy
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[QUOTE="late347, post: 5482904, member: 531743"] [h2]Homework Statement [/h2] ship arrives from the Atlantic to the Baltic sea. How much does the draft of the ship deepen due to the saltyness of the water... density of atlantic water = 1027kg/m^3 density of baltic water = 1005kg/m^3 at these depths the crosssectional area of the ship can be assumed to be 4000m^2 ship and cargo mass = 10000 metric tons g= 9,81m/s^2 [h2]Homework Equations[/h2] N= ρ V g G=mg [h2]The Attempt at a Solution[/h2] I have no idea what ship draft is. I have only once been on a ship in my life. I have no idea why the the cross-sectional area would be relevant in this problem. But I decided to calculate the volumes of water, which the ship displaces in atlantic, And then displaces in the baltic. I think those volumes will be different because water density changes. N[SUB]atlantic[/SUB] = ρ[SUB]atlantic[/SUB] x V[SUB]atlantic[/SUB] x g V[SUB]atlantic[/SUB]= 9737, 0983 m^3N[SUB]baltic[/SUB] = ρ[SUB]baltic[/SUB] x V[SUB]baltic[/SUB]x g V[SUB]baltic[/SUB] = 9950,2487 m^3I was later on thinking that maybe if I were to take all the water into a huge box Then you could divide the displaced water volumes, by the area Therefore you end up with height ( I think) Or something to this effect could be done. If this is calculated so, then the difference between the heights of the displaced water is (9950,2487m^3) / 4000m^2 = 2,4875 m = height baltic 9737,0983 m^3 / 4000m^2= 2,4342 m height atlantic. calculate Δheight = 0,0533m the deepening effect ought to be about 5,3 cm please notify whether or not errors of understanding of judgement were made. [/QUOTE]
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Ship draft (draught) word problem buoyancy
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