A ship, carrying freshwater to a desert island in the Caribbean, has a horizontal cross-sectional area of 2700 m^2 at the waterline. When unloaded, the ship rises 6.00 m higher in the sea. How much water was delivered?
There was an identical one in my book with a cross-sectional area of 2650 m^2 and the ship rose 8.5 m high. In which the answer was 2.44 x 10^7. However when i try to solve it i always get 2.25 x 10^7. Isn't it supposed to be 2650*8.5*1000? If not could someone show me what i'm doing wrong? Also, show steps and how you got to your answer. Thanks.
2007-01-05 14:17:50 · 4 answers · asked by leon27607 3 in Science & Mathematics ➔ Physics
btw they want the answer in kg
2007-01-06 04:36:33 · update #1
omfg i tried 1.76E7 kg but it still didn't work -_-
2007-01-06 04:43:54 · update #2
The mass of the fresh water that was offloaded would be equal to the mass of seawater that the boat stopped displacing. As was mentioned, seawater has a different density from freshwater. The density of seawater is about 1025 kg/m^3. So the area of the boat that rose out of the water in the problem from your book would have been (2650 m^2)*(8.5 m) = 22,525 m^3. Multiply that by the density of seawater to get (22525 m^3)*(1025 kg/m^3) = 2.31 x 10^7 kg. This is not the answer the book gave you. I could not find any resource that says seawater has a density of 1085 kg/m^3, but using this value does get you an answer of (22525 m^3)*(1085 kg/m^3) = 2.44 x 10^7 kg in this problem, so it must be the value you're supposed to use.
Applying the same formula to your new problem, the answer should be (2700 m^2)(6 m)(1085 kg/m^3) = 1.76 x 10^7 kg.
2007-01-13 05:16:49 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
Yes Leon! The density of sea water is some 8.5% greater than that of fresh water. So if waterline rose by h meters then the level of fresh water had been h*1.085 meters; thus V=2700*6*1.085 = 17577 m^3 or 17.57 tons
2007-01-05 16:59:30 · answer #2 · answered by Anonymous · 0⤊ 0⤋
You need to account for the density difference between fresh water and salt water.
2007-01-05 14:33:06 · answer #3 · answered by Anonymous · 0⤊ 0⤋
SG is a ratio of an merchandise's density to that of water. An merchandise with a SG = a million.00 will drift with a hundred% of its quantity under water. A block of timber with density = 0.seventy 9 floats with seventy 9% of its quantity under water. ANS-a million 550/19.3 = 28.5 cm³ ANS-2 quantity of block of timber = 2500/0.fifty 5 ? 4545 cm³ in water this block floats with fifty 5% of its quantity under water in oil this block floats with 2500/0.eighty 5 = 2941 cm³ under water ANS-3 10.3 m of unpolluted water = a million.01E5 Pa 750/10.3 = seventy two.8 atmospheres ANS
2016-12-16 03:17:24 · answer #4 · answered by ? 4 · 0⤊ 0⤋