A ship, carrying fresh water to a desert island in the Caribbe...

Question

A ship, carrying fresh water to a desert island in the Caribbean, has a horizontal cross-sectional area of 2240 m² at the waterline. When unloaded, the ship rises 8.55 m higher in the sea. How much water (m³) was delivered?

Accepted Answer

Hey, everyone in this problem, we're told that a barge delivering vegetable oil to a remote coastal village has a horizontal cross sectional area of 2250 m squared. At the waterline. We asked how much oil is delivered when the oil was unloaded? And the bar rose 9.5 m higher in the seawater. We're told that the surface seawater density is 1025 kg per meter cubed and the density of vegetable oil is 915 kg per meter cubed. We're given four answer choices all in meters cubed. Option A 2.39 multiplied by 10 to the exponent four. Option B 4.06 multiplied by 10 to the exponent four. Option C 5.5 multiplied by 10 to the exponent four and option D 6.71 multiplied by 10 to the exponent four. What we're looking for is the volume of oil delivered. OK. And what we're told about is this rise in this bar. He so if we think about what we have going on, we can think about Archimedes principle and Archimedes Principle tells us that the buoyant force, the upward buoyant force exerted on some object that's immersed in a water or in some fluid is gonna be equal to the weight of the fluid that that body displaces. And you know, we're told a little bit about the height of water. So the amount of water that's displaced by this barge. And so we want to draw a free body diagram of what Archimedes principle tells us. We have this buoyant force pointing upwards and we have this force of gravity pointing downwards of the oil of the water displaced. All right. So let's think about this in terms of the equations. And we know that the buoyant force FB is equal to the density of the seawater. So row sw multiplied by the volume of the seawater was multiplied by the acceleration due to gravity. OK. And because we know that that buoyant force has to be equal to that force of gravity. From Archimedes Principle, we can say that this equation row swvswg must be equal to the mass of the oil multiplied by the acceleration due to gravity. So we can see we have G on both sides, we can divide that out. And what this tells us is that the mass of the oil is going to be equal to the density of the seawater multiplied by the volume of the seawater. Yeah, that's the volume that is displaced. Wait, so we have this equation, we have to remember what we're looking for, we're looking for the volume of the oil, not the mass of the oil. And so what we're gonna do, we're gonna use what we've written so far to find the mass of the oil, then we can relate the mass of the oil to the volume that we were looking for through our density equation. OK. So this is step one and that is finding that mass of the oil. OK. So let's go ahead and think about filling in some of these terms. And so the density of the seawater we're told in the problem. So we're gonna leave that for. Now, we're gonna be able to substitute that indirectly. Tell me the volume of the seawater display. Well, we know the cross sectional area of the barge, ok? So we can look at the volume as just that cross sectional area multiplied by the height. OK? And both of those values were given. So what we can find is that the mass of the oil is going to be equal to 1000 and 25 kg per meter cubed multiplied by 2250 m squared multiplied by 9.5 m. And this gives us a mass of the oil of 21,909,375 kg. And that's a lot of oil. All right. So we have the mass, that's part one. Now, we want to move on to finding what we are actually looking for which is the volume. OK. So let's recall that the density of the oil can be written as the mass of the oil divided by the volume of the oil. And that goes for anything, the density is the mass divided by the volume. So we're looking for the volume of the oil. We can write this as the mass divided by the density. Well, the mass we know 21,909,007 375 kg. And the density we were given in the problems we divide by 915 kg per meter cubed. And what we end up with is a volume of oil of 23,944 0.67 m cubed. OK. So that is the volume of oil that was delivered. We were able to figure that out just by looking at that barge and seeing how high it rose after the oil was unloaded. And we had to use Archimedes principle there to help us out. We're gonna round our answer to three significant digits. And when we compare it to the answer choices, we can see that the correct answer corresponds with answer choice. A 2.39 multiplied by 10 to the exponent 4 m cubed. Thanks everyone for watching. I hope this video helped see you in the next one.

A ship, carrying fresh water to a desert island in the Caribbean, has a horizontal cross-sectional area of 2240 m² at the waterline. When unloaded, the ship rises 8.55 m higher in the sea. How much water (m³) was delivered?

Key Concepts

Buoyancy

Volume Displacement

Cross-Sectional Area

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