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Problem 37 Medium Difficulty

The tip of the iceberg. Icebergs consist of freshwater ice and float in the ocean with only about 10$\%$ of their volume above water (the "tip of the iceberg," so to speak). This percentage can vary, depending on the condition of the ice. Assume that the ice has the density given in Table $13.1,$ although, in reality, this can vary considerably, depending on the condition of the ice and the amount of impurities in it. (a) What does this 10$\%$
observation tell us is the density of seawater? (b) What percentage of the icebergs' volume would be above water if they were floating in a large freshwater lake such as Lake Superior?

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Video Transcript

Okay, so in this problem, we have seawater. This is the seawater, and we have only 10% of an ice bag floating above the surface of the water. So what we want to calculate in this problem is, what is the density off the seawater knowing? Onley? This this this problem? First of all, we know that we have a mess, the weight off the ice back here. And we have a buoyant force that keep the system in the Cleveland. So this is the boy and force. And this is what you're gonna call the mess of the ice Times G. Okay, so, using the arguments principle, we know that this system in sick liberal because the buoyant force should be cool. The mass of the ice times G but documented principal say's that the buoyant force is just density off the sea water. In this case, the volume off the object that is submerged times gravity deceleration. So just need to be equal mass off the ice times G. We came across the gravity here and we simplify or problem to density off the seawater times. The Fordham submerge, of course, the mass of the ice. But the problem doesn't say what is the mass of the ice. So we're going to change the mess of the ice to intensity off the ice times the total volume off the ice. Okay, so finally, our equation is going to be the density of the sea water times divorce him. That is submerge going to be equal their city off the ice times that Toto volume with the iceberg. But what we know, we know that the volume that is submerge is equal 90% of the total boredom. So it's just zero point nine off the So let's use this relation to solve our problem. So we have that density of the sea water times 0.9 v Going to be cool the density of the Weiss V. So again we can cross this right? I'm here and finally we know that the density off the seawater is just the density of the ice divided by 0.9. So if Riko Cletus we know that we're going to find one point zero to times 10 23 kilograms per meter cubed X. So that's the answer for the volume off the seawater. Okay, now let's calculate the second I think of this problem that what it should be the volume soup emerge if the ice was on a normal lake, so going to use the same equation off the first night. And so we have that the density off the water, not seawater just I'm just time is normal water, uh, times the volumes submerge is going to be equal. Did then city off the ice times the volume off the ice. She's just be okay, so diffraction off the volume that will be submerged. Let's see, it's going to be Fordham's merge, going to be divided by the volume going to be equal density off the ice, divided by the density of the water. So this is just, you know, point 92 times 10 to the three, which is the density off the ice. Zero points 92 divided by the density of the water just one times 10 to 3. Therefore discovered that the volume that will be submerged in a normal lake off the ice is just 92% off the total Voyager. And that's the final answer. Thanks for watching

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Problem

At $20^{\circ} \mathrm{C},$ the surface tension o…

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