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A hollow cylindrical copper pipe is 1.50 $\mathrm…

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Problem 7 Easy Difficulty

A cube of compressible material (such as Styrofoam or balsa wood) has a density $\rho$ and sides of length $L .$ (a) If you keep its mass the same, but compress each side to half its length, what will be its new density, in terms of $\rho ?$ (b) If you keep the mass and shape the same, what would the length of each side have to be (in terms of $L )$ so that the density of the cube was three times its original value?

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Video Transcript

this problem deals with some algebraic manipulation of our density equation. So to start off with, we're told we have a compressible cube with a density of roe and a side length of L. And we're saying that we keep the mass of this cube of the same, but we're going to compress the length of each side to half of its original length. So we want to know what is the new density in terms of the old density. To do that, let's write what are old density is We know that the density of this cube will be equal to its mass, divided by its volume. And in this case, the volume of our cube is l a times L A times L or weaken, right? L cubed. We know that it's all cube because this is a cube. The length, width and height are all the same. And we can express that as l. Now what we're going to say is, we're going to deal with this new row, this new density. I'm gonna call it Ro Prime. I'm gonna write it in red. This new density, the way we're going to get it, is we're keeping mass the same. But instead of having the original side length, our side length is now half of what it used to be. So instead of El Times, all times L it is 0.5 l a times 0.5 l a times 0.5 l which I can also right as 0.1 to 5 l cubed and what I can see right away is I know that original Roe Black Row is equal to em over l cubed in this expression, I have em over l cubed. So instead of writing that I can just say that Row prime is equal to one over 10.1 to 5 times original roe and instead of dealing with that fraction and a decimal one divided by 10.125 is equal to eight. So what we have now is our new density rope rhyme is equal to eight times our old density. And that makes sense because if we make our cube smaller, its density should go up. Now. Our second question says we're going to keep the the same shape and mass, But we're asking, what would we need to do to l in order to result in three times the density. So again we know that four. This cube density is equal to mass over l cubed length. Angeles Times height is L Times old times l. We can rearrange this equation to solve for L on its own. It's a little bit messy. What we can do first begin multiply by El cubed on both sides. So we end up with l Cubed Times. Row is equal to em than what we can do is you can divide by row on each side That would leave us with El Cubit is equal to m divided by row and then last but not least, to get l completely by itself, we would need to cube root both sides. The opposite of Cuban would be Q brooding something so that would take care of that. So what we would end up with if we solved for L all on its own was l is equal to the cube root of em over Rome, ass over density. That is our original case in our original cube. The length is equal to the cube root of the mass of the cube, divided by the density of the Cube. Now we're going to talk about this new length that we're dealing with. So l prime, which all right in red, just like last time is equal to the cube Root of mass is saying the same. But instead of this original density, we're going to be doing three times that density. So what we can do now? Just like last time, we can see that we have these old units of l still in this new definition of El Prime for l. We had Q brew of em over p and l prime. We have Q brew of em overpay. So instead of writing all that out, we can say that L prime is equal to original L the cube root of having overpay. And then it is multiplied by a factor of what would be left and what would be left. Once we take out that l would be the cube root of one third times R l. So we get that 1/3 because we're dividing by three on the inside. So it's kind of like an invisible one in the numerator, and we still have to take that cube root of 1/3. Even though we're also que brooding M overpay. So again, what we can say is we can say that this new length is equal to the cube root of one third time's the old length.

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