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A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?

1.6

Physics 101 Mechanics

Chapter 13

Fluid Mechanics

Temperature and Heat

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Sheffield

Lectures

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this problem tells us that we have a lead sphere and an aluminum sphere of the same mass. And our job is to find the ratio of the radius of the aluminum sphere to the lead sphere. Bathroom. I'm going to use a L the atomic symbol for aluminum represent aluminum and PB, the atomic symbol for lead to represent lead. Now, our general game plan here is first. I'm going to use a mass of one kilogram for both of these fears just to make calculations a little bit easy and also to avoid having too much of a symbolic mess. So I'm choosing this arbitrary mass of one kilogram. Our general game plan is going to be to take the mass of each of these fears, use the densities of both of these materials, which we have. We can look up, use that to turn it into the volume of those fears, and then we're going to use the volume to find the radius of those fears. In order to do that, we need to know that density is equal to mass over volume. We need to know that the volume of a sphere is equal to 4/3 pi radius cubed and we also need to know the densities of these two materials. The density of lead is equal to 1.13 times 10 to the three kilograms per cubic meter. And the density of aluminum is equal to 2.7 times 10 to the three kilograms per cubic meter. Um so let's get started first will deal with finding the radius of this lead sphere. So we will start there. So find the radius of our lead sphere again. We know that our density is equal. Thio 11.3 times 10 to the three kilograms per cubic meter. We know that because density is equal to mass over volume, we can rearrange. That's a volume is equal to mass over density, which means that our volume is equal to one. We chose that arbitrary mass 1/11 300. I just moved it from scientific notation to standard notation and that would be in meters cubed and what we can do once we know that volume, I'm not gonna put that into a decimal form for now, I know that that volume is equal to 4/3 pi r cubed. I got that from knowing that the volume of a sphere is equal to 4/3 pi r cubed. What I'm going to do now is I'm going to rearrange that equation to solve for our and that's going to be the first thing we're looking for. If we rearrange that, we know that our is equal to the cube root of 1/11 300 times, 3/4 times, one over pi. We do that calculation and we find that the radius of our lead sphere is equal to 0.27645 meters. Not going to do any rounding quite yet. What's now find the radius of our aluminum sphere. The density of aluminum is equal to 2.7 times 10 to 3 kilograms per cubic meter again from before. I know that volume is equal to mass over density. So what I can do is I can use my arbitrary mass of one over density of 2700 again moving from scientific notation into standard notation. What I can do then is I can say that that volume is equal to 4/3 pi r cubed our volume first fear I will rearrange and I will end up with the cube root of our volume times 3/4 times one over pi. The way I get from our original volume expression to this new radius centric expression is I multiply by the reciprocal 4/3 which is 3/4 multiple by the reciprocal of pie which is one over pi. And then to isolate that r the R is being cubes, I will cue brute the entire expression. When I calculate that I find that the radius of our aluminum spear you go 0.4455 meters is the second piece of information that we need. And now we're going to solve for a final answer. Remember, we're looking for the radius of the aluminum spear to the radius of the lead sphere now, because we chose an arbitrary mass, we don't know what these radio I actually are. But we know that we use the same mass and both, so we can still set this up as a proportion. So, using arbitrary mess, we found that the radius of our lead sphere was equal to 0.27645 meters. We found in the radius of our aluminum spear was equal to 0.4455 meters. Just a quick check. It would make sense that the aluminum spear is larger than the lead sphere because aluminum has a lower density, so it's going to be bigger to have the same mess. We are looking for the ratio of the radius of the aluminum sphere to the radius of the lead sphere, so consent that up. So it's our radius for the aluminum on the left radius for the lead on the right. That is our ratio. We can compute that down to a final number by dividing the radius of our aluminum sphere by the radius of our lead sphere, and we will end up with a final answer of 1.6115 That is the ratio of aluminum to our lead radius

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