00:01
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.
00:12
Now throughout, i'm going to use al, the atomic symbol for aluminum, to represent aluminum, and pb, the atomic symbol for lead to represent lead.
00:20
Now, our general game plan here is, first, i'm going to use a mass of one kilogram for both of these spheres, just to make the calculations a little bit easy, and also to avoid having too much of a symbolic mess.
00:36
So i'm choosing this arbitrary mass of one kilogram.
00:38
Our general game plan is going to be to take the mass of each of these spheres, use the densities of both of these materials, which we can look up, use that to turn it into the volume of those spheres, and then we're going to use the volume to find the radius of those spheres.
00:59
In order to do that, we need to know that density is equal to mass over volume.
01:06
We need to know that the volume of a sphere is equal to four -thirds pi radius cubed.
01:16
And we also need to know the densities of these two materials.
01:19
The density of lead is equal to 1 .13 times 10 to the 3 kilograms per cubic meter.
01:28
And the density of aluminum is equal to 2 .7 times 10 to the 3 kilograms per cubic meter.
01:38
So let's get started.
01:40
First, we'll deal with finding the radius of this lead sphere.
01:45
So we will start there.
01:47
So to find the radius of our lead sphere, again, we know that our density is equal to 11 .3 times 10 to the 3 kilograms, per cubic meter.
02:01
We know that because density is equal to mass over volume, we can rearrange that.
02:08
Say volume is equal to mass over density, which means that our volume is equal to one.
02:13
We chose that arbitrary mass 1 over 11300.
02:17
I just moved it from scientific notation to standard notation.
02:22
And that would be in meters cubed.
02:27
And what we can do once we know that volume, i'm not going to put that into a decimal form for now.
02:32
I know that that volume is equal to 4 thirds pi r cubed.
02:40
I got that from knowing that the volume of a sphere is equal to 4 thirds pi r cubed.
02:47
What i'm going to do now is i'm going to rearrange that equation to solve for r, and that's going to be the first thing we are looking for.
02:55
If we rearrange that, we know that r is equal to the cube root of 1 over 11300 times 3 quarters, times 1 over pi.
03:09
We do that calculation and we find that the radius of our lead sphere is equal to 0 .0 -27645 meters...