Density of Geometric Objects - Video Tutorials & Practice Problems

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The Density of geometric objects generally includes spheres, cubes and cylinders.

Calculating Density

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concept

Density of Geometric Objects Concept

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So we understand the density is just mass per volume. Now we're gonna take that idea of density and apply to geometric objects Were going to say here when given the mass of a geometric object you can related to its volume and density are typical. Geometric objects are a sphere, a cube and a cylinder. Each one of them has their own volume equation which when later on, relate to density. If we wish so here we have a sphere. Now, in this sphere we have our radius. And remember, the radius is just the distance from the center to the edge of the sphere. When it comes to a spear, its volume equation is volume equals 4/ times pi times radius. Cute notice. None of these form those were going to write are in purple boxes, which means you don't need to commit it to memory. Typically, when it comes to volumes of geometric objects, your professor will give it to you within the question or on a formula sheet. Now, when it comes to a cube, Cube has all these sides which we label a in a cube. We assume that they're all of equal length. As a result of this, the voluble cube is equal to a cube where again a is equal to the length or the edge of that cube. Finally, we have a cylinder, and in the cylinder we have to take into account two variables. We have the height of the cylinder, and we have, of course, again, the radius of the cylinder. Taking these two into perspective. When it comes to a cylinder, we have volume equals pi times, radius squared, times the height. So now we're going to take a look at density questions which relate to these different types of geometric objects.

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example

Density of Geometric Objects Example

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So here in this example question, it says the density of silver is 10.5 g per centimeters. Cute. What is the mass in kilograms off a cube of silver that measures 0.56 m on each side. All right, although we're talking about density here, we're talking about a cube weaken still, apply the rules that we learned when it comes to giving amounts, converting factors and end amounts. So in this question, our end amount or given amount actually is the one that has only one unit connected to it. And that would be our 10.56 m. So this is our given amount. Then we have to think about where is our end amount? Where do we need to go and where do we stop Our end amount here would be kilograms, so we need kilograms from the Silver Cube. Next we look to see what kind of conversion factors are available. Here. We have the density of the of the silver. Remember density itself is a conversion factor because it's tying together two different units. So that would be 10.5 g per one centimeters. Cute. All right, so now let's put it all together. So we have given here given amount. We have to get to end them out. Okay? And remember to connect them together. We use our conversion factors. Now we see our conversion factor. There is 10.5 g per centimeters. Cute. We can try toe. Isolate these grams here. If I can isolate those grams, I can convert them later into kilograms. But to be able to isolate those grams, I have to cancel out these centimeters. Cube, Remember, Centimeters Cube is just a unit for volume. Sometimes were given an amount we have are given amount. But sometimes we have to change it up. We have to change it so that we can use it within our dimensional analysis set up. Since this whole section is on the density off geometric objects and we have a cube. I can determine the volume of the cube. Remember, volume equals length. Cute. So that be 56 m cube. So that comes out to be 0.1756161 m, cube. So this is my actual given amount that I'm gonna utilize within this set up, and I need to change it. 2 m cube because from there I can convert it to Centimeters Cube and then from Centimeters Cube I'll be able to cancel out the centimeters from the density and be left with grams. All right, so our conversion factor one is just a metric prefix conversion meters, go here on the bottom Centimeters go here on top. Remember, the coefficient of one goes on the same side with the metric prefix. So one senti is 10 to the negative too. Here these meters Air Cube. But these were not so I would have to cube the whole thing. Then we're going to have our leaders cubes cancel out. And now we have Centimeters Cube because it's in centimeters cubes. I can now use my conversion factor too. So my conversion factor to I bring this down 10.5 g one centimeters cubes on the bottom, centimeters cubes will cancel out. And now I have grams off my substance, but I don't want it in grams I needed in kilograms. So one mawr conversion factor is actually another metric prefix conversion. So we're gonna put here grams on the bottom, kilograms on top. Remember, Kilo is a metric prefix, so 1 kg is 10 to the three. So grams here cancel out and I'll be left with kilograms at the end. So when you plug this in initially it'll be 0.715 point times 10.5 divided by 10 to the negative to which is gonna be cubed, divided by also tend to the three. If you plug this incorrectly, you'll get 1 968 kg. But looking back on the original question, 0.56 has to sig figs and 10.5 has three sig figs. So we have to go with the least number of significant figures. So getting this to to sick figs weaken right it either as kg or 1.8 times 10 to the 3 kg. Either one is, ah, reliable, reasonable answer. So just remember, sometimes our dimensional analysis set up. The given amount has to be changed a little bit in order to fit and cancel. I with other present units. By doing that, were able to isolate our kilograms within this example. Now that we've done this question, let's move on to our practice question

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Problem

Problem

A copper wire (density = 8.96 g/cm^{3}) has a diameter of 0.32 mm. If a sample of this copper wire has a mass of 21.7 g, how long is the wire?

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Problem

Problem

If the density of a certain spherical atomic nucleus is 1.0 x 10^{14} g/cm^{3} and its mass is 3.5 x 10^{-23} g, what is the radius in angstroms? (Å= 10^{−10} m)

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