Solving Density Problems - Video Tutorials & Practice Problems

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1

concept

Problems with Mass, Volume, & Density

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4m

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Hey, guys. So you may run into some problems involving density. So in the studio, I'm just gonna give you a quick refresher on how we solve density problems just in case. It's been some time since you've seen this stuff. So guys, remember that density is defined as mass, which is the amount of stuff that something has divided by the volume, which is the amount of space that it takes up. The symbol for density is this Greek letter. Roach kind of looks like a P, but it's a little curved and it's defined as mass over volume em over V. Now, in the S I system, the units are kilograms per meter cubed and really guys, all these problems are just gonna relate these three variables inside of this equation density mass and volume for just different geometric shapes like rectangular prisms, cubes, fear, cylinders, all that stuff. And then eventually you may. You may also have to convert some units, so that's really all there's to it. Let's just jump straight into an example. So we've got the average density of the earth is 5500. We're gonna assume that's the sphere which might be news to some of you. Contrary to what you see on the Internet, the earth is actually around. The radius is 39 60 and we're gonna calculate with the mass is so first things first, I'm just gonna draw a quick little sketch of the Earth. It's a sphere, so I'm gonna draw dry like this. So here's the sphere and the radius, which I'm gonna call capital are we already know what that is? It's 3900 and 60. So I know the density of the earth is 5500. And I know the radius of the earth is 39 60 but it's in miles and the density is in kilograms per meter. Cute. So I'm going to use this information to find out what the mass of the Earth is. This is my target variable. So here's what gonna dio and these problems involving density you're always just gonna start off just with by writing the density equation. So remember that the density equation row is just equal to mass over volume. So if I'm trying to figure out what the mass is, basically all I need to do is just rearrange the equation so I could move this velocity term up here and just say that row times the volume is equal to the mass. Now, I already know what the density is and it's in the right units, kilograms per meter cubed so I can use this. Basically figure out the mass is all I have to do is just figure out what the volume of the earth is. So notice All we have is the radius. We don't actually have the volume, so let's just go over here and figure out what that volume is. So if I can figure out the volume of the earth which just just modeled after a sphere So we're gonna use this volume equation up here, which is four thirds pi times radius of the Earth Cube. So we can say that this is the radius of the Earth. If I can figure out this volume here, I can plug it back into my equation for density and figure out the mass. Unfortunately, what happens is that if I do this and if I used the unit that I have for Radius, this is given to me in miles, which unfortunately is not the S I unit. So first I actually have to do a unit conversion. So I'm told here that one mild is equal to approximately 16 09 m. So what that means is I can set up a little unit conversion here. I can say that 39 60 times a conversion factor. Well, give me meters. So to do that, I want to set up my conversion factors so that the miles is on the bottom. So little cancel out. And so one mile, I'm told here is equal to 16 09 m. And so if I work this out, the miles cancel and I'm just left with meters, which is the S I unit that I want. So if you work this out in your calculators, you're gonna get 637 or 6.37 Let me just write this out a little bit. 6.37 times 10 to the six, and that's in meters. You could also just displayed as a as a full number. That's totally fine. I just chose to do in scientific notation because it's a little bit easier that way. So here my units and meters. So now I've got the radius. I'm just gonna plug it back into this equation and figure the volume. So the volume of the earth is just equal to four thirds times pi times 6.37 times 10 to the sixth. You have to cube that. Make sure that you plug this into your calculator, do this first and then multiplied by this raised to the third power on your calculator should be handling the rest. And when you do that, you're gonna get the volume of the earth is equal to 1.0. That's actually what's 1.8 times 10 to the 21 and centimeters cubes. Now we just plug it back to this equation. So now we have the density, which is 5500, and we have the volume, which is 1.8 times 10 to the 21 you're gonna get the mass of the volume. And our mass is just 5.94 times 10 to the 24 kg. So that's what we get for the mass of the earth. It's actually pretty close. Um, might be just a slightly off from the rial value just in case, you go look it up, and that's because we rounded a bit. But that's basically the right answer for the mass of the Earth. That's it for this one, guys, let me know if you have any questions.

2

Problem

Problem

A wooden cylinder has a radius of 3.5 cm and a height of 6 cm. If the mass is 161 g, what is the density of the wooden cylinder?

A

222 kg/m^{3}

B

3.767×10^{3} kg/m^{3}

C

697 kg/m^{3}

D

2440 kg/m^{3}

3

example

Finding the Side Length of a Cube

Video duration:

3m

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Hey, guys, hopefully get a chance to check out this example. Problem. We've got an iron cube with the mass. We're told, with the density of this Iron Cube is and we're gonna figure out the length of the sides of the Cube. So the first thing I'd like to do is just draw a quick little sketch are diagram. We're told we have an iron cubes. I'm gonna draw a little cube here. And a cube is basically, like a special case of rectangular Prism in which all the lengths of the sides are the same. So I'm gonna call The side s here. You don't have to say l length, width and height because they're all the same exact number. So what else do I know? I'm told the mass the mass is just equal to 0.515 It's in kilograms. So it's the right unit. Also told the density of iron 7.87 times 10 to the third. This is also in the correct units, kilograms per meters cube. And so I'm gonna use this information to figure out what the side lengths of the Cube are. That's just the letter s So how do I use this to figure out what s is? That's my target Variable. Well, like we said, every time we have a density like densities, masses and volumes involved just started with the density equation. So I brought My row is equal to mass divided by volume. So how does this get me any closer to figuring out what the side links are? Well, I already know what the what the density is. And I also would already know what the masses. So it's gonna have to do something with this volume term. So the volume of a rectangular prison, remember, is just given by this equation length, times with times height. But in a special case of a cube where all these letters are the same length, width and height, it's actually an even simpler equation. It's just the side length cube. So this is my target variable. So this is actually what I have to figure out so I can relate this back to the volume and the volume I could get from using the density equation. So that's what I'm gonna dio. So I'm gonna use this density equation to figure out my volume and then basically, just pass it back into this equation and then figure out the side length. So let's go ahead and do that. If I'm looking for the volume, I just need to move it over to one side. So what I can do is I can trade places with this row with this density, and I can say that volume is equal to mass divided by the density. So my volume is just my mass, which is 0.515 divided by the density 7.87 times, 10 to the third and notice how I've already checked the units. And I don't have to you do any union conversions because they're all in the same sort of like they all compatible with each other. So I just go ahead and plug this in and what I get is 6.54 times 10 to the minus 5 m cube. So now basically exactly like what we said, we can pass this number back into this equation and then figure out the side length. So that means my volume now which is 6.54 times 10 to the minus fifth, is equal to the side length cube. So how do I get rid of this cubed? And how do I just get s? Well, I have to take the cube root of both sides. So I have to take the cube root, and I'm gonna take the cube root here. So what happens is the cube root and the cube will cancel out, leaving me just the side length. And so, if you plug in the cube roots of this number 6.54 times 10 to the minus five, you're just gonna get 0.4 m or that would just be also four centimeters. So either one of these is correct. Just depends on which unit you would have to express it in. Um, but that's really it. So let me know if you guys have any questions, that's about this one.

4

Problem

Problem

Copper has a density of 8.96 g/cm^{3}. If a single copper atom as a mass of 1.055×10^{-25} kg, what is the volume of a copper atom?

A

1.18×10^{-26} m^{3}

B

9.45×10^{-25} m^{3}

C

1.18×10^{-29} m^{3}

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