Problem

A certain audio compact dise (CD) contains 783.21…

00:59

Problem 50 Hard Difficulty

How big is a ton? That is, what is the volume of something that weighs a ton? To be specific, estimate the diameter of a 1-ton rock, but first make a wild guess. will it be 1 ft across. 3 ftr, or the size of a car? [Hint. Rock has mass per volume about 3 times that of water, which is 1 $\mathrm{kg}$ per liter $\left(10^{3} \mathrm{cm}^{3}\right)$ or 62 $\mathrm{lb}$ per cubic foot.

Answer

3 $\mathrm{ft}$

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 1

Introduction, Measurement, Estimating

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Video Transcript

Our question says that how big is a ton? That is. What is the volume of something that weighs a ton? To be specific, estimate the diameter of a one ton rock. But first, just make a wild guess. Will it be a foot a foot across three feet across or the size of a car? Hint. Rock, Hint. Rock has mass per volume about three times that of water, which is one kilogram per leader. Tend to third centimeters cubed or 62 pounds per cubic foot. So I wrote down what we were given in this problem. We were given that the density of water, which is mass per value that's density. So I wrote density as row. So the density of water, which is rows of W, is equal to 62 pounds per cubic foot. And it says that the didn't the density of rock, which I wrote his Rosa are is three times the density of water, so three times roast of W is 186 pounds for per cubic foot. It also tells us the mass of the rock in question in the mass of the rock is one time well by definition. One ton is 2,000 pounds. The question then is what is the diameter of this rock If we assume that it's ah perfect sphere. So the diameter which I wrote his D is equal to what? In order to do this, we're going to make use of the fact that the volume of the sphere is equal to 4/3 pi R to the third 4/3 pi r Who forgot pie once, right? Well, yeah, Let's write pie first to keep it consistent with what I said in common notation. Hi R to the third. Okay, well, we can rearrange us to solve for R because we want our since really what we want is distance or the diameter, but diameter is two times the radius, so solving for the radius will be just fine. In the radius, then, is three times of volume divided by floor Kai. But the radius was cube. So this entire thing is going to be too cute group or to the 1/3. Okay, well, all we need then his volume and we can get volume using the fact that we know the density of the object in question and we know the mass of the object in question because volume is also equal to the density of the object in question. I'm sorry, the mass of the object in question, which is the rock so imps of our divided by the density of the object in question. And you can check real quick with just using dimensional analysis. So the mass is in pounds and the density is in pounds per cubic foot. The pounds is going to cancel all you're going to be left with this cubic foot and volumes is in cubic foot. So we know the dimensions workout. So let's go ahead and carry out this operation. We have 2,000 pounds. Excuse me, 2,000 pounds divided by the density, which is 186 pounds per cubic foot. And if you do that, that comes out to be a 10.8 cubic feet. Okay, so now we know the volume in question of the rock so we can solve for the distance or the diameter. Excuse me. The diameter diameter is two times are Okay, so this is going to be too. Times equals two times three multiplied by 10 0.8 cubic feet divided by for pie, and all of that is raised to the 1/3 and again, the dimensions. Check out because we have cubic feet here inside the parentheses raised to the 1/3 you're going to be left with just feet and you expect the diameter to be in feet, so it works out. If you carry out this operation, you find this is equal to 2.74 feet and in assets. In the beginning, if we thought it would be approximately one foot three foot or the size of the car will, 2.74 feet is approximately three feet, so there's your answer him box set in.

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