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How big is a million dollars? At the time this problem was written, the price of gold was about $\$ 1239$ per ounce, while that of platinum was about $\$ 1508$ an ounce. The "ounce" in this case is the troy ounce, which is equal to 31.1035 g. The more familiar avoirdupois ounce is equal to 28.35 g.) The density of gold is 19.3 $\mathrm{g} / \mathrm{cm}^{3}$ and that of platinum is 21.4 $\mathrm{g} / \mathrm{cm}^{3} .$ (a) If you find a spherical gold nugget worth 1.00 million dollars,what would be its diameter? (b) How much would a platinum nugget of this size be worth?

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a)diameter $=2 \times 6.771=13.542 \mathrm{cm}$b) $1.349 \times 10^{6} \mathrm{dollars}$

Physics 101 Mechanics

Chapter 13

Fluid Mechanics

Temperature and Heat

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

McMaster University

Lectures

03:45

In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.

09:49

A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas. Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion. Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids. Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.

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this question asked us to find the diameter of a spherical gold nugget that is worth $1 million. The first thing that we're going to need to do is figure out the information we need in order to solve this problem, and much of it is given. We're told the value of gold is $1239 per ounce. We're told this ounces equal to 28.35 grams, were given the density of gold, which is 19.3 grams per cubic centimeter. We're also going to need to know some sphere geometry, so we should know that the volume of a sphere is you put a 4/3 pi r cubed and we're going to eventually need to find the diameter of this nugget. So we need to be able to take the radius from the previous equation and turn it into a diameter by doubling it. Now, our pathway for kind of solving this problem is first, we're going to take our sphere worth $1 million and we're going to figure out it's weight in ounces. I never want to take that way, announces we're going to convert it into a Mass in grams. We're going to use that information and its density to find its volume. And then finally, once we know the volume of the sphere we can solve, four are so to start off with in order to find the weight announces of our $1,000,000 sphere of gold, we're going to take that $1 million. And we know that for every 12 $39 of that $1 million there is one ounce of gold. We got that from the information right here that I've circled in red. Now what we can do is if we divide by that value, we will end up with the amount of ounces of gold that we have, and that is 807.103 ounces. I'm going to do my best, not round until we get to the very end of our problem. So we've turned our valiant ounces. Now we need to turn our ounces into Graham's. We have 807.103 ounces. We are told that there are 28.35 grams in one of these ounces and if we multiply, we see that are 807.1 of three ounces is equal to 2281.4 grams. So now we've converted that value into Graham's. Now we need to turn that into a volume, and we're going to use the provided density for that. Luckily, our density is already dealing with grams. We don't need to convert into kilograms. So what we're going to do, we know that the density of our gold is equal to the mass of that gold divided by the volume of that gold. So we can rearrange that equation Algebraic lee to solve for the volume of the gold is equal to the mass of the gold which we have divided by the density of the gold which we have so to find my volume that is equal to my mass 22881.4 grams over my density which is provided 19.3 grams per cubic centimeter. I can double check dimensionally that this will end up with a volume and it does because our grams will cancel out. We'll be left with a centimeter cubed in the numerator. And so the volume of my sphere is 1185.56 centimeters. Cute. So now we have figured out our volume. Our next step is to figure out the radius of this sphere. Do that. I'm gonna grab a new white board to find the radius of my sphere. Well, first, I know the volume of a sphere is equal to 4/3 pi r cubed are being the radius. If I rearrange that equation, I can say that it is also equal to the cubed root of three V over four pie where V is my volume. If I substituted the volume that I know, I'm going to need to take the cube root of three times 1185.56 Again, that's my volume and cubic centimeters divided by four pi and that will be equal to my radius. I compute that and I find that the radius of my sphere is 6.5656 centimeters and then doubling It will give me my diameter. I could remember that diameter is equal to two times my radius. So to find my diameter, I just double it. And my diameter of my sphere worth $1 million is 13.13 cent of years. That is the diameter of my sphere. My second question then I need to work with now is what is the value of a platinum sphere of the same size. And to find that I'm gonna need a few more pieces of information from the problem. We need the information about platinum. So I'm gonna need to know the value of platinum, which, the problem tells me, is 15 $08 per ounce. And I'm also going to need to know the density of platinum which is provided, and we're told it is 21.4 grams per cubic centimeter. Now to solve this portion, the problem, We're pretty much going to go backwards. So instead of starting with the value and working down to the diameter, we're starting with the diameter and working up to the value. Now, what's nice about this problem was we can skip a few steps. So we're told the the platinum sphere is going to have the same size as our gold sphere, so we know that its volume is going to be the same. And the volume of our gold sphere was 11. 85.56 cubic centimeters. I can use this information to figure out the mass of this platinum sphere. If I rearrange our density equation. I know that mass is equal to our volume times air density and again, our volume 11 85.56. And the density of platinum provided from the problem is 21.4. So, again, my, um, volume that I found is in cubic centimeters. My density is in grams per cubic centimeter. So if I multiply those two things, I will be left with a mass in grams and that is what I get. The mass of this platinum sphere is 25 3 71 rams. Now what I'm going to do is I'm going to convert it back into this ounces unit. I need that ounces unit in order to turn it into the value because my value is described in ounces. So I'm gonna take my mass in grams. And I know that there are 28.35 grams in one of these ounces. If I divide my mass by 28.35 grams to get the number of ounces in this sample, that is E 94.92 ounces. And then finally, my last step. I know that I have 8 94.92 ounces and I'm told the value per ounce is 15 $08. So if I multiply those two things, I will get the value of this sphere in dollars, which is 1.3 for nine times 10 to the six dollars. If I want Thio round a little bit here, I think are large. Our most significant value head, rather at least significant value has three significant digits. Some went around this to the third significant digit. So it around this appropriately 1.35 times 10 to the $6 is my answer. So we found the diameter of a gold sphere with a value of $1 million we found the value of a platinum sphere of the same size

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