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Chapter 12: Problem 4

You win the lottery and decide to impress your friends by exhibiting a million-dollar cube of gold. At the time, gold is selling for \(\$ 426.60\) per troy ounce, and 1.0000 troy ounce equals 31.1035 g. How tall would your million-dollar cube be?

Short Answer

Expert verified
The cube would be approximately 15.45 cm tall.

Step by step solution

01

Determine the Total Weight of Gold in Troy Ounces

First, calculate how many troy ounces of gold you would get for a million dollars. Since the price of gold is \( \\(426.60 \) per troy ounce, the number of troy ounces you can purchase with \( \\)1,000,000 \) is calculated as: \[ \text{Total weight in troy ounces} = \frac{1,000,000}{426.60} \approx 2344.34 \text{ troy ounces} \]
02

Convert Troy Ounces to Grams

Now, convert the total weight from troy ounces to grams using the conversion factor (1 troy ounce = 31.1035 grams). Thus, the total weight in grams is calculated by: \[ \text{Total weight in grams} = 2344.34 \times 31.1035 \approx 72909.74 \text{ grams} \]
03

Calculate the Volume of Gold

Given that the density of gold is approximately 19.32 grams per cubic centimeter, you can find the volume by using the formula \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \). Thus: \[ \text{Volume} = \frac{72909.74}{19.32} \approx 3774.78 \text{ cm}^3 \]
04

Determine the Size of the Cube

Assuming the volume corresponds to a cube, one can find the length of one side by finding the cube root of the volume. Hence, the side length \( s \) is: \[ s = \sqrt[3]{3774.78} \approx 15.45 \text{ cm} \]
05

Convert the Size to Inches or Feet if Necessary (Optional)

If you prefer expressing the size in inches (given 1 inch is 2.54 cm), convert the length as follows: \[ \text{Side length in inches} = \frac{15.45}{2.54} \approx 6.08 \text{ inches} \]. Alternatively, each side of the cube is about 0.51 feet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Converting Troy Ounces to Grams
Converting the weight of gold from troy ounces to grams is essential for understanding and working with measurements in the metric system. A troy ounce is a unit of measure commonly used for precious metals like gold. When working with such measurements, we often need to convert between units.

The conversion from troy ounces to grams is simple but precise. The conversion factor is:
  • 1 troy ounce = 31.1035 grams.
To convert troy ounces to grams, multiply the number of troy ounces by 31.1035. Take the case of converting 2344.34 troy ounces to grams:
  • The calculation would be: 2344.34 脳 31.1035 = 72909.74 grams.
This step is crucial for accurately determining the mass of a material in grams, which is the standard unit in science.

By converting to grams, you can also easily calculate the volume, density, and other properties of the material, since the metric system units are interrelated and easier to work with in mathematical formulas.
Volume Calculation for Cubes
Understanding how to calculate the volume of a cube is an essential geometric skill. A cube is a three-dimensional shape with six equal square faces. To find the volume of a cube, use the formula:
  • Volume of a cube = side length 脳 side length 脳 side length (or side length cubed, denoted as \(s^3\)).
Given the context of calculating the volume of a gold cube, once you have the mass and density, you can find the volume using the formula:
  • Volume = Mass/Density
For example, with a gold mass of 72909.74 grams and a density of 19.32 grams per cubic centimeter, the volume would be:
  • Volume = \(\frac{72909.74}{19.32} \approx 3774.78 \text{ cm}^3\)
Next, to find the side length of the cube (since it is a cube, all sides are equal), you determine the cube root of the volume:
  • Side length = \(\sqrt[3]{3774.78} \approx 15.45 \text{ cm}\)
Calculating the volume of a cube from its material properties allows for practical applications, from crafting to scientific research. It shows the relationship between physical size and mass properties.
Understanding the Density of Gold
The density of gold is a key property that defines its behavior and applications. Density is defined as the mass per unit volume, commonly expressed in grams per cubic centimeter (g/cm鲁) for solids. For gold, this density is approximately 19.32 g/cm鲁.

Why is understanding the density of gold important?
  • Indentification: It helps identify pure gold by comparing measured densities.
  • Design and Manufacturing: Helps in planning the mass and volume when designing items, ensuring proper weight distribution.
  • Valuation: Since gold is often weighed and estimated for value, knowing its density allows for precise calculations of worth and weight.
To calculate the density, you use:
  • Density (\(\rho\)) = \(\frac{Mass}{Volume}\)
Using the density of gold when calculating the volume helps derive the correct cube size in exercises such as creating an expensive gold cube. By understanding the density, one can convert mass measurements into practical size estimates.

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Most popular questions from this chapter

Advertisements for a certain small car claim that it floats in water. (a) If the car's mass is 900 \(\mathrm{kg}\) and its interior volume is \(3.0 \mathrm{m}^{3},\) what fraction of the car is immersed when it floats? You can ignore the volume of steel and other materials. (b) Water gradually leaks in and displaces the air in the car. What fraction of the interior volume is filled with water when the car sinks?

A pressure difference of \(6.00 \times 10^{4} \mathrm{Pa}\) is required to maintain a volume flow rate of 0.800 \(\mathrm{m}^{3} / \mathrm{s}\) for a viscous fluid flowing through a section of cylindrical pipe that has radius 0.210 \(\mathrm{m} .\) What pressure difference is required to maintain the same volume flow rate if the radius of the pipe is decreased to 0.0700 \(\mathrm{m} ?\)

A swimming pool is 5.0 \(\mathrm{m}\) long, 4.0 \(\mathrm{m}\) wide, and 3.0 \(\mathrm{m}\) deep. Compute the force exerted by the water against (a) the bottom and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth \(h,\) and integrate this over the end of the pool.) Do not include the force due to air pressure.

The surface pressure on Venus is 92 atm, and the acceleration due to gravity there is 0.894\(g .\) In a future exploratory mission, an upright cylindrical tank of benzene is sealed at the top but stillpressurized at 92 atm iust above the benzene. The tank has a diameter of \(1.72 \mathrm{m},\) and the benzene column is 11.50 \(\mathrm{m}\) tall. Ignore any effects due to the very high temperature on Venus. (a) What total force is exerted on the inside surface of the bottom of the tank? (b) What force does the Venusian atmosphere exert on the outside surface of the bottom of the tank? (c) What total inward force does the atmosphere exert on the vertical walls of the tank?

A liquid flowing from a vertical pipe has a definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed \(v_{0}\) and the radius of the stream of liquid is \(r_{0}\) . (a) Find an equation for the speed of the liquid as a function of the distance \(y\) it has fallen. Combining this with the equation of continuity, find an expression for the radius of the stream as a function of \(y\) . (b) If water flows out of a vertical pipe at a speed of 1.20 \(\mathrm{m} / \mathrm{s}\) , how far below the outlet will the radius be one-half the original radius of the stream?
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