Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 $1282 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 of gold would be approximately 10.78 cm (or 4.24 inches) tall.

Step by step solution

01

Calculate Mass of Gold in Troy Ounces

First, determine how many troy ounces of gold you can buy with one million dollars. Since gold costs $1282 per troy ounce, you can find the number of troy ounces by dividing the total money by the price per ounce: \[ \text{Troy ounces} = \frac{1,000,000}{1282} \approx 780.03 \text{ troy ounces} \]
02

Convert Troy Ounces to Grams

Each troy ounce is equivalent to 31.1035 grams. Convert the total gold weight from troy ounces to grams using this conversion factor:\[ \text{Grams of gold} = 780.03 \times 31.1035 \approx 24,272.50 \text{ grams} \]
03

Calculate Volume of Gold in Cubic Centimeters

Find the volume of the gold by dividing its mass in grams by the density of gold (19.32 g/cm³):\[ \text{Volume} = \frac{24,272.50 \text{ grams}}{19.32 \text{ g/cm}^3} \approx 1,256.82 \text{ cm}^3 \]
04

Determine Side Length of the Cube

The volume of a cube is given by the cube of its side length. Therefore, solve for the side length using:\[ s^3 = 1,256.82 \]Thus, the side length is:\[ s = \sqrt[3]{1,256.82} \approx 10.78 \text{ cm} \]
05

Convert Side Length to Inches

Convert the side length from centimeters to inches by dividing by 2.54 (since there are 2.54 cm in an inch):\[ s \approx \frac{10.78}{2.54} \approx 4.24 \text{ inches} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Troy Ounce
A troy ounce is a unit of measure often used in the precious metals market, including gold, silver, and platinum. It's crucial to understand that a troy ounce is not the same as a regular ounce. While a troy ounce equals approximately 31.1035 grams, a standard ounce (often referred to as an avoirdupois ounce) is only about 28.35 grams.
This distinction is important since pricing and weight calculations in the context of precious metals rely on the troy ounce. Knowing the weight difference helps avoid confusion and ensures precise conversions and calculations in financial contexts.
Density
Density is a key concept in physics and chemistry that connects mass and volume. It is defined as the mass of an object divided by its volume, usually expressed in grams per cubic centimeter (g/cm³) for solids and liquids. In this scenario, the density of gold is provided as 19.32 g/cm³.
Understanding density allows us to determine how much space a certain mass of material will occupy. It can also be helpful in identifying substances based on their mass-to-volume ratio, as different materials have distinct densities depending on their atomic structure and composition.
  • High density means a substance is heavy for its size.
  • Low density indicates a substance is light for its size.
Volume of a Cube
The volume of a cube is calculated by taking the cube of its side length. This is expressed mathematically as: \[ V = s^3 \] where \( V \) represents volume and \( s \) denotes the side length of the cube.
Solving this equation allows us to find any missing dimension, provided we have the other two. In this exercise, knowing the volume of gold in cubic centimeters enables us to solve for the side length of the billion-dollar gold cube by applying the cube root to the volume. This emphasizes why spatial reasoning is essential in geometry and real-world applications.
Unit Conversion
In many scientific problems, converting units accurately is indispensable for obtaining correct results. Units conversion involves transforming one unit of measurement into another without changing the original value or quantity of the measurement.
For example, in the exercise: - Gold's weight in troy ounces was first converted to grams. - Then, dimensions were converted from centimeters to inches (used for the final side measurement of the cube).
These conversions ensure that we are working with units that are appropriate and consistent within calculations, bridging different systems of measurement effectively. Converting units helps in understanding and comparing different types of measurements without altering the significance of the data involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Scientists have found evidence that Mars may once have had an ocean 0.500 km deep. The acceleration due to gravity on Mars is 3.71 m/s\(^2\). (a) What would be the gauge pressure at the bottom of such an ocean, assuming it was freshwater? (b) To what depth would you need to go in the earth's ocean to experience the same gauge pressure?

(a) What is the \(difference\) between the pressure of the blood in your brain when you stand on your head and the pressure when you stand on your feet? Assume that you are 1.85 m tall. The density of blood is 1060 kg/m\(^3\). (b) What effect does the increased pressure have on the blood vessels in your brain?

An ore sample weighs 17.50 N in air. When the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 11.20 N. Find the total volume and the density of the sample.

Ballooning on Mars. It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atmosphere would keep the balloon aloft. The density of the Martian atmosphere is 0.0154 kg/m\(^3\) (although this varies with temperature). Suppose we construct these balloons of a thin but tough plastic having a density such that each square meter has a mass of 5.00 g. We inflate them with a very light gas whose mass we can ignore. (a) What should be the radius and mass of these balloons so they just hover above the surface of Mars? (b) If we released one of the balloons from part (a) on earth, where the atmospheric density is 1.20 kg/m\(^3\), what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down? (c) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

When an open-faced boat has a mass of 5750 kg, including its cargo and passengers, it floats with the water just up to the top of its gunwales (sides) on a freshwater lake. (a) What is the volume of this boat? (b) The captain decides that it is too dangerous to float with his boat on the verge of sinking, so he decides to throw some cargo overboard so that 20% of the boat's volume will be above water. How much mass should he throw out?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free