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Chapter 13: Problem 3

A standard gold bar stored at Fort Knox, Kentucky, is 7.00 inches long, 3.63 inches wide, and 1.75 inches tall. Gold has a density of \(19,300 \mathrm{kg} / \mathrm{m}^{3} .\) What is the mass of such a gold bar?

Short Answer

Expert verified
The mass of the gold bar is 14.09 kg

Step by step solution

01

Convert Dimensions from Inches to Meters

Each dimension of the gold bar needs to be converted from inches to meters, as the density of gold is given in kg/m鲁. Using the conversion factor \(1 inch = 0.0254 meters\), multiply each dimension by it: \n length = 7.00 inches * 0.0254 m/inch = 0.178 m \n width = 3.63 inches * 0.0254 m/inch = 0.092 m \n height = 1.75 inches * 0.0254 m/inch = 0.0445 m
02

Calculate the Volume of the Gold Bar

The volume of a rectangular prism (which the gold bar is in this case) can be calculated by multiplying the length, width, and height. Using the meters measurements from step 1 calculate the volume: \n volume = length * width * height = 0.178 m * 0.092 m * 0.0445 m = 0.00073 m鲁
03

Calculate the Mass of the Gold Bar

Finally, use the density of gold to find the mass of the gold bar. The formula to use here is: \n mass = density * volume \n Plug in the given density of gold (19,300 kg/m鲁) and the calculated volume from step 2 (0.00073 m鲁): \n mass = 19,300 kg/m鲁 * 0.00073 m鲁 = 14.09 kg

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Key Concepts

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

Density Calculation
Density helps us understand how mass is distributed in a specific volume. It is a property that describes how much mass is present in a given volume of a substance. The formula for density is:\[\text{Density} = \frac{\text{Mass}}{\text{Volume}}\]This means that if you know the density and volume of an object, you can find its mass by rearranging the formula to:\[\text{Mass} = \text{Density} \times \text{Volume}\]Gold has a high density because its atoms are closely packed, making it heavier than other less dense materials. When dealing with problems involving density, always remember to use the correct units as they can affect the accuracy of your calculations.
Unit Conversion
Unit conversion is essential when working with measurements in different units. In physics, you often need to switch between units. For instance, the dimensions of the gold bar are initially given in inches. However, converting them to meters is necessary for consistency when calculating volume and mass with a density given in kg/m鲁.
  • 1 inch is equivalent to 0.0254 meters.
  • Multiply each dimension by 0.0254 to convert it to meters.
By converting all units to a consistent system, calculations become straightforward and accurate. Improper unit conversions can lead to errors in physical calculations, so always double-check your conversion factors.
Volume Calculation
Next, let's focus on calculating volume, which is the amount of space an object occupies. For a rectangular prism like the gold bar, volume is found using the formula:\[\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}\]Suppose we have converted measurements to meters, plug in the values:
  • Length = 0.178 meters
  • Width = 0.092 meters
  • Height = 0.0445 meters
The volume thus comes out to be:\[0.178 \times 0.092 \times 0.0445 = 0.00072909 \text{ m}^3\]Volume is a crucial component in density and mass calculations because it directly affects the mass of the object for a given density.
Mass Calculation
Mass calculation involves using the density and volume to find the mass of an object. Once you have the volume in cubic meters, use the given density to determine the mass.
  • Given density of gold = 19,300 kg/m鲁
  • Calculated Volume = 0.00072909 m鲁
Use the mass formula:\[\text{Mass} = \text{Density} \times \text{Volume}\]So, substitute the numbers:\[\text{Mass} = 19,300 \times 0.00072909 = 14.08 \text{ kg}\]Understanding mass calculations helps us not only in physics but also in real-world applications, like determining how much a gold bar actually weighs, reflecting its value accurately.

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Most popular questions from this chapter

A flat slab of styrofoam, with a density of \(32 \mathrm{kg} / \mathrm{m}^{3},\) floats on a lake. What minimum volume must the slab have so that a \(40 \mathrm{kg}\) boy can sit on the slab without it sinking?

\(\mathrm{A} 2.0 \mathrm{mL}\) syringe has an inner diameter of \(6.0 \mathrm{mm},\) a needle inner diameter of \(0.25 \mathrm{mm},\) and a plunger pad diameter (where you place your finger) of \(1.2 \mathrm{cm} .\) A nurse uses the syringe to inject medicine into a patient whose blood pressure is \(140 / 100\). Assume the liquid is an ideal fluid. a. What is the minimum force the nurse needs to apply to the syringe? b. The nurse empties the syringe in 2.0 s. What is the flow speed of the medicine through the needle?

During typical urination, a man releases about \(400 \mathrm{mL}\) of urine in about 30 seconds through the urethra, which we can model as a tube \(4 \mathrm{mm}\) in diameter and \(20 \mathrm{cm}\) long. Assume that urine has the same density as water, and that viscosity can be ignored for this flow. a. What is the flow speed in the urethra? b. If we assume that the fluid is released at the same height as the bladder and that the fluid is at rest in the bladder (a reasonable approximation), what bladder pressure would be necessary to produce this flow? (In fact, there are additional factors that require additional pressure; the actual pressure is higher than this.)

A 35-cm-tall, 5.0-cm-diameter cylindrical beaker is filled to its brim with water. What is the downward force of the water on the bottom of the beaker?

A patient has developed a narrowing of the coronary arteries that provide blood to the heart. If the rate of blood flow through the arteries is to stay the same, A. The length of the arteries must decrease as well. B. The speed of the flow must not change. C. The pressure difference across the arteries must stay the same, requiring a decrease in blood pressure. D. The pressure difference across the arteries must increase, requiring an increase in blood pressure.
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