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Chapter 12: Problem 2

A cube 5.0 \(\mathrm{cm}\) on each side is made of a metal alloy. After you drill a cylindrical hole 2.0 \(\mathrm{cm}\) in diameter all the way through and perpendicular to one face, you find that the cube weighs 7.50 \(\mathrm{N}\) . (a) What is the density of this metal? (b) What did the cube weigh before you drilled the hole in it?

Short Answer

Expert verified
(a) The density of the metal is approximately 7004 kg/m鲁. (b) The original weight of the cube was about 8.59 N.

Step by step solution

01

Calculate the Volume of the Cube

A cube has a simple volume calculation. Since each side of the cube is 5.0 cm, we calculate the volume using the formula for the volume of a cube: \[ V_{ ext{cube}} = ext{side}^3 = (5.0 \ ext{cm})^3 = 125 \ ext{cm}^3 \]
02

Calculate the Volume of the Cylindrical Hole

First, find the radius of the cylindrical hole, which is half the diameter. The diameter is 2.0 cm, so the radius is 1.0 cm. The height of the cylinder is the same as the side of the cube (5.0 cm).Calculate the volume of the cylinder: \[ V_{ ext{cylinder}} = \pi imes ext{radius}^2 \times ext{height} = \pi \times (1.0 \ ext{cm})^2 \times 5.0 \ ext{cm} = 5\pi \ ext{cm}^3 \] approximately \(15.71\ \text{cm}^3 \).
03

Determine the Volume of the Drilled Cube

Subtract the volume of the cylindrical hole from the volume of the cube to find the remaining volume of the drilled cube:\[ V_{ ext{drilled}} = 125 \ ext{cm}^3 - 5\pi \ ext{cm}^3 \approx 125 \ ext{cm}^3 - 15.71 \ ext{cm}^3 \approx 109.29 \ ext{cm}^3 \]
04

Calculate the Density of the Metal

The density \( \rho \) can be found using the formula \( \rho = \frac{m}{V} \), where \( m \) is the mass and \( V \) is the volume. Since weight \( W = m \times g \), we have:\[ m = \frac{W}{g} = \frac{7.50 \ ext{N}}{9.81 \ ext{m/s}^2} \approx 0.765 \ ext{kg} \]Thus, the density is:\[ \rho = \frac{0.765 \ ext{kg}}{109.29 \ \text{cm}^3} \times 1000 \ \frac{kg}{m^3/cm^3} \approx 7004 \ \text{kg/m}^3 \]
05

Determine the Original Weight of the Cube

Find the original volume of the cube before drilling the hole, which was 125 \( \text{cm}^3 \). Using the density, calculate the original mass:\[ m_{ ext{original}} = \rho \times V_{ ext{cube}} = 7004 \ \frac{kg}{m^3} \times 125 \ \text{cm}^3 \times 10^{-6} \ ext{m}^3/ ext{cm}^3 \approx 0.8755 \ \text{kg} \]Convert mass to weight:\[ W_{ ext{original}} = m_{ ext{original}} \times g = 0.8755 \ ext{kg} \times 9.81 \ ext{m/s}^2 \approx 8.59 \ ext{N} \]

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Key Concepts

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

Volume of Cube
To calculate the volume of a cube, we use the formula \( V_{\text{cube}} = \text{side}^3 \). Each side of the cube is the same length, simplifying the process. In our scenario, the cube has a side length of 5.0 cm. Plugging that into the formula gives us:
  • \( V_{\text{cube}} = (5.0 \ \text{cm})^3 = 125 \ \text{cm}^3 \)
Cube volume calculations are straightforward due to their equal side lengths. This foundational geometric concept is critical for finding the volume of simple solid forms.
It helps in various applications like determining space availability within a container or computing physical properties like density.
Cylindrical Volume
The volume of a cylinder is slightly more complex than that of a cube. It requires knowledge of both the cylinder's radius and its height. The formula is \( V_{\text{cylinder}} = \pi \times \text{radius}^2 \times \text{height} \).
In our example, the cylinder's diameter is 2.0 cm, giving it a radius of 1.0 cm (since radius is half the diameter). The height is the same as the cube's side, 5.0 cm.
  • \( V_{\text{cylinder}} = \pi \times (1.0 \ \text{cm})^2 \times 5.0 \ \text{cm} = 5\pi \ \text{cm}^3 \)
  • \( V_{\text{cylinder}} \approx 15.71 \ \text{cm}^3 \)
Understanding how to derive such volumes is crucial in tasks involving subtracting solid shapes to form complex structures, which is often seen in manufacturing.
Mass and Weight Relationship
Mass and weight are related but distinct concepts in physics. Mass refers to the amount of matter in an object, measured in kilograms. Weight, however, is the force exerted by gravity on that mass and is measured in newtons.
In our exercise, the relationship between mass \( m \) and weight \( W \) is given by \( W = m \times g \), where \( g \) is the gravitational acceleration, approximately 9.81 \( \text{m/s}^2 \) on Earth.
  • To find mass when you have weight: \( m = \frac{W}{g} \)
  • After drilling, the mass was 0.765 kg.
Understanding this relationship helps us solve problems where mass must be derived from weight, such as calculating material properties like density.
Conversion of Units
Unit conversion is essential in scientific calculations, allowing consistent and accurate measurements across dimensions and scales. In density calculations, this involves converting volume from cubic centimeters to cubic meters for compatibility with kilograms in the SI unit system.
Since 1 \( \text{cm}^3 \) is \( 10^{-6} \ \text{m}^3 \), densities in kg/m鲁 require us to account for this small magnitude change:
  • Density in \( \frac{\text{kg}}{\text{m}^3} \) = \( \frac{\text{mass in kg}}{\text{volume in m}^3} \)
  • Here, the drilled cube's volume conversion is: \( 109.29 \ \text{cm}^3 \times 10^{-6} \ \text{m}^3/\text{cm}^3 \)
Such conversions are vital in science and engineering, ensuring accurate calculations and maintaining safety standards in product design and testing.

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Most popular questions from this chapter

In intravenous feeding, a needle is inserted in a vein in the patient's arm and a tube leads from the needle to a reservoir of fluid (density 1050 \(\mathrm{kg} / \mathrm{m}^{3} )\) located at height \(h\) above the arm. The top of the reservoir is open to the air. If the gauge pressure inside the vein is 5980 \(\mathrm{Pat}\) is the minimum value of \(h\) that allows fluid to enter the vein? Assume the needle diameter is large enough that you can ignore the viscosity (see Section 12.6\()\) of the fluid.

Viscous blood is flowing through an artery partially clogged by cholesterol. A surgeon wants to remove enough of the cholesterol to double the flow rate of blood through this artery. If the original diameter of the artery is \(D\) , what should be the new diameter (in terms of \(D\) ) to accomplish this for the same pressure gradient?

A hot-air balloon has a volume of 2200 \(\mathrm{m}^{3} .\) The balloon fabric (the envelope) weighs 900 \(\mathrm{N} .\) The basket with gear and full propane tanks weighs 1700 \(\mathrm{N} .\) If the balloon can barely lift an additional 3200 \(\mathrm{N}\) of passengers, breakfast, and champagne when the outside air density is \(1.23 \mathrm{kg} / \mathrm{m}^{3},\) what is the average density of the heated gases in the envelope?

A cubical block of wood 0.100 \(\mathrm{m}\) on a side and with a density of 550 \(\mathrm{kg} / \mathrm{m}^{3}\) floats in a jar of water. Oil with a density of 750 \(\mathrm{kg} / \mathrm{m}^{3}\) is poured on the water until the top of the oil layer is 0.035 \(\mathrm{m}\) below the top of the block. (a) How deep is the oil layer? (b) What is the gauge pressure at the block's lower face?

A single ice cube with mass 9.70 g floats in a glass completely full of 420 \(\mathrm{cm}^{3}\) of water. You can ignore the water's surface tension and its variation in density with temperature (as long as it remains a liquid). (a) What volume of water does the ice cube displace? (b) When the ice cube has completely melted, has any water overflowed? If so, how much? If not, explain why this is so. (c) Suppose the water in the glass had been very salty water of density 1050 \(\mathrm{kg} / \mathrm{m}^{3} .\) What volume of salt water would the \(9.70-\mathrm{g}\) ice cube displace? (d) Redo part (b) for the freshwater ice cube in the salty water.
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