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Chapter 12: Problem 6

A solid cube of aluminum is \(2.00 \mathrm{~cm}\) on each edge. The density of aluminum is \(2700 \mathrm{~kg} / \mathrm{m}^{3}\). Find the mass of the cube. $$ \text { Mass of cube }=\rho V=\left(2700 \mathrm{~kg} / \mathrm{m}^{3}\right)(0.0200 \mathrm{~m})^{3}=0.0216 \mathrm{~kg}=21.6 \mathrm{~g} $$

Short Answer

Expert verified
The mass of the cube is 21.6 grams.

Step by step solution

01

Convert Edge Length to Meters

The edge length of the cube is given as 2.00 cm. Convert this to meters by dividing by 100, since there are 100 cm in a meter. Thus, you get 0.0200 meters.
02

Find Volume of the Cube

The volume of a cube is calculated using the formula \( V = a^3 \), where \( a \) is the edge length. Substitute 0.0200 meters for \( a \). \[ V = (0.0200 ext{ m})^3 = 8.00 imes 10^{-6} ext{ m}^3 \]
03

Use Density Formula to Find Mass

Use the formula relating mass, density, and volume \( m = \rho V \). Given the density \( \rho = 2700 \text{ kg/m}^3 \), multiply it by the volume found in step 2.\[ m = (2700 \text{ kg/m}^3)(8.00 imes 10^{-6} \text{ m}^3) \]
04

Calculate the Mass

Perform the multiplication, \[ m = 21.6 imes 10^{-3} \text{ kg} = 0.0216 \text{ kg} \].Convert kg to grams by multiplying by 1000, resulting in 21.6 grams.

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Key Concepts

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

Volume Calculation
To find the volume of a solid object, you can use formulas specific to its shape. For a cube, the volume is calculated as the cube of its edge length, represented mathematically as \( V = a^3 \), where \( a \) is the edge length.
In this exercise, the cube's edge length is 2.00 cm, which needs to first be converted to meters before substituting into the formula, since utilizing consistent units is crucial for accuracy.When the edge length is 0.0200 meters, the volume of the cube is:
  • Calculate \( V = (0.0200 \, \mathrm{m})^3 \)
  • The result is \( V = 8.00 \times 10^{-6} \, \mathrm{m}^3 \)
A proper understanding of volume calculation allows accurate determination of other physical properties, such as mass, when combined with additional parameters.
Unit Conversion
Unit conversion is an essential skill in solving physics problems, as it ensures all measurements are consistent. Here, converting centimeters to meters is necessary because the density of aluminum is in \( \, \mathrm{kg/m}^3 \), requiring length measurements in meters for compatibility.
To convert 2.00 cm to meters:
  • Recall the conversion factor: 1 meter = 100 centimeters
  • Divide the length in centimeters by 100
  • \( 2.00 \mathrm{\,cm} = \frac{2.00}{100} \mathrm{\,m} = 0.0200 \mathrm{\,m} \)
Performing these conversions accurately allows for precise calculations of volume and, consequently, mass.
Mass Calculation
In physics, the mass of an object can be determined using its density and volume, with the relationship given by the formula \( m = \rho V \). This equation links these important quantities, where:
  • \( m \) represents mass
  • \( \rho \) symbolizes density
  • \( V \) denotes volume
To find the mass of the aluminum cube:
  • Use the calculated volume \( V = 8.00 \times 10^{-6} \, \mathrm{m}^3 \)
  • Multiply by the density \( \rho = 2700 \mathrm{\,kg/m}^3 \)
  • \( m = 2700 \, \times 8.00 \times 10^{-6} \mathrm{\,kg} = 0.0216 \mathrm{\,kg} \)
  • Convert mass to grams by multiplying by 1000: \( 0.0216 \mathrm{\,kg} \times 1000 = 21.6 \mathrm{\,g} \)
Calculating mass accurately requires understanding these interrelated concepts, and converting units correctly impacts the final result. With practice, you can confidently apply these principles to find the mass of various objects.

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Most popular questions from this chapter

A vertical wire \(5.0 \mathrm{~m}\) long and of \(0.0088 \mathrm{~cm}^{2}\) cross- sectional area has a modulus \(Y=200 \mathrm{GPa}\). A \(2.0-\mathrm{kg}\) object is fastened to its end and stretches the wire elastically. If the object is now pulled down a little and released, the object undergoes vertical SHM. Find the period of its vibration. The force constant of the wire acting as a vertical spring is given by \(k=F / \Delta L\), where \(\Delta L\) is the deformation produced by the force (weight) \(F\). But, from \(F / A=Y\left(\Delta L / L_{0}\right)\), $$ k=\frac{F}{\Delta L}=\frac{A Y}{L_{0}}=\frac{\left(8.8 \times 10^{-7} \mathrm{~m}^{2}\right)\left(2.00 \times 10^{11} \mathrm{~Pa}\right)}{5.0 \mathrm{~m}}=35 \mathrm{kN} / \mathrm{m} $$ Then for the period we have $$ T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{2.0 \mathrm{~kg}}{35 \times 10^{3} \mathrm{~N} / \mathrm{m}}}=0.047 \mathrm{~s} $$

A \(60-\mathrm{kg}\) motor sits on four cylindrical rubber blocks. Each cylinder has a height of \(3.0 \mathrm{~cm}\) and a crosssectional area of \(15 \mathrm{~cm}^{2}\). The shear modulus for this rubber is \(2.0 \mathrm{MPa} .(a)\) If a sideways force of \(300 \mathrm{~N}\) is applied to the motor, how far will it move sideways? (b) With what frequency will the motor vibrate back and forth sideways if disturbed?

A solid cylindrical steel column is \(4.0 \mathrm{~m}\) long and \(9.0 \mathrm{~cm}\) in diameter. What will be its decrease in length when carrying a load of \(80000 \mathrm{~kg}\) ? \(Y=1.9 \times 10^{11} \mathrm{~Pa}\). First find the Cross-sectional area of column \(=\pi r^{2}=\pi(0.045 \mathrm{~m})^{2}=6.36 \times 10^{-3} \mathrm{~m}^{2}\) Then, from \(Y=(F / A) /\left(\Delta L / L_{0}\right)\), $$ \Delta L=\frac{F L_{0}}{A Y}=\frac{\left[\left(8.00 \times 10^{4}\right)(9.81) \mathrm{N}\right](4.0 \mathrm{~m})}{\left(6.36 \times 10^{-3} \mathrm{~m}^{2}\right)\left(1.9 \times 10^{11} \mathrm{~Pa}\right)}=2.6 \times 10^{-3} \mathrm{~m}=2.6 \mathrm{~mm} $$

A thin sheet of gold foil has an area of \(3.12 \mathrm{~cm}^{2}\) and a mass of \(6.50 \mathrm{mg}\). How thick is the sheet? The density of gold is \(19300 \mathrm{~kg} / \mathrm{m}^{3}\). One milligram is \(10^{-6} \mathrm{~kg}\), so the mass of the sheet is \(6.50 \times 10^{-6} \mathrm{~kg}\). Its volume is $$ V=(\text { area }) \times(\text { thickness })=\left(3.12 \times 10^{-4} \mathrm{~m}^{2}(d)\right. $$ where \(d\) is the thickness of the sheet. We equate this expression for the volume to \(m / \rho\) to get $$ \left(3.12 \times 10^{-4} \mathrm{~m}^{2}\right)(d)=\frac{6.50 \times 10^{-6} \mathrm{~kg}}{19300 \mathrm{~kg} / \mathrm{m}^{3}} $$ from which \(d=1.08 \times 10^{-6} \mathrm{~m}=1.08 \mu \mathrm{m}\).

A box-shaped piece of gelatin dessert has a top area of \(15 \mathrm{~cm}^{2}\) and a height of \(3.0 \mathrm{~cm}\). When a shearing force of \(0.50 \mathrm{~N}\) is applied to the upper surface, the upper surface displaces \(4.0 \mathrm{~mm}\) relative to the bottom surface. What are the shearing stress, the shearing strain, and the shear modulus for the gelatin? $$ \begin{array}{l} \sigma_{s}=\frac{\text { Tangential force }}{\text { Area of face }}=\frac{0.50 \mathrm{~N}}{15 \times 10^{-4} \mathrm{~m}^{2}}=0.33 \mathrm{kPa} \\ \varepsilon_{s}-\frac{\text { Displacement }}{\text { Height }}-\frac{0.40 \mathrm{~cm}}{3.0 \mathrm{~cm}}-0.13 \\ S=\frac{0.33 \mathrm{kPa}}{0.13}=2.5 \mathrm{kPa} \end{array} $$
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