Math Problem Statement

One cubic meter (1.00 m3) of aluminum has a mass of 2.70E+3 kg, and 1.00 m3 of copper has a mass of 8.92E+3 kg. Find the radius of an aluminum sphere whose mass is the same as that of an copper sphere of radius 1.85 cm.

Solution

To solve the problem, we need to find the radius of an aluminum sphere with the same mass as that of a copper sphere with a known radius of 1.85 cm.

Step-by-Step Solution:

  1. Find the mass of the copper sphere:

    • The volume VV of a sphere is given by: V=43πr3V = \frac{4}{3} \pi r^3
    • The density of copper is given as ρCu=8.92×103kg/m3\rho_{\text{Cu}} = 8.92 \times 10^3 \, \text{kg/m}^3.
    • The radius rCur_{\text{Cu}} of the copper sphere is 1.85 cm = 0.0185 m.
    • The mass of the copper sphere mCum_{\text{Cu}} is: mCu=ρCu×VCu=ρCu×(43πrCu3)m_{\text{Cu}} = \rho_{\text{Cu}} \times V_{\text{Cu}} = \rho_{\text{Cu}} \times \left( \frac{4}{3} \pi r_{\text{Cu}}^3 \right) Substituting the values, we get: mCu=8.92×103×43π(0.0185)3m_{\text{Cu}} = 8.92 \times 10^3 \times \frac{4}{3} \pi (0.0185)^3
  2. Set the mass of the aluminum sphere equal to the mass of the copper sphere:

    • Let rAlr_{\text{Al}} be the radius of the aluminum sphere.
    • The density of aluminum is ρAl=2.70×103kg/m3\rho_{\text{Al}} = 2.70 \times 10^3 \, \text{kg/m}^3.
    • The mass of the aluminum sphere mAlm_{\text{Al}} is: mAl=ρAl×(43πrAl3)m_{\text{Al}} = \rho_{\text{Al}} \times \left( \frac{4}{3} \pi r_{\text{Al}}^3 \right)
    • Since mAl=mCum_{\text{Al}} = m_{\text{Cu}}, we equate the masses: ρAl×(43πrAl3)=ρCu×(43πrCu3)\rho_{\text{Al}} \times \left( \frac{4}{3} \pi r_{\text{Al}}^3 \right) = \rho_{\text{Cu}} \times \left( \frac{4}{3} \pi r_{\text{Cu}}^3 \right)
  3. Solve for rAlr_{\text{Al}}:

    • Simplifying the equation: ρAlrAl3=ρCurCu3\rho_{\text{Al}} r_{\text{Al}}^3 = \rho_{\text{Cu}} r_{\text{Cu}}^3 rAl3=ρCuρAlrCu3r_{\text{Al}}^3 = \frac{\rho_{\text{Cu}}}{\rho_{\text{Al}}} r_{\text{Cu}}^3 rAl=(ρCuρAl)1/3rCur_{\text{Al}} = \left( \frac{\rho_{\text{Cu}}}{\rho_{\text{Al}}} \right)^{1/3} r_{\text{Cu}}
    • Substituting the values: rAl=(8.92×1032.70×103)1/3×0.0185r_{\text{Al}} = \left( \frac{8.92 \times 10^3}{2.70 \times 10^3} \right)^{1/3} \times 0.0185

Let's calculate the final value.The radius of the aluminum sphere is approximately 0.0276m0.0276 \, \text{m} or 2.76 cm.

Would you like further details or have any questions?

Here are five related questions you might find useful:

  1. How does the mass of a sphere change with its radius?
  2. What is the formula for the volume of a sphere?
  3. How do different materials' densities affect the size of objects with the same mass?
  4. What is the relationship between mass, volume, and density?
  5. How do you convert between cubic meters and cubic centimeters?

Tip: When comparing objects of the same mass but different materials, remember that the one with lower density will have a larger volume.

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Math Problem Analysis

Mathematical Concepts

Geometry
Density
Volume of Spheres

Formulas

Volume of a sphere: V = (4/3)πr^3
Density: ρ = mass/volume
Relation between the radii of two spheres with the same mass: r_Al = (ρ_Cu/ρ_Al)^(1/3) * r_Cu

Theorems

Principle of density and mass equivalence in spheres

Suitable Grade Level

Grades 10-12