Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across international borders by disguising it as lumps of iron ore. He commands his engineer minions to form the gold into little spheres with a diameter of exactly and paint them black. However, his chief engineer points out that customs officials will surely notice the unusual weight of the "iron ore" if the balls are made of solid gold (density=19.3g/cm^3 ). She suggests forming the gold into hollow balls instead (see sketch at right), so that the fake "iron ore" has the same density as real iron ore (5.15g/cm^3). One of the balls of fake "iron ore," sliced in half. Calculate the required thickness of the walls of each hollow lump of "iron ore." Be sure your answer has a unit symbol, if necessary, and round it to significant digits.

Respuesta :

Answer:

0.295cm thick wall

Explanation:

In the complete question, the diameter of the gold sphere is said to be 6cm.

Step 1: Determine the apparent volume of 6cm diameter sphere (including wall and hollow inside)

[tex]V=\frac{4}{3}pir^3[/tex]

[tex]V=\frac{4}{3}pi(3)^3[/tex]

[tex]V=113.04 cm^3[/tex]

Step 2: Determine mass of gold required for sphere to have density of iron ore if the apparent volume is as calculated in Step 1

[tex]mass=density*volume[/tex]

[tex]mass=5.15*113.04[/tex]

[tex]mass=582.16g[/tex]

Step 3: Determine volume of gold based on mass calculated in Step 2

[tex]volume=\frac{mass}{density}[/tex]

[tex]volume=\frac{582.16}{19.3}[/tex]

[tex]volume=30.16cm^3[/tex]

Step 4: Determine thickness of wall of a hollow sphere for volume calculated in Step 3

[tex]V=\frac{4}{3}pi(r_{outer}^3-r_{inner}^3)[/tex]

[tex]30.16=\frac{4}{3}pi(3^3-(3-t)^3)[/tex]

[tex](3-t)^3=19.796[/tex]

[tex](3-t)=2.705[/tex]

[tex]t=0.295cm[/tex]