Answer:
thickness is 0.29 cm
Explanation:
To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.
Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).
We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.
The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.
This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.
Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.
So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.
Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].
Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].
30.1635 * 3/(4 * 3.14) = 27 - r^3.
Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.
Therefore, r = 2.7051 cm.
This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.
Rounding to two significant figures yields
the thickness = 0.29 cm.
