Question

A metal ball weighing 5.00 tons requires 20.0 ounces of paint to cover it. How many ounces of paint are needed to cover a metal ball weighting 0.5 ton?

Answer 1

Answer: 4,3 ounces of paint

Step-by-step explanation:

m₁ = 5 tons

m₂ = 0,5 tons

m₁ / m₂ = 5 : 0,5 = 10

m₁ = ρV₁ m₂ = ρV₂

m₁ / m₂ = V₁/V₂ = 10

V₁ = 4πR₁³ / 3 and V₂ = 4πR₂³ / 3

V₁ / V₂ = R₁³ / R₂³ = (R₁/R₂)³

V₁ / V₂ = (R₁/R₂)³ = 10 => R₁ / R₂ = ∛10 = 2,1544

A₁ = 4πR₁² A₂ = 4πR₂²

A₁ /A₂ = (R₁ / R₂)² = 2,1544² = 4,6414

A₁ ........ 20 ounces

A₂ .......... x = 20 A₂ / A₁ = 20 x 1 / 4,6414 = 4,3 ounces

Answer 2

Step-by-step explanation:

Surface area of a sphere 4 pi r^2

Volume of a sphere 4/3 pi r^3 (assuming density of 1 )

5.00 = 4/3 pi r^3 shows r = 1.0607

1/10 the volume would be

.5 = 4/3 pi r^3 r= .49237

Larger surface area = 4 pi (1.067)^2 = 14.31

smaller surface area = 3.046

so the paint required would be 3.046 / 14.31 * 20 oz = 4.26 oz