A 1938 nickel has a diameter of 21.21 mm, a thickness of 1.95 mm, and weighs 0.04905 N. What is its density?

Question

A 1938 nickel has a diameter of  21.21 mm, a thickness of  1.95 mm, and weighs  0.04905 N. What is its density?

Quizplus · Accepted Answer

To find the density, use the formula: Density = Mass/Volume. First, convert the weight (force) to mass using $ m = \frac{F}{g} $, where $ g = 9.81 \, \text{m/s}^2 $. So, mass $ m = \frac{0.04905}{9.81} \approx 0.005 \, \text{kg} $. The volume of the cylinder (nickel) is $ V = \pi r^2 h $, where $ r = \frac{21.21}{2} \times 10^{-3} \, \text{m} $ and $ h = 1.95 \times 10^{-3} \, \text{m} $. Calculate the volume and then the density, which results in approximately $ 7.26 \times 10^3 \, \text{kg/m}^3 $.

A 1938 Nickel Has a Diameter of 21 3.63×103 kg/m33.63 \times 10 ^ { 3 } \mathrm {~kg} / \mathrm { m } ^ { 3 }3.63×103 kg/m3

A 1938 Nickel Has a Diameter of 21 $3.63 \times 10 ^ { 3 } \mathrm {~kg} / \mathrm { m } ^ { 3 }$