Question

On July 16, 1882, a massive thunderstorm over Dubuque, Iowa, produced huge hailstones. The diameter of some of the hailstones was 17 in. Ice weighs about 0.033 lb/in3. What is the approximate weight of a 17 in. hailstone?

Answer 1

It's simple, first we have to find the volume of a sphere (the hailstones) with a diameter 17 in. The formula is:\[V=\frac{4\pi*r^3}{3}=\frac{\pi*d^3}{6}\]I'll use the second one:\[V=\frac{\pi*d^3}{6}\]\[V=\frac{\pi*17^3}{6}\]\[\boxed{V=1637.67\pi~in^3}\]Now we have to use this formula:\[\rho=\frac{m}{V}\]We have:\[\rho=0.033~lb/in^3\\V=1637.67\pi~in^3\]Then\[m=\rho*V\]\[m=0.033*1637.67\]\[\boxed{m\approx169.78~lb}\]