On July 16, 1882, a massive thunderstorm of Dubuque, Iowa, produced huge hailstones. The diameter of some of the hailstones was 17 inches. Ice weighs about 0.033 pounds/inches cubed. What was the approximate weight of these hailstones to the nearest pound?
Find the volume of the hailstone.
Sphere
Solve for volume
V=4
3πr3
Multiply the volume by 0.033
V=pi/6 D^3
=pi/6 (17)^3
=2572.44in^3 multiply by 0.033
=84.89lb
r = 17/2 = 8.5
V = (4/3)π r^3
= (4/3)π(8.5)^3 = appr 2575.44 inches^3
mass = 2575.44(.033) = 84.89 pounds
seems very unreasonable.
A hailstone with a diameter of 17 inches ? ABsurd
To calculate the approximate weight of the hailstones, we need to find the volume of each hailstone and then multiply it by the weight of ice per cubic inch.
First, we need to find the volume of a hailstone. The hailstone is assumed to be spherical since its diameter is given. The formula to calculate the volume of a sphere is:
V = (4/3)πr³
where V is the volume and r is the radius of the sphere. Since we are given the diameter, which is twice the radius, we can use a radius value of (17 inches / 2) = 8.5 inches.
Now we can calculate the volume of the hailstone:
V = (4/3)π(8.5)^3
Using the value of π ≈ 3.14159, we can substitute the radius into the formula and solve for V.
V ≈ (4/3)π(8.5)^3
V ≈ (4/3) * 3.14159 * (8.5)^3
V ≈ 4.18879 * 614.125
V ≈ 2563.33 cubic inches
Now that we have the volume of each hailstone, we can multiply it by the weight of ice per cubic inch, which is given as 0.033 pounds/inch³:
Weight of hailstone ≈ volume * weight of ice
Weight of hailstone ≈ 2563.33 * 0.033
Calculating this multiplication:
Weight of hailstone ≈ 84.47 pounds
Therefore, to the nearest pound, the approximate weight of these hailstones would be 84 pounds.