Over July 16, 1882, a massive thunderstorm over Dubuque, Iowa, produced huge hailstones. The diameter of some of the hailstones was 17 inches. Ice weighs about 0.033 lb/in^3. What was the approximate weight of these hailstones to the nearest pound?
The book has the answer of 85 lbs .
I need help on trying to solve it , please help ???
r = 8.5
Vol = (4/3)pi(r^3)
= (4/3)pi(8.5)^3)
= 2572.44 inches^3
mass = 2572.44(.033) = 84.89 or 85 lbs
Weight = density x volume
Just multiply the density (in lb/in^3) by the hailstone volume in cubic inches.
You will need to remember that volume of a sphere of diameter D is
V = (pi/6)D^3
just cry
To solve this problem, we need to calculate the volume of the hailstones and then multiply it by the weight of ice per unit volume.
First, let's calculate the volume of the hailstones. The diameter given is 17 inches, which means the radius (half of the diameter) is 17/2 = 8.5 inches.
The volume of a hailstone can be determined using the formula for the volume of a sphere:
V = (4/3)πr³
Substituting in the radius value, we get:
V = (4/3)π(8.5)^3
Using the approximation of π as 3.14:
V = (4/3)(3.14)(8.5)^3
Calculating this value gives us the volume of each hailstone.
Once we have the volume, we can multiply it by the weight of ice per unit volume to find the approximate weight. According to the problem, ice weighs about 0.033 lb/in³.
Weight = Volume × Weight per unit volume
Weight = Volume × 0.033 lb/in³
Now, let's put it all together and calculate the weight of the hailstones:
V = (4/3)(3.14)(8.5)^3
Weight = V × 0.033 lb/in³
By plugging these values into a calculator, you will get the weight in pounds.
Therefore, the approximate weight of these hailstones to the nearest pound is 85 lbs, as stated in the book.