Asked by Joy
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 ???
The book has the answer of 85 lbs .
I need help on trying to solve it , please help ???
Answers
Answered by
Reiny
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
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
Answered by
drwls
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 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
Answered by
michele
just cry
Answered by
ken
the "Circumference" of the hailstone was 17" not the Diameter. No wonder your answer was catastrophic (85lb.)! This question was unanswered so long -- it's melted by now.
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