SOLUTION: One ton of lead is used to make pipe, 2 in. outside diameter, 1/4 in. thick. Determine pipe length. 5% is allowed for waste. Lead weighs 710 lbs. per cu. ft. One ton

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Question 1161568: One ton of lead is used to make pipe, 2 in. outside diameter, 1/4 in. thick.
Determine pipe length.
5% is allowed for waste.
Lead weighs 710 lbs. per cu. ft.
One ton = 2000 lbs.
1900 lbs. available.
Pipe is a hollow cylinder.
Unsure how to solve.
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
If this is cut open, the pipe becomes a rectangle x in long and 2pi*2 in wide (the circumference of a circle with radius 2 inches, and 1/4 " thick.
This is x*(1/4)*4 pi or x*pi cubic inches. One ft of pipe is 12 π in^3

Lead is 710 lb per cubic foot and 1900 lbs is available or 2.676 ft^3 of lead. dividing 1900 lbs/710 lb/ft^3
the pipe uses pi*x cubic inches of lead, or pi*x/1728 ft^3 of lead, since (12 inches)^3=1728 in^3=1 ft^3

12 π in^3* 1 ft^3/1728 in^3=0.02817 ft^3 of lead per foot length. Note, we have ft^3 of lead and ft as a unit of pipe length
There are 2.676 ft^3 of lead available. divide that by 0.02817 ft^3 lead/foot of pipe and units will be feet of pipe.
Rounding at the end, it is 108.77 feet of pipe or 109 feet.


Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.

            The tutor  @Boreal  made almost uncountable number of errors on the way,  starting from his first line.

            He was so hurry that even did not check his calculations,  that have nothing common with reality.

            So, I came to bring the correct solution. 



Outer radius of the pipe is R_out = 1 inch.

Interior radius of the pipe is 1/4 inch less: R_in = 3/4 in.


The transverse section area of the pipe is  pi%2A%28R_out%5E2+-+R_in%5E2%29 = pi%2A%281%5E2+-+%283%2F4%29%5E2%29 = pi%2A%287%2F16%29 square inches.


The volume of the one foot of the pipe is  pi%2A12%2A%287%2F16%29%7D%7D+in%5E3+=+%7B%7B%7Bpi%2A%2821%2F4%29 in^3 = 3.14%2A%2821%2F4%29 in^3 = 16.485 in^3.


The weight of 1 foot of the pipe is  %28710%2F%2812%5E3%29%29%2A16.485 = 6.773 lbs.


The length of the pipe = 1900%2F6.773 = 280.52 ft.


Take  5%  off for waste :   the final length is  0.95*280.52 = 266.5 ft.    ANSWER

Solved.


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For many years working with engineers and scientists, I learned that there are 1000 ways to make mistakes
and to get wrong answers at the end in the result, and there is only ONE WAY to do the job correctly.