SOLUTION: Ref Question 1059604;
A coil of steel wire weighs 50 lbs., diameter 0.025 in. Steel weighs .283 lbs. / cu. in. Determine length of wire.
The wire is a regular cylinder.
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A coil of steel wire weighs 50 lbs., diameter 0.025 in. Steel weighs .283 lbs. / cu. in. Determine length of wire.
The wire is a regular cylinder.
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Question 1059671: Ref Question 1059604;
A coil of steel wire weighs 50 lbs., diameter 0.025 in. Steel weighs .283 lbs. / cu. in. Determine length of wire.
The wire is a regular cylinder.
Area of circle; pi * r^2.
3.1416 * 0.0125 * 0.0125 = 0.0049 sq. in.
0.0049 * 1 = 0.0049 cu. in. (volume of 1 ft.)
50 / .283 = 176.678 = 177 cu. in. (vol. of wire).
177 / 0.0049 = 36122.48 ft.
Answer given is less. Where am I incorrect ?
You can put this solution on YOUR website! 0.0049 * 1 = 0.0049 cu. in. (volume of 1 ft.) <---- Your mistake is here.
You should multiply by 12 inches not by 1 foot to avoid mixing units.
