SOLUTION: Ref. Question 1059604 Follow-up:
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 c
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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 c
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Question 1059621: Ref. Question 1059604 Follow-up:
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 ? Thanks.
You can put this solution on YOUR website! 0.0049 * 1 = 0.0049 cu. in. (volume of 1 ft.) <--- Mistake in this line
1 foot equals 12 inches, you mixed your units up.
Should be,
0.0049*12
