Asked by angie
if a tape is wound around a cylinder at a rate of 1 rad/s for a half hour, if each rotation deposits one layer of tape and if the final difference between inner ad outer radii of the winding in one cm, what is the thickness of the tape?
Answers
Answered by
bobpursley
how many layers?
n= rotationsi30min/2PI= 1800sec*1rad/sec/2pi
= 1800/2PI
thicknesstape=1cm/1800/2PI= 2PIcm/1800
n= rotationsi30min/2PI= 1800sec*1rad/sec/2pi
= 1800/2PI
thicknesstape=1cm/1800/2PI= 2PIcm/1800
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