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Screen shot 2011-06-21 at 11.50.45 PM

Screen shot 2011-06-21 at 11.50.45 PM - [Ifll[part 1 of I...

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Unformatted text preview: [Ifll [part 1 of I} 10.1] points A. dentisth drill starts from rest. After 4.23 s of constant angular acceleration it turns at a rate of 14541] rcvfrnin. Find the drill’s angular acceleration. Correct answer: 35535-4 radfsfi. Explanation: Let: t=-'-1.Efls and w; = 14541] revfrnin. Since all] = I], _w.r_wfl'_wf '1— t _ t _1¢15-"='5revflnin 2—11 1min _ 4.235 ‘m' 555 = 355.:54 radfsz l. DD: [part 2 of I} lflufl points Throughout what angle does the drill rotate during this period? Correct answer: 3255.42 rad. Explanation: 1 H=wi+§fltfi 1 = n + E [355.1'54 was?) {4.23 s]: =|3255.£12rad|. ...
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