The spot speed on the edge of the millstone with 0.075 m diameter, continuously reach from 12 m/s to 25 m/s within 6/2 s. how much is the angle acceleration (corner hurry)of the millstone within this time?
2006-06-24 09:22:42 · 5 answers · asked by h_hooryar82 1 in Science & Mathematics ➔ Physics
4?
2006-06-24 09:25:48 · answer #1 · answered by hellion210 6 · 2⤊ 0⤋
Angular acceleration is A=(v^2)/r
(144m^2/s^2)/0.075m = 1920 m/s^2
(625m^2/s^2)/0.075m = 8333.333 m/s^2
8333.33 - 1920 =6413.33 m/s^2
(6413.33 m/s^2)/(6s/2)=2137.78 m/s^3
2006-06-24 17:22:02 · answer #2 · answered by SteveA8 6 · 0⤊ 0⤋
I can
2006-06-24 17:23:13 · answer #3 · answered by CoolBuddy 2 · 0⤊ 0⤋
8741.66 rad/sec
2006-06-24 16:44:14 · answer #4 · answered by Dr. House 2 · 0⤊ 0⤋
Not me. Uh?
2006-06-24 16:32:20 · answer #5 · answered by wefields@swbell.net 3 · 0⤊ 0⤋