dsaefong00
24.12.2019 •
Physics
Acord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. a steady horizontal pull of 40.0 n to the right is exerted on the cord, pulling it off tangentially from the wheel. the wheel is mounted on frictionless bearings on a horizontal axle through its center. (a) compute the angular acceleration of the wh
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Ответ:
The angular acceleration of the wheel is 4.33m/s^2
Explanation:
F=40N, r=0.25m, m=9.20kg
V = √(Fr ÷ m) = √(40×0.25 ÷ 9.2) = √1.09 = 1.04m/s
Angular acceleration (a) = V^2/r = 1.04^2/0.25 = 1.0816/0.25 = 4.33m/s^2
Ответ:
12
Explanation: