Question

5:22 AA session.masteringphysics.com Problem 8.22 Mastering Physics 6 of 8 Constants The tires of a car make 93 revolutions as the car reduces its speed uniformly from 95.0 km/h to 63.0 km/h. The tires have a diameter of 0.88 m. Part A What was the angular acceleration of the tires? Express your answer using two significant figures. a = 1.7 rad/s2 Submit Request Answer Incorrect; Try Again; 9 attempts remaining Part B If the car continues to decelerate at this rate, how much more time is required for it to stop? Express your answer to two significant figures and include the appropriate units. t = 35 Units Submit Request Answer

Answer 1

0.44 0.44 sel: 6948 ) Giron: N=93 gousletion, = gs.okm/h = 26.28 mis wir uns = 26.3.96 : Crevet no 88m? W = 59.97 gadis N2 = 63.0km/n = 17.5 mis Wz= 175 = 39.77 seadis W = Wynlig - 20:2 cuadls w'= 2AN = Ne oo - 2012 N'=3.21 berolution I see time for N=93 reustutions =) = 93 It = 28.91 sec) @ angulo aueleration & angular oudleratul WoW, tot (28.91) 39.77- 59.97 + a = -0.698 had/32 | .ne shows retardati az 0.698 od 182 6 W₂=0 & Now, w, = 39,77 -wtat =0 = 59.77 + (-0.69874 t = 39.77 0-698 t= 56.97 see - = 251.22.cd. A Ww=27N 3 W= 27*2400 sol: 3 of & N= 2400 rpon 27x2400 = 251-2rolls 60 A was rew=251.2 gadis B VWU z 25162x 9.40 = 100.48 mis U2100.48m/s