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The long, slender rod (1 = 1/12ml^2) has a mass of 8 kg and a length of 10 m. The disc attached (I=1/2mr^2) has a mass of 4 k

Angular impulse is described as the sum of torques multiplied by the time of application. How long were the torques applied i

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kk - Gm r=2m 7200W Im k em z 3m Wang = 8x9.8 W₂ = mg - 784N Fux9.8 lom - tom torque ?=-(0)3-(Wa) 10+ F (6) 39.2N - om - netIslendey = ML? +M ( 3)2 - 5x8x610) + 863)2 - Islendy - 138.67 kg.mz Itotal = Islender + Idisia = 138.67(leg m2) + 408 (69.774

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