solution_6

# solution_6 - ouseph(jao888 HW06 tsoi(92940 This print-out...

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ouseph (jao888) – HW06 – tsoi – (92940) 2 of constant angular acceleration it turns at a rate of 20240 rev / min. Find the drill’s angular acceleration. Correct answer: 1655 . 88 rad / s 2 . Explanation: Let : t = 1 . 28 s and ω f = 20240 rev / min . Since ω 0 = 0, α = ω f ω 0 t = ω f t = 20240 rev / min 1 . 28 s · 2 π rev · 1 min 60 s = 1655 . 88 rad / s 2 . 007 (part 2 of 2) 10.0 points Throughout what angle does the drill rotate during this period? Correct answer: 1356 . 5 rad. Explanation: θ = ω 0 t + 1 2 α t 2 = 0 + 1 2 (1655 . 88 rad / s 2 ) (1 . 28 s) 2 = 1356 . 5 rad . 008 10.0 points An automobile moves at constant speed down one hill and up another hill along the smoothly curved surface as shown below. car Which of the following diagrams best rep- resents the directions of the velocity and the acceleration of the automobile at the instant that it is at the lowest position as shown? 1. v a correct 2. v a 3. v a 4. v a 5. v a Explanation: At the lowest position, the instantaneous velocity of the automobile is directed horizon- tally to the right. Since the speed of the automobile is un- changed, there is no tangential accelera- tion. Also since the path of the automobile is curved, there is centripetal acceleration, pointing upward. 009 (part 1 of 5) 10.0 points A car travels at a speed of 22 m / s around a curve of radius 42 m.
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solution_6 - ouseph(jao888 HW06 tsoi(92940 This print-out...

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