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Homework 6

# Homework 6 - homework 06 BAUTISTA ALDO Due Feb 3 2006 4:00...

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homework 06 – BAUTISTA, ALDO – Due: Feb 3 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 0 points. An athlete swings a 6 . 72 kg ball horizon- tally on the end of a rope. The ball moves in a circle of radius 0 . 899 m at an angular speed of 0 . 558 rev / s. What is the tangential speed of the ball? Correct answer: 3 . 15191 m / s (tolerance ± 1 %). Explanation: ω = 0 . 558 rev / s = 3 . 50602 rad / s . v t = r ω = (0 . 899 m)(3 . 50602 rad / s) = 3 . 15191 m / s . Question 2 Part 2 of 3. 0 points. What is its centripetal acceleration? Correct answer: 11 . 0507 m / s 2 (tolerance ± 1 %). Explanation: a r = r ω 2 = (0 . 899 m)(3 . 50602 rad / s) 2 = 11 . 0507 m / s 2 . Question 3 Part 3 of 3. 0 points. If the maximum tension the rope can with- stand before breaking is 156 N, what is the maximum tangential speed the ball can have? Correct answer: 4 . 56833 m / s (tolerance ± 1 %). Explanation: F = m v 2 r . If the maximum value of T is 156 N, then we have 156 N = (6 . 72 kg) v 2 0 . 899 m . From which, v max = 4 . 56833 m / s . Question 4 Part 1 of 3. 10 points. The figure below represents, at a given in- stant, the total acceleration of a particle mov- ing clockwise in a circle of radius 3 . 3 m. The magnitude of the total acceleration of the par- ticle at the given instant is 17 . 7 m / s 2 ; the an- gle between the vector of total acceleration and the radius-vector of the particle is 30 . 3 . 3 m v 17 . 7 m / s 2 30 At this instant of time, find the centripetal acceleration of the particle.

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Homework 6 - homework 06 BAUTISTA ALDO Due Feb 3 2006 4:00...

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