hw16 - Overton, Mays Homework 16 Due: Mar 2 2005, 4:00 am...

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Overton, Mays – Homework 16 – Due: Mar 2 2005, 4:00 am – Inst: Turner 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points The initial kinetic energy imparted to a 0 . 3 kg bullet is 1955 J. The acceleration oF gravity is 9 . 81 m / s 2 . Neglecting air resistance, fnd the range oF this projectile when it is fred at an angle such that the range equals the maximum height attained. Correct answer: 623 . 727 m. Explanation: Let : m = 0 . 3 kg , K = 1955 J , and g = 9 . 81 m / s 2 . Assume that the fring height is negligible and that the bullet lands at the same elevation From which it was fred. The range R is given by R = v 2 0 g sin2 θ ±rom kinematics, x = ( v 0 cos θ ) t y = ( v 0 sin θ ) t - 1 2 g t 2 R = h max , so v 2 0 g sin2 θ = ( v 0 sin θ ) 2 2 g v 2 0 g (2 sin θ cos θ ) = ( v 0 sin θ ) 2 2 g tan θ = 4 θ = 76 . The kinetic energy is K = 1 2 mv 2 0 v 2 0 = 2 K m , so R = v 2 0 g sin2 θ = 2 K mg sin2 θ = 2(1955 J) (0 . 3 kg)(9 . 81 m / s 2 ) sin152 = 623 . 727 m . 002 (part 1 oF 1) 10 points Two atoms oF equal mass interact with a po- tential energy U ( r ) = A r 8 - B r 4 A > 0 , B > 0 At what separation is the Force zero? 1. 4 r A B 2. 4 r 2 A B correct 3. 4 r B A 4. 4 r B 2 A Explanation: U ( r ) = A r 8 - B r 4 F = - dU dr = - d dr ( Ar - 8 ) + d dr ( Br - 4 ) = +8 Ar - 9 - 4 B r - 5 F = 0 thereFore 8 A r 9 = 4 B r 5 2 A B = r 4 r = 4 r 2 A B 003 (part 1 oF 1) 10 points AsoFtballpitcherrotatesa0 . 29kg ballaround a vertical circular path oF radius 0 . 599 m be- Fore releasing it. The pitcher exerts a 34 . 3 N
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Overton, Mays – Homework 16 – Due: Mar 2 2005, 4:00 am – Inst: Turner 2 force directed parallel to the motion of the ball around the complete circular path. The speed of the ball at the top of the circle is 12 . 7 m / s. The acceleration of gravity is 9 . 8 m / s 2 . If the ball is released at the bottom of the circle, what is its speed upon release? Correct answer: 25 . 0982 m / s. Explanation: Let U = 0 at the bottom of the circle. Given :
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hw16 - Overton, Mays Homework 16 Due: Mar 2 2005, 4:00 am...

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