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Unformatted text preview: kaplan (hmk378) Homework 2 Weathers (17104) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A bullet is fired straight up from a gun with a muzzle velocity of 288 m / s. The acceleration of gravity is 9 . 8 m / s 2 . Neglecting air resistance, what will be its displacement after 1 . 1 s? Correct answer: 310 . 871 m. Explanation: Displacement is defined by s = s o + v o t + 1 2 a t 2 . The initial velocity of the bullet is v o = v and its initial displacement is s o = 0 . The gravitational acceleration is negative, so the displacement is s = s o + v o t 1 2 g t 2 = v t 1 2 g t 2 = (288 m / s) (1 . 1 s) 1 2 g (1 . 1 s) 2 = 310 . 871 m . 002 (part 1 of 2) 10.0 points An electron in the cathode ray tube of a tele vision set enters a region where it accelerates uniformly from a speed of 34400 m / s to a speed of 4 . 93 10 6 m / s in a distance of 2 . 9 cm. What is its acceleration? Correct answer: 4 . 1903 10 14 m / s 2 . Explanation: Basic Concepts: Constant acceleration The kinematics equations v f = v o + at v 2 f = v 2 o + 2 as Solution: Taking the direction of motion to be along the x axis, the second equation yields a = v 2 f v 2 o 2 x = ( 4 . 93 10 6 m / s ) 2 (34400 m / s) 2 2(2 . 9 cm) 100 cm 1 m = 4 . 1903 10 14 m / s 2 003 (part 2 of 2) 10.0 points For what length of time is the electron in this region where it accelerates? Correct answer: 1 . 16832 10 8 s. Explanation: The first equation yields t = v f v o a = 4 . 93 10 6 m / s 34400 m / s 4 . 1903 10 14 m / s 2 = 1 . 16832 10 8 s Although the acceleration is very large in this example, it occurs over a very short time...
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This document was uploaded on 10/26/2011 for the course PHYS 1710 at North Texas.
 Fall '08
 Weathers
 mechanics, Work

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