A force F = (2.5 + 8.4j + 6.4k) N acts on a 3.6kg object that moves in 2.3s from an initial position 1 = (3.4i + 3.6j + 3.2k ) m to a final position 72 = (5.31 + 4.0) + 8.2k): m. Find (a) the work done on the object by the force in that time interval, (b) the average power due to the force during that time interval, and (c) the angle between the vectors ri and r 2. (a) Number Units (b) Number Units (c) Number Units (degrees) Click if you would like to Show Work for this question: Open Show Work

A force F = (2.5 + 8.4j + 6.4k) N acts on a 3.6kg object that moves in 2.3s from an initial position 1 = (3.4i + 3.6j + 3.2k ) m to a final position 72 = (5.31 + 4.0) + 8.2k): m. Find (a) the work done on the object by the force in that time interval, (b) the average power due to the force during that time interval, and (c) the angle between the vectors ri and r 2. (a) Number Units (b) Number Units (c) Number Units (degrees) Click if you would like to Show Work for this question: Open Show Work

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Description: Physics, 1leItaisHelp | System AnnouncementsPRINTER VERSION4 ВАСKNEXTChapter 07, Problem 084Z Your answer is partially correct. Try again.A force F = (2.51 + 8.4 + 6.4k) N :acts on a 3.6kg object that moves in 2.3s from an initial position 1 = (3.4i

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter7: Energy Of A System

Section: Chapter Questions

Problem 7.4OQ: A cart is set rolling across a level table, at the same speed on every trial. If it runs into a...

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