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

College Physics

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Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Physics, 1le
Itais
Help | System Announcements
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Chapter 07, Problem 084
Z 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 + 3.6j + 3.2
3.28)
m to a final position
T2 = (5.3i + 4.0j + 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 T1 and T2.
(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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