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Edit View History Bookmarks People Window Help Mail Tools Mail-jbrya45@isu.edu Course: 2017 Fall CM 2 11 3 , WileyPLUS myLSU × × re https://edugen.wileyplus.com/edugen/student/mainfr.uni ssignment PRINTER VERSION BACK CES Chapter 04, Problem 015 When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down.Suppose the weight of the sky diver is 851 N and the drag force has a magnitude of 1044 N. The mass of the sky diver is 86.8 kg. Take upward to be the positive direction. What is his acceleration, including sign? Number Units the tolerance is +/-2% 14 LINK TO TEXT 20 Question Attempts: Unlimited SAVE FOR LATER SUBMIT ANSWER 29 41 52 Version 4.24.1.20 2ohn Wiley & Sons, om edygen/shared assignment /test/aglist.uniid ttrem/hered (assignment /test/ aglist.unind- Inc, All Rights Reserved

Explanation / Answer

As given in problem,

Drag force, F1 = 1044 N in upward direction

Weight, W = 851 N in downward direction

So, net force will be upward direction, of (1044 - 851) N = 193 N

So his acceleration would be, a = F/m = 193 N / 86.8 kg = 2.22 m/s2 (upward direction)

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