The acceleration due to gravity is given as g = 32.125 fl/s² and that its variati /s³ per 1000 feet ascent. Compute for the height in feet and in miles above vhich (a) the local gravity becomes g = 30.012 fl/s³, (b) the weight of a g decreased by 6.85% and, (c) the weight of a 225 Jom man at top of the Mt. E Note: The height of the mountain is 5,642 meters.

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter5: More Applications Of Newton’s Laws
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The acceleration due to gravity is given as g = 32.125 fl/s? and that its variation is 0.0047
fl/s? per 1000 feet ascent. Compute for the height in feet and in miles above this point for
which (a) the local gravity becomes g = 30.012 fl/s, (b) the weight of a given man is
decreased by 6.85% and, (c) the weight of a 225 Ibm man at top of the Mt. Elbrus.
Note: The height of the mountain is 5,642 meters.
Transcribed Image Text:The acceleration due to gravity is given as g = 32.125 fl/s? and that its variation is 0.0047 fl/s? per 1000 feet ascent. Compute for the height in feet and in miles above this point for which (a) the local gravity becomes g = 30.012 fl/s, (b) the weight of a given man is decreased by 6.85% and, (c) the weight of a 225 Ibm man at top of the Mt. Elbrus. Note: The height of the mountain is 5,642 meters.
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