1. The acceleration due to gravity is given as g = 32.125 ft/s? and that its variation is 0.0033 ft/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 = 30.151 ft/s?, (b) the weight of a given man is decreased by 6.35% and, (c) the weight of a 225 lbm man at top of the Mt. Denali. Note: The height of the mountain is 6,190 meters.
1. The acceleration due to gravity is given as g = 32.125 ft/s? and that its variation is 0.0033 ft/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 = 30.151 ft/s?, (b) the weight of a given man is decreased by 6.35% and, (c) the weight of a 225 lbm man at top of the Mt. Denali. Note: The height of the mountain is 6,190 meters.
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![1. The acceleration due to gravity is given as g = 32.125 ft/s? and that its variation is 0.0033 ft/s² per 1000
feet ascent. Compute for the height in feet and in miles above this point for which (a) the local gravity
becomes g = 30.151 ft/s?, (b) the weight of a given man is decreased by 6.35% and, (c) the weight of a
225 lbm man at top of the Mt. Denali. Note: The height of the mountain is 6,190 meters.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F87e49072-06dc-417d-8b40-79dce212e798%2Fccc72fd0-ef18-4ad6-9a30-0f73dea2a048%2F92mw0kg_processed.png&w=3840&q=75)
Transcribed Image Text:1. The acceleration due to gravity is given as g = 32.125 ft/s? and that its variation is 0.0033 ft/s² per 1000
feet ascent. Compute for the height in feet and in miles above this point for which (a) the local gravity
becomes g = 30.151 ft/s?, (b) the weight of a given man is decreased by 6.35% and, (c) the weight of a
225 lbm man at top of the Mt. Denali. Note: The height of the mountain is 6,190 meters.
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