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Question 448589: I have a problem here, r^-2 = wC^-1, r=distance, w=weight, C=constant. Apparently we can use 1 mile = 5,280 feet for your conversions. I need to work this out for "r" first then "C" then "w". Can you help me? This is the entire question off the appendix.
Hint for this assignment: Pay attention to the units of measure. You may have to convert from feet to miles in this assignment. You can use 1 mile = 5,280 feet for your conversions.
1. Many people know that the weight of an object varies on different planets, but did you know that the weight of an object on earth also varies according to the elevation of the object? In particular, the weight of an object follows this equation: , where C is a constant, and r is the distance that the object is from the center of the earth.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I have a problem here, r^-2 = wC^-1, r=distance, w=weight, C=constant.
1/r^2 = w/C
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r = sqrt[C/w]
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C = wr^2
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w = C/r^2
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Apparently we can use 1 mile = 5,280 feet for your conversions. I need to work this out for "r" first then "C" then "w". Can you help me? This is the entire question off the appendix.
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Hint for this assignment: Pay attention to the units of measure. You may have to convert from feet to miles in this assignment. You can use 1 mile = 5,280 feet for your conversions.
1. Many people know that the weight of an object varies on different planets, but did you know that the weight of an object on earth also varies according to the elevation of the object? In particular, the weight of an object follows this equation:w = C/r^2 , where C is a constant, and r is the distance that the object is from the center of the earth.
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Cheers,
Stan H.
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