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 = Cr^-2.where C is a constant, and r is the distance that the object is from the center of the earth.
A. Suppose that an object is 100 pounds when it is at sea level. Find the value of C that makes the equation true. (Sea level is 3,963 miles from the center of the earth.)
B. Use the value of C you found in the previous question to determine how much the object would weigh in
i.Death Valley (282 feet below sea level)
ii.The top of Mt McKinley (20,430 feet above sea level)
2007-09-04 07:45:30 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You don't have to convert the miles into feet. You can convert the feet into miles 1 mile = 5280 feet.
So C= 100 lbs * (3963 miles )^2 = 1570536900 lbs*miles^2
Death Valley distance from the center of the Earth=
3963 miles - 282feet/(5280 feet/mile)) or (3962.94659 miles)
W(Death Valley) =
(1570536900 lbs*miles^2)/(3962.94659)^2 =100.00269 lbs
McKinley distance from the center of Earth =
3963 miles + 20430 feet /(5280 feet/mile)=3966.86931818..
W(McKinley) =
(1570536900 lbs*miles^2)/(3966.86931818)^2 =99.8050Lbs
2007-09-04 08:27:37 · answer #1 · answered by alints_2000 4 · 0⤊ 0⤋
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2016-12-12 17:59:10 · answer #2 · answered by ciprian 4 · 0⤊ 0⤋
Hi. It would make more sense if you convert miles to feet. Then everything just plugs in.
2007-09-04 07:53:52 · answer #3 · answered by Cirric 7 · 0⤊ 0⤋
100 = C/3963^2, and C = 1.57 billion (rounded)
A. w = 1.57bn/3962.946^2 = 100.003 lbs
B. w = 1.57bn/3966.869^2 = 99.805. lbs
2007-09-04 07:56:56 · answer #4 · answered by John V 6 · 0⤊ 0⤋