Question 137176: This is homework.
Pay attention to th units of measure. You may have to convert from feet to miles several times in this assignment. You can use 1 mile= 5,280 feet for your conversions. Many people know that the weight of an object varies on different planets, but did you know that the weight of an 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. Solve the equation w=Cr^-2 for r.
b. 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.)
c. Use the value of C you found in the previous question to determine how much the objec would weigh in
i. Death Valley (282 feet below sea level)
ii. The top o Mt McKinley (20,430 feet above sea level)
Thanks,
MT
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! This is homework.
Pay attention to th units of measure. You may have to convert from feet to miles several times in this assignment. You can use 1 mile= 5,280 feet for your conversions.
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Many people know that the weight of an object varies on different planets, but did you know that the weight of an earth also varies according to the elevation of the object?
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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. Solve the equation w=Cr^-2 for r.
Multiply both sides by r^2 to get:
wr^2 = C
r^2 = C/w
r = sqrt[C/w]
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b. 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.)
3963 = sqrt[C/100]
C/100 = 3963^2
C = 100*3963^2 = 1570536900
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c. 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)
Distance from center = 3963 - (282/5280) = 3962.946591 miles = 20,924,358 ft
w=Cr^-2
w = 1570536900*3962.946591^-2= 100.0026954 lbs.
ii. The top o Mt McKinley (20,430 feet above sea level)
Distance from center = 3963+(20430/5280) = 3966.869318 miles
w= Cr^-2
w = 1570536900*3966.869318^-2 = 99.80501344 lbs
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Cheers,
Stan H.
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