Posted by asd on Sunday, September 19, 2010 at 4:39pm.
Could you please check my work on these word problems? My solutions just don't seem right, could you please let me know where I may have messed up.
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 Earth.
Hint: Pay attention to the 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."
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 object would weigh in Death Valley (282 feet below sea level) 282ft=0.053409mi??
My solutions:
a. r=sqrt(C/w)
b. w=Cr^-2 or w=C*1/(r^2) ??
100=C(1/3,963^2)
100=C(1/15,705,369)
100*15,705,369=C(1/15,705,369)*15,705,369
C=1,570,536,900 ??
c. w=1,570,536,900(1/0.053409^2)
w=1,570,536,900(1/0.002853)
w=1,570,536,900(350.508237)
???? w=550,486,119,962.4453??
If you could let me know if and where I may have made any mistakes I'm sure I could complete the rest of the assignment.
Thank you!
- Algebra - bobpursley, Sunday, September 19, 2010 at 4:54pm
a is wrong. r=sqrt(W/C)
b. You need units on C.
c. nope, r would be below sea level,
r=3963-.053209
now find that r, and rework.
- Algebra - asd, Sunday, September 19, 2010 at 5:45pm
How did you come up with your answer to a? Everytime I try to rework it I get r=sqrt(C/w)
- Algebra - PsyDAG, Monday, September 20, 2010 at 1:27pm
asd, I get the same answer as you do to a.
W = Cr^-2 = C(1/r^2)
Multiply both sides by r^2 and divide both sides by W.
r^2 = C/W
r = √(C/W)
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