Posted by Sue Bonds on Saturday, August 13, 2011 at 1:11am.
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=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 150 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
i. Death Valley (282 feet below sea level)
ii. The top of Mt McKinley (20,430 feet above sea level)
2. After an accident, how do the police determine the speed at which the car had be travelling prior to the accident?
The formula s = 2√(5L) can be used to approximate the speed s, in miles per hour of a car that has left skid marks of length L, in feet.
a. How far will a car skid at 55 mph?
b. How far will a car skid at 100 mph?
3. The equation D= 1.2√h gives the distance, D, in miles that a person can see to the horizon from a height, h, in feet.
a. How far can you see from an airplane window at 32,000 feet?
b. If you can see 13 miles to the horizon from the top of an observation
tower, what is the height of the tower?

Algebra  Anonymous, Saturday, August 13, 2011 at 6:53pm
do ur own work

Algebra  Henry, Saturday, August 13, 2011 at 11:03pm
1a. W = Cr2,
Cr = w + 2,
r = (W+2) / C.
b. C = (w+2) / r,
C = (150+2) / 3963 =0.03835.
2a. S = 2sqrt(5L) = 55mi/h,
sqrt(5L) = 27.5.
Square both sides:
5L = 756.25,
L = 151.3 Ft.
b. Same procedure as a.
3a. Solve the given Eq for D.
b. D = 1.2sqrt(h) = 13mi,
sqrt(h) = 10.83,
h = 117.4 Ft.
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