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The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth. A particular person weighs 192 pounds on the surface of the earth and the radius of the earth is 3900 miles. Determine the equation that relates weight, W, to the distance from the center of the earth, d, for this person.
Answer choices on picture.

The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth A particul class=

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hisroyal1
The question tells us that the weight of a person varies with the square of their distance to the center of the Earth.

Weight = 192 pounds
Radius of the earth = 3900  miles 

Let's introduce the constant, k
192 = [tex] \frac{k}{3900 ^{2} } [/tex]
k = 192 * 3900²
k = 192 * 15,210,000
k = 2,290,320,000

Putting the constant k back into the equation, W = [tex] \frac{k}{d ^{2} } [/tex]

W = [tex] \frac{2,290,320,000}{d ^{2} } [/tex]

(OPTION D)
The weight of a person on or above the surface of the earth varies inversely as the square of the distance the person is from the center of the earth.

The distance the person is from the center of the earth: d

W=K/d^2

W=192 pounds, d=R: Radius of the earth=3900 miles
Replacing in the equation above:
192=K/3900^2

Solving for K:
192=K/15,210,000
192*15,210,000=K
2,920,320,000=K
K=2,920,320,000

Then the equation is:
W=K/d^2 and K=2,920,320,000

W=2,920,320,000/d^2

Answer: Option d. W=2,920,320,000/d^2 


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