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January 30, 2015

January 30, 2015

Posted by **Dee** on Thursday, August 19, 2010 at 5:42pm.

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.

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: , where C is a constant, and r is the distance that the object is from the center of Earth.

a. Solve the equation 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

i. Death Valley (282 feet below sea level).

ii. the top of Mount McKinley (20,320 feet above sea level).

2. The equation gives the distance, D, in miles that a person can see to the horizon from a height, h, in feet.

a. Solve this equation for h.

b. Long’s Peak in Rocky Mountain National Park, is 14,255 feet in elevation. How far can you see to the horizon from the top of Long’s Peak? Can you see Cheyenne, Wyoming (about 89 miles away)? Explain your answer.

- MAT/117- Appendix E -
**Ms. Sue**, Thursday, August 19, 2010 at 6:19pmHow would you like us to help you with this assignment?

- MAT/117- Appendix E -
**Damon**, Thursday, August 19, 2010 at 6:34pmI can not see your equation but assume it is of the form

W = C/r^2

then

r^2 = C/W

r = sqrt (C/W)

Now sea level in feet is

r = 3963 miles (5280 ft/mile) = 20.9 * 10^6 feet

so

100 = C/(20.9*10^6)^2

C = 4.37*10*16

In death valley r = 20.9*10^6 - 282

r = 20.899*10^6

r^2 = 436.8*10*12

W = 100.05 tiny bit heavier

do the same thing on the mountain. It will be slightly lighter

For the second problem I also can not see your formula

It will be of form

D =sqrt (h(2R+h))

if h <<r (which it is here)

D = sqrt (2 R h)

then

D^2 = 2 R h

h = D^2/(2R)

Now

see how far you can see the horizon from Long's peak (find D for h = height of Long's peak)

then see how far you can see from Cheyenne (You will have to look up how high the city is)

add those two to find out if it is more than 89 miles.

We will have to assume that nothing sticks up between them.

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