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:w=Cr^-2 , where C is a constant, and r is the distance that the object is from the center of Earth.

a. Solve the equation w=Cr^-2 for C.

b. Suppose that an object is 500 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 (382 feet below sea level).

ii. the top of Mount McKinley (30,320 feet above sea level).

(Round answers to the nearest hundredth ).

2. The equation D= 1.2 sqrt h 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 15,345 feet in elevation. How far can you see to the horizon from the top of Long’s Peak? (Round the answer to the nearest hundredth). Can you see Cheyenne, Wyoming (about 92 miles away)? Explain your answer.

To solve these problems, we will go step by step and explain the process along the way.

1.
a. Solve the equation w=Cr^-2 for C:
To solve for C, we need to isolate it on one side of the equation. We can do this by rearranging the equation:
w = C/r^2
Multiply both sides by r^2:
w * r^2 = C
So, the value of C is equal to w times r^2.

b. Find the value of C that makes the equation true:
We are given that the object weighs 500 pounds when at sea level, which is 3,963 miles from the center of the Earth. Using the equation, we can substitute the values:
500 = C/(3,963)^2
To solve for C, we multiply both sides by (3,963)^2:
C = 500 * (3,963)^2

c. Determine the weight of the object in Death Valley (382 feet below sea level):
To find the weight of the object in Death Valley, we need to calculate the distance from the center of the Earth. Sea level is 3,963 miles from the center, and Death Valley is 382 feet below sea level.
Since we are given the conversion of 1 mile = 5,280 feet, we can convert the distance to miles:
382 feet / 5,280 feet/mile ~= 0.0724 miles
Now, we can substitute this value into the equation:
w = C/r^2
w = C/(0.0724)^2
Using the value of C found in the previous question, calculate w.

ii. Determine the weight of the object at the top of Mount McKinley (30,320 feet above sea level):
Similar to the previous question, we need to calculate the distance from the center of the Earth. Sea level is 3,963 miles from the center, and Mount McKinley is 30,320 feet above sea level.
Convert the distance to miles:
30,320 feet / 5,280 feet/mile ~= 5.7576 miles
Substitute this value into the equation:
w = C/r^2
w = C/(5.7576)^2
Using the value of C found earlier, calculate w.

2.
a. Solve the equation D= 1.2 sqrt h for h:
To solve for h, we need to isolate it on one side of the equation. Rearrange the equation:
D = 1.2 sqrt h
Divide both sides by 1.2:
D/1.2 = sqrt h
Now, square both sides of the equation:
(D/1.2)^2 = h
So, the value of h is equal to (D/1.2)^2.

b. Determine the distance you can see to the horizon from the top of Long’s Peak:
Using the equation, substitute the given height of Long's Peak (15,345 feet) for h. Solve for D.

Can you see Cheyenne, Wyoming (about 92 miles away)? Explain your answer:
Compare the calculated distance of how far you can see to the horizon from Long's Peak with the distance to Cheyenne, Wyoming (92 miles). If the calculated distance is greater than or equal to the distance to Cheyenne, then it is possible to see Cheyenne from the top of Long's Peak. Otherwise, it is not possible.