Could someone look over this paper before I turn it in? Thanks
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 w=cr^-2 for r.
w = Cr^-2,
w/C = r^-2
w/C = 1/r^2
(w/C)^2 = 1/r
r = 1/(w/C)^2
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.)
100 = C(3,963)^-2
100 = C(62.950)
C = 100 / 62.95
C = 1.58
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).
5280 feet = 1 mile
282 feet = 282 / 5280 = 0.053 miles
Therefore r = 3,963 - 0.053 = 3962.94 miles
Since C = 1.58, therefore
w = 1.58(3962.94)^-2
w = 1.58(62.95)
w = 99.46 pounds
ii. the top of Mount McKinley (20,320 feet above sea level).
5280 feet = 1 mile
20,320 feet = 20,320 / 5280 = 3.84 miles
Therefore r = 3963 + 3.84 = 3966.84 miles
Since C = 1.58, therefore
w = 1.58(3966.84)^-2
w = 1.58(62.98)
w = 99.50 pounds
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.
H=(D/1.2)^2
H=D^2/1.44
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.
14,255=D^2/1.44
D^2=14,525*1.44
D^2=20,916
D=144.62
Yes Cheyenne, Wyoming can be seen because 89 miles is less than the 144.6
To solve this paper, there are a few steps you can follow:
1. Read the instructions and prompt carefully to understand the problem you need to solve.
2. For part 1a, you need to solve the equation w = cr^-2 for r. To do this, first divide both sides of the equation by C, which gives w/C = r^-2.
3. Next, take the reciprocal of both sides of the equation, so 1/(w/C) = r^2. Simplify further to (w/C)^-2 = r^2.
4. Finally, take the square root of both sides of the equation to find r: r = 1/(w/C)^2.
5. For part 1b, you are given the object's weight at sea level (100 pounds) and the distance from the center of the Earth at sea level (3,963 miles). You need to find the value of C that makes the equation true.
6. Plug in the given values into the equation: 100 = C(3,963)^-2.
7. Solve for C by dividing both sides of the equation by (3,963)^-2: C = 100 / (3,963)^-2.
8. Calculate the value of C using a calculator or by manually doing the calculations.
9. For part 1c, you are asked to determine how much the object would weigh in Death Valley (282 feet below sea level) and on top of Mount McKinley (20,320 feet above sea level).
10. Convert the given heights to miles by using the conversion factor 1 mile = 5,280 feet. For example, Death Valley is 282 feet, which is equal to 282 / 5280 = 0.053 miles.
11. Use the value of C you calculated in step 8, and plug it into the equation w = 1.58 * r^-2 to find the weight of the object in Death Valley and on top of Mount McKinley.
12. Calculate the weight for each location using the respective distances from the center of the Earth (r) and the value of C you found. Make sure to convert distances from feet to miles when necessary.
13. Present your calculated weights with appropriate units and rounding according to the instructions.
14. For part 2a, you need to solve the equation H=(D/1.2)^2 for h. Divide both sides of the equation by 1.44 to get H/1.44 = (D/1.2)^2.
15. Take the square root of both sides of the equation to solve for D: D = sqrt(H / 1.44).
16. For part 2b, you are given the height of Long's Peak (14,255 feet) and asked to determine how far you can see to the horizon from there. Use the equation D = sqrt(H / 1.44) and plug in the given height.
17. Calculate D using a calculator or manually perform the calculation.
18. Compare the result with the distance to Cheyenne, Wyoming (89 miles) to determine if it can be seen. State your answer and explanation clearly.
Remember to double-check your calculations and final answers to ensure accuracy before submitting your paper.