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
1a. W/C = 1/r^2.
Invert both sides:
C/W = r^2/1,
sqrt(C/W) = r.
r = sqrt(C/W).
1b. sqrt(C/100) = 3963,
C/100 = (3963)^2,
C = 100*(3963)^2 = 1,570,536,900.
To solve this question on weight variation based on elevation, the following steps were followed:
1. The equation w = cr^-2 was given, where w is the weight of an object and r is the distance of the object from the center of the Earth. To solve for r, the equation was rearranged.
2. The equation w/C = 1/r^2 was obtained by dividing both sides of the original equation by C.
3. Taking the square root of both sides, the equation (w/C)^2 = 1/r was obtained.
4. By reciprocating both sides of the equation, the value of r was found to be r = 1/(w/C)^2.
Moving on to the next part:
1. The object was given to be weighing 100 pounds when at sea level, which is 3,963 miles from the center of the Earth. To find the value of C, the equation C(3,963)^-2 = 100 was solved.
2. Solving the equation, C was found to be equal to 1.58.
Continuing to part c:
To determine the weight of the object in Death Valley (282 feet below sea level):
1. The given elevation of Death Valley was converted from feet to miles by dividing it by 5280 (1 mile = 5280 feet), resulting in 0.053 miles.
2. Subtracting the converted value from the sea level distance (3,963 miles), the new distance from the center of the Earth (r) was calculated to be approximately 3962.94 miles.
3. Using the obtained value of C (1.58), the weight of the object (w) was determined to be 99.46 pounds.
To find the weight of the object at the top of Mount McKinley (20,320 feet above sea level):
1. The given elevation of Mount McKinley was converted from feet to miles by dividing it by 5280, resulting in 3.84 miles.
2. Adding the converted value to the sea level distance (3,963 miles), the new distance from the center of the Earth (r) was calculated to be approximately 3966.84 miles.
3. Using the obtained value of C (1.58), the weight of the object (w) was determined to be 99.50 pounds.
Moving on to the second question on the distance to the horizon:
1. The equation given was h = (D/1.2)^2, where h represents the height in feet and D represents the distance in miles.
2. To solve for h, the equation was rearranged to obtain h = D^2/1.44.
Moving to the next part:
To find the distance to the horizon from the top of Long's Peak (elevation of 14,255 feet):
1. The given elevation was substituted into the equation, resulting in h = (D^2)/(1.44) = (14,255^2)/(1.44).
2. The equation was then simplified, finding D to be approximately 144.62 miles.
Regarding the question of whether Cheyenne, Wyoming (about 89 miles away) can be seen from the top of Long's Peak:
1. Comparing the distance to the horizon (144.62 miles) with the distance to Cheyenne (89 miles), it was concluded that Cheyenne can be seen from the top of Long's Peak since 89 miles is less than 144.62 miles.