Hint for this assignment: Pay attention to the units of measure. You may have to convert from feet to miles 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 the earth.
a. Solve the equation for r.
b. Suppose that an object is 30 pounds when it is at sea level. Find the value of C that makes the equation true. (Sea level is approximately 4,000 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 Mt McKinley (20,320 feet above sea level)
2. The equation gives the distance, D, in kilometers that a person can see to the horizon from a height, h, in meters.
a. Solve this equation for h.
b. Mount Evans in the Rocky Mountain National Park, is 4,375 meters in elevation. How far can you see to the horizon from the top of Mount Evans? Can you see Fort Collins (about 220 kilometers away)? Explain your answer.
1a. equation is r^-2=w/c
2. equation is d=3.6sqrt(h)
I forgot to include the equation into the problems.
To solve these problems, we'll follow the given instructions and use the provided conversion factor.
1. Weight of an object on earth:
a. The equation is given as W = C / r, where W is the weight, C is a constant, and r is the distance from the center of the earth.
To solve for r, we can rearrange the equation to r = C / W.
b. We are told that the object weighs 30 pounds at sea level, which is approximately 4,000 miles from the center of the earth.
Using the given conversion factor of 1 mile = 5,280 feet, we can convert the distance to feet: 4,000 miles * 5,280 feet/mile = 21,120,000 feet.
Substituting the values into the equation r = C / W, we have r = 21,120,000 feet.
c. Now, using the value of r obtained in the previous question, we can determine the weight of the object at different locations.
i. Death Valley (282 feet below sea level):
The distance from the center of the earth can be calculated as follows:
Distance = Sea level distance - Death Valley elevation
= 21,120,000 feet - (-282 feet) (minus because it is below sea level)
= 21,120,000 feet + 282 feet
= 21,120,282 feet
Now, using the equation r = C / W, and the weight at sea level (30 pounds), we can solve for C:
21,120,282 feet = C / 30 pounds
C = 21,120,282 feet * 30 pounds
C = 633,608,460 feet/pound
ii. The top of Mt McKinley (20,320 feet above sea level):
The distance from the center of the earth can be calculated as:
Distance = Sea level distance + Mt McKinley elevation
= 21,120,000 feet + 20,320 feet
= 21,140,320 feet
Using the equation r = C / W and the weight at sea level (30 pounds), we can solve for the weight at the top of Mt McKinley:
21,140,320 feet = C / Weight
C = 21,140,320 feet * 30 pounds
C = 634,209,600 feet/pound
Therefore, the object would weigh 633,608,460 pounds in Death Valley and 634,209,600 pounds on the top of Mt McKinley.
2. The distance a person can see to the horizon:
a. The equation is given as D = 3.57 √h, where D is the distance in kilometers and h is the height in meters.
To solve for h, we can rearrange the equation as h = (D / 3.57)².
b. Mount Evans is 4,375 meters in elevation. To calculate the distance you can see to the horizon from the top of Mount Evans, we substitute this elevation into the equation:
h = (D / 3.57)²
4,375 = (D / 3.57)²
Solving for D:
D = 3.57 √4,375
D ≈ 3.57 * 66.11
D ≈ 235.77 kilometers
Therefore, from the top of Mount Evans, you can see approximately 235.77 kilometers to the horizon. Fort Collins, which is about 220 kilometers away, would not be visible as it falls within the range of visibility.