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.
What is the equation?
To solve these equations and answer the given questions, we'll need to use the given information and apply the relevant formulas. Let's go through each question step by step.
1. Equation for weight variation with elevation:
The equation for weight variation with elevation is given as W = C/r, where W is the weight of an object, C is a constant, and r is the distance from the center of the Earth.
a. Solving for r:
To solve for r, we need to rearrange the equation. Multiply both sides of the equation by r:
W * r = C
Divide both sides by W:
r = C/W
b. Finding the value of C:
In this question, we're given that the weight of an object is 100 pounds when at sea level (r = 3,963 miles). We can substitute these values into the equation and solve for C.
100 * 3,963 = C
c. Calculating weight at different elevations:
Now that we have the value for C, we can use it to calculate the weight of the object at different elevations.
i. Death Valley (282 feet below sea level):
To calculate the weight at Death Valley, we can substitute the distance from the center of the Earth as -3,963 - 282 miles (negative because it's below sea level).
Weight = C / (-3,963 - 282)
ii. Top of Mount McKinley (20,320 feet above sea level):
To calculate the weight at the top of Mount McKinley, we can substitute the distance from the center of the Earth as 3,963 + 20,320 miles (positive because it's above sea level).
Weight = C / (3,963 + 20,320)
2. Equation for distance to the horizon:
The equation for the distance, D, to the horizon from a height, h, in feet is given as D = √(1.5h).
a. Solving for h:
To solve for h, we need to rearrange the equation. Square both sides of the equation:
D^2 = 1.5h
Divide both sides by 1.5:
h = D^2 / 1.5
b. Distance to the horizon from Long’s Peak:
We're given the elevation of Long's Peak as 14,255 feet. We can substitute this value into the equation and calculate the distance, D, to the horizon.
D = √(1.5 * 14,255)
Based on the distance calculated, you can check if it allows you to see Cheyenne, Wyoming, which is approximately 89 miles away. The curvature of the Earth might obstruct the view, so you can conclude whether it's possible based on the given information.