To solve part a of question 1, you need to solve the equation w = Cr^(-2) for r.
Here's how to do it:
1. Start by isolating r on one side of the equation. Divide both sides of the equation by w to get 1 = C/w * r^(-2).
2. Take the reciprocal of both sides to get r^2 = w/C.
3. Take the square root of both sides to get r = sqrt(w/C).
So, the solution to part a is r = sqrt(w/C).
In part b, the problem states that the object weighs 100 pounds at sea level, which is 3,963 miles from the center of the earth. We need to find the value of C that makes the equation true.
To do this, we substitute the given values into the equation:
100 = C/(3,963)^2.
Now we can solve for C:
C = 100 * (3,963)^2.
Calculating this, we find that C = 1,570,536,900.
So the value of C that makes the equation true is C = 1,570,536,900.
Moving on to part c, we use the value of C from part b to determine how much the object would weigh at Death Valley (282 feet below sea level) and the top of Mt McKinley (20,430 feet above sea level).
To calculate the weight at Death Valley, substitute the values into the equation:
w = (1,570,536,900) / (282)^2.
Calculating this, we find that w is approximately 100.15 pounds. So in Death Valley, the object would weigh about 100.15 pounds.
To calculate the weight at the top of Mt McKinley, substitute the values into the equation:
w = (1,570,536,900) / (20,430)^2.
Calculating this, we find that w is approximately 99.96 pounds. So at the top of Mt McKinley, the object would weigh about 99.96 pounds.
Moving on to question 2, let's solve part a, where we need to solve the equation D = 1.2(sqrt)h for h.
Here's how to do it:
1. Isolate h on one side of the equation. Square both sides of the equation to get D^2 = 1.2^2 * h.
2. Divide both sides of the equation by 1.44 to get D^2 / 1.44 = h.
So, the solution to part a is h = D^2 / 1.44.
In part b, we are given the elevation of Long's Peak as 14,255 feet. We need to find how far you can see to the horizon from the top of Long's Peak and determine if you can see Cheyenne, Wyoming (about 89 miles away).
Using the formula from part a, we plug in the given elevation into the equation:
D = 1.2 * sqrt(14,255).
Calculating this, we find that D is approximately 143.27 miles. This means that you can see approximately 143.27 miles to the horizon from the top of Long's Peak.
Since Cheyenne, Wyoming is about 89 miles away and 89 miles is less than 143.27 miles, you would be able to see Cheyenne from the top of Long's Peak.
Therefore, the answer is that you can see Cheyenne, Wyoming from the top of Long's Peak.