Many people know that the weight of an objectvaries on different planets,butdid you know that the wieght of an object on earth also varies according to the elevation oj an object.in particualr ,the weight of an equation follows this equation:w=cr -2 squared, where c is constant and r is the distancethat the object is from the center of the earth. solve the followingA) solve the equation w=cr -2squared.B)suppose that the weight of an object is 100 pds when at seal level. find the value of cthat makes the equation true at 3,963 miles from the center of the earth. C)use the value of cin prior question todetermine how much object would wweigh in death valley(282 ft below sea level). D)the top of mt mckinly at20,320 ft above sea level. Hint;1 mile =5380 ft

A) To solve the equation w = cr - 2 squared, we need to substitute the given values of r and solve for w. However, there seems to be a missing term in the equation, as it should be w = cr - 2^2 (squared). Please clarify if there is an additional term in the equation.

B) To find the value of c that makes the equation true at 3,963 miles from the center of the earth, we need to substitute the given values of w (weight at sea level) and r (distance) in the equation.

Given:
w = 100 pounds (weight at sea level)
r = 3,963 miles (distance from the center of the earth)

Using the equation w = cr - 2^2, we substitute the values:
100 = c * 3,963 - 2^2

Simplifying:
100 = c * 3,963 - 4

Rearranging:
c * 3,963 = 100 + 4
c * 3,963 = 104

Dividing both sides by 3,963:
c = 104 / 3,963

So the value of c that makes the equation true at 3,963 miles from the center of the earth is approximately 0.0262.

C) To determine how much the object would weigh in Death Valley (282 ft below sea level), we need to calculate the new distance from the center of the earth.

Given:
Distance from sea level to Death Valley = 282 ft.

To convert feet to miles, we divide by 5,280 (since 1 mile = 5,280 feet):
Distance from sea level to Death Valley = 282 / 5280 miles (approximately 0.0534 miles).

Using the equation w = cr - 2^2, and substituting the values:
w = 0.0262 * 0.0534 - 2^2

Calculating:
w = 0.00138909 - 4

w = -3.99861091 pounds (approximately)

Therefore, the object would weigh approximately -3.9986 pounds in Death Valley.

Note: Negative weight indicates that the object experiences an upward force in Death Valley due to the change in elevation.

D) To determine the weight of an object at the top of Mt. McKinley (20,320 ft above sea level), we need to calculate the new distance from the center of the earth.

Given:
Height of Mt. McKinley = 20,320 ft.

To convert feet to miles, we divide by 5,280:
Height of Mt. McKinley = 20,320 / 5,280 miles (approximately 3.8485 miles).

Using the equation w = cr - 2^2, and substituting the values:
w = 0.0262 * 3.8485 - 2^2

Calculating:
w = 0.1007577 - 4

w = -3.8992423 pounds (approximately)

Therefore, the object would weigh approximately -3.8992 pounds at the top of Mt. McKinley.

Again, negative weight indicates an upward force due to the change in elevation.