Calculate the percent weight reduction of the plane as it moves from the sea level to 11,719.342 m.

%(∆W) = 0.37%

Explanation:

According to Newton's law of gravitation which states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

F = Gm1m2/r^2

Where

F = force between the masses

G universal gravitational constant

m1 and m2 = mass of the two particles

r = distance between the centre of the two mass

Therefore, weigh of an object on earth is inversely proportional to the square of its distance from the centre of the earth

W₁/W₂ = r₂²/r₁² 1

W₂ = W₁r₁²/r₂²

At sea level the weight of the plane is W1 and at distance r₁ from the centre of the earth which is equal to the radius of the earth.

The radius of the earth is = 6378.1km

r₁ = radius of the earth = 6 378.1km = 6,378,100m

r₂ = r₁ + 11,719.342m = 6,378,100m + 11,719.342m

r₂ = 6,389,819.342m

W₂ = W₁r₁²/r₂²

W₂ = W₁[(6378100)²/(6,389,819.342)²]

W₂ = W₁[0.996335234422]

W₂/W₁ = 0.9963

fraction reduction of the weight is

ΔW/W₁ = 1 - W₂/W₁ = 1 - 0.9963 = 0.0037

percentage change :

%(∆W) = 0.0037 × 100% = 0.37%

Therefore, the percentage weight loss is 0.37%

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation: