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%

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

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