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A spherical raindrop 2.5 mm in diameter falls through a vertical distance of 3900 m. Take the cross-sectional area of a raindrop = πr2, drag coefficient = 0.45, density of water to be 1000 kg/m3, and density of air to be 1.2 kg/m3. (a) Calculate the speed (in m/s) a spherical raindrop would achieve falling from 3900 m in the absence of air drag. m/s

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Answer:

276.62 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s² (positive downward and negative upward)

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 3900+0^2}\\\Rightarrow v=276.62\ m/s[/tex]

Neglecting air drag the velocity of the spherical drop would be 276.62 m/s

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