A truncated cone of length L cm is traveling through a fluid at a speed of 8.71

Question

A truncated cone of length L cm is traveling through a fluid at a speed of 8.71 m/s. The narrow end of the cylinder has a diameter of 2.42 cm, whereas the wider end has a diameter of 3.68 cm. The drag force acting on the cylinder is 27.6 newtons. The drag force "Fd" acting on an object passing through a fluid is given by Fd=1/2pA?v^2, where A is the effective cross-sectional area of the object, ? (the Greek letter "gamma") is the drag coefficient, ? is the density of the fluid, and v is the speed of the object. Calculate the drag coefficient ? of the object if the fluid is water of density 1000.00 kg/m3

Explanation / Answer

F = 1/2 D p A v2

D = 2 F / (p A v2) = 2F / (p ( pi r2) v2)

= 2F / ( p (pi (d / 2)2) v2 )= 8 F / p pi d2 v2

= 8 * 27.6 / (1000.00 * pi * 0.03682 * 8.712

= 0.684

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