Shown to the right is a truncated cone of length L cm traveling through a fluid

Question

Shown to the right is a truncated cone of length L cm traveling through a fluid at a speed of 8.50 m/s. The narrow end of the cylinder has a diameter of 2.32 cm, whereas the wider end has a diameter of 3.53 cm. The drag force acting on the cylinder is 25.6 newtons. The drag force Fd acting on an object passing through a fluid is given by...

Shown to the right is a truncated cone of length L cm traveling through a fluid at a speed of 8.50 m/s. The narrow end of the cylinder has a diameter of 2.32 cm, whereas the wider end has a diameter of 3.53 cm. The drag force acting on the cylinder is 25.6 newtons. The drag force Fd acting on an object passing through a fluid is given by Fd=-p.ATv 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 coefficientof the object if the fluid is water of density 1000.00 kg/m3 Number .0036

Explanation / Answer

We use the larger diameter for the cross-sectional area because if u look at the thing from the front view, that is the area u see.

F = 1/2 DAv^2
Where D is the drag coefficient
D = 2F/(Av^2)
= 2F/((r^2)v^2)
= 2F/(((d/2)^2)v^2)
= 8F/d^2v^2
= 8(25.6)/(1000)()(0.0353)^2(8.5)^2 = 0.724

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