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 8.74 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 28.6 newtons.

Shown to the right is a truncated cone of length L cm traveling through a fluid at a speed of 8.74 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 28.6 newtons. The drag force Fd acting on an object passing through a fluid is given by Fd = 1/2 rhoAGammav2 where A is the effective cross-sectional area of the object, Gamma (the Greek letter "gamma") is the drag coefficient, rho is the density of the fluid, and v is the speed of the object. Calculate the drag coefficient Gamma of the object if the fluid is water of density 1000.00 kg/m3

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 D?Av^2
D = 2F/(?Av^2)
= 2F/(?(?r^2)v^2)
= 2F/(?(?(d/2)^2)v^2)
= 8F/??d^2v^2
= 8(28.6)/(1000.00)(?)(0.0368)^2(8.74)^2 = 0.704

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