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Bas_tet [7]
3 years ago
6

If you could embark on Jules Verne’s journey to the center of the earth you would discover that the force of gravity would at fi

rst increase as you and
Physics
1 answer:
jekas [21]3 years ago
8 0

i dont know my dude but thanks for the free points XD

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The pressure difference, , across a partial blockage in an artery (called a stenosis) is approximated by the equation where is t
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The question is incomplete. The complete question is  :

The pressure difference, Δp, acK_uross a partial blockage in an artery (called a stenosis) is approximated by the equation :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Where V is the blood velocity, μ the blood viscosity {FT/L2}, ρ the blood density {M/L3}, D the  artery diameter, A_0 the area of the unobstructed artery, and A1 the area of the stenosis.  Determine the dimensions of the constants K_v and K_u. Would this equation be valid in any  system of units?

Solution :

From the dimension homogeneity, we require :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Here, x means dimension of x. i.e.

$[ML^{-1}T^{-2}]=\frac{[K_v][ML^{-1}T^{-1}][LT^{-1}]}{[L]}+[K_u][1][ML^{-3}][L^2T^{-2}]$

                    $=[K_v][ML^{-1}T^{-2}]+[K_u][ML^{-1}T^{-2}]$

So, $[K_u]=[K_v]=[1 ]=$ dimensionless

So, K_u and K_v  are dimensionless constants.

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