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Sagot :

Answer:

\frac{4}{3}\pi R^3 \rho_s g = \frac{4}{3}\pi R^3 \rho g + \frac{1}{2} C_d \rho \pi R^2 w_s^2

\frac{4}{3}\pi R^3 \rho_s g = deposition of non-cohesive sediments

\pi = approx 3.14

R = radius of the spherical object being deposited

\rho = density of spherical object

g = gravitational acceleration

C_d = drag coefficient

w_s = particle's settling velocit