A metal sphere of density ρ_s and volume V_s is placed inside a liquid of density ρ_L. The sphere starts accelerating down under the influence of weight, buoyancy, and drag (with proportionality constant B to the speed u).  Find the acceleration of the sphere when {force_of_buoyancy + force_of_drag}= ρ_s · V_s · g. Solve for ρ_s=1.5 Kg/m^3, V_s=0.001 m^3, B=0.05 Kg/s and ρ_L=0.5 Kg/m^3. If the problem has no solution input -999 otherwise your answer in X.XX m/s^2

The given equation is {force_of_buoyancy + force_of_drag}= ρ_s · V_s · g.

The force of buoyancy is given by the equation F_b = ρ_L · V_s · g, where ρ_L is the density of the liquid, V_s is the volume of the sphere, and g is the acceleration due to gravity.

The force of drag is given by the equation F_d = B · u, where B is the drag coefficient and u is the speed of the sphere.

Substituting these equations into the given equation, we get:

ρ_L · V_s · g + B · u = ρ_s · V_s · g

We are asked to find the acceleration of the sphere, which is given by the equation a = F/m, where F is the net force on the sphere and m is the mass of the sphere. The mass of the sphere is given by the equation m = ρ_s · V_s.

Substituting the given values into these equations, we get:

0.5 Kg/m^3 · 0.001 m^3 · 9.8 m/s^2 + 0.05 Kg/s · u = 1.5 Kg/m^3 · 0.001 m^3 · 9.8 m/s^2

Solving for u, we get:

u = (1.5 Kg/m^3 · 0.001 m^3 · 9.8 m/s^2 - 0.5 Kg/m^3 · 0.001 m^3 · 9.8 m/s^2) / 0.05 Kg/s

u = 0.2 m/s

Substituting this value into the equation for acceleration, we get:

a = (1.5 Kg/m^3 · 0.001 m^3 · 9.8 m/s^2 - 0.5 Kg/m^3 · 0.001 m^3 · 9.8 m/s^2) / (1.5 Kg/m^3 · 0.001 m^3)

a = 6.6 m/s^2

Therefore, the acceleration of the sphere is 6.6 m/s^2.