An Object moving in a liquid experiences a linear drag force: D= (bv, direction opposite the motion), where b is constant called the drag coefficient. For a sphere of radius R, the drag constant can be computed as b = 6πηR, where η is the viscosity of the liquid.

a. Use what you've learned in calculus to prove that
ax= vx . dvx/dx
b. Find an algebraic expression for , Vx (x) the x-component of velocity as a funtion of distance traveled, for a spherical particle of radius R and mass m that is shot horizontally with initial speed Vo through a liquid viscosity η.
c. Water at 20°C has viscosity η= 1.0 x 10^-3 Ns/m^2.Suppose a 1.0-cm-diameter, 1.0 g marble is shot horizontally into a tank of 20°C water at 10 cm/s. How far will it travel before stopping?

The rule of the chain of derivatives and Newton's second law allows to find the results for the questions about the motion in a viscous liquid are:

a) the acceleration is: a = [tex]v \ \frac{dv}{dx}[/tex]

b) the velocity of the marble is: [tex]v =v_o - \frac{6\pi \eta R}{m} \ x[/tex]

c) the distance traveled before stopping is: x = 1.06 m

a) The velocity is defined as the variation of the position with respect to the time and the acceleration of a body is defined as the variation of the velocity with the time.

[tex]v= \frac{dx}{dt} \\a= \frac{dv}{dt}[/tex]

Let's use the chain rule of derivatives.

[tex]a = \frac{dv}{dx} \ \frac{dx}{dt}[/tex]

Let'ssubstitute

[tex]a = v \ \frac{dx}{dt}[/tex]

b) Newton's second law states that the net force is proportional to the product of the mass and the acceleration of the body.

Σ F = m a

Where F is the force, m is the mass and a the acceleration. Indicates that the only force is the friction force.

fr = - b v

Let's substitute.

- b v = [tex]m \ \ v \frac{dv}{dx}[/tex]

[tex]- \frac{b}{m} \ dx = dv[/tex]

let's integrate and evaluate.

[tex]- \frac{b}{m} \ ( x-x_o) = v-v_o[/tex]

let us fix the reference system at the initial point, therefore x₀ = 0 for the initial velocity v₀.

[tex]- \frac{b}{m} x = v - v_o \\v =v_o - \frac{b}{m} \ x[/tex]

They indicate that the constant b, for a sphere has the form

b = 6π ηR

let's substitute.

[tex]v =v_o - \frac{6\pi \eta R}{m} \ x[/tex]

c) Ask the displacement of the marble before stopping.

Indicates that the liquid is water with a viscosity of η = 1 10⁻³ N s, a marble of mass m = 1.0 g = 10⁻³ kg is thrown with an initial velocity of v₀ = 10 cm/s = 0,10 m/s and has a diameter of 1.0 cm, so its radius is R = 0.5 cm = 5 10⁻³ m.

Since they ask for the distance traveled to a stop, the final speed is zero.

[tex]0 = v_o - \frac{6\pi \eta R}{m} \ x \\ \\ x = \frac{m}{6\pi \eta Ry} \ v_o[/tex]

let's calculate

x = [tex]\frac{1 \ 10^{-3} }{6\pi \ 10^{-3} 5 \ 10^{-3}} \ 0.10[/tex]

x = 1,061 m

In conclusion using the chain rule of derivatives and Newton's second law we can find the results for the questions about the motion of the marble in a viscous liquid are:

a) the acceleration is: [tex]a = v \ \frac{dv}{dx}[/tex]

b) the velocity of the marble is: [tex]v =v_o - \frac{6\pi \eta R}{m} \ x[/tex]

c) the distance traveled before stopping is: x = 1.06 m

Learn more here: brainly.com/question/20575411