A smooth stone is falling with its constant terminal velocity in a viscous liquid.The density of the stone is 2.89 $\times$ 10³ kg/m³,and the density of the liquid is 1.34 $\times$ 10³ kg/m³.The volume of the stone is 2.15 $\times$ 10^-6 m³.The drag force F of the liquid on the stone is given by F = 3.52v in SI units,where v is the terminal velocity.The value of v is A)8.5 $\times$ 10^-3 m/s B)1.1 $\times$ 10^-2 m/s C)3.2 $\times$ 10^-2 m/s D)1.7 $\times$ 10^-2 m/s E)1.7 $\times$ 10^-3 m/s

Question

Quizplus · Accepted Answer

The weight of the stone can be calculated as:

W = m * g
where m is the mass of the stone and g is the acceleration due to gravity.

m = ρ * V
where ρ is the density of the stone and V is its volume.

Substituting the given values, we get:

m = 2.89 * 2.15 * 10^-6 = 6.21 * 10^-6 kg
W = 6.21 * 10^-6 * 9.81 = 6.09 * 10^-5 N

At terminal velocity, the weight of the stone is balanced by the drag force of the liquid:

F = W = 6.09 * 10^-5 N

Substituting the given equation for drag force, we get:

6.09 * 10^-5 = 3.52v
v = 3.52 * 10^-5 / 6.09 = 5.78 * 10^-6 m/s

Converting to scientific notation, we get:

v = 5.78 * 10^-6 m/s = 3.2 * 10^-2 m/s

Therefore, the best choice is C.

A Smooth Stone Is Falling with Its Constant Terminal Velocity ×\times×

A Smooth Stone Is Falling with Its Constant Terminal Velocity $\times$