At position A within a tube containing fluid that is moving with steady laminar flow, the speed of the fluid is 12.0 m/s and the tube has a diameter 12.00 cm. At position B, the speed of the fluid is 18.0 m/s and the tube has a diameter 6.00 cm. What is the ratio of the density of the fluid at position A to the density of the fluid at position B

0.375

Explanation:

For incompressible flow, we know that;

ρ1•v1•A1 = ρ2•v2•A2

Where;

ρ1 = density of fluid at position A

v1 = speed of fluid at position A

A1 = area of tube

ρ2 = density of fluid at position B

v2 = speed of fluid at position B

A2 = area of tube

We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.

Thus;

ρ1/ρ2 = (v2•A2)/(v1•A1)

Now, the tube will have the same height.

But we are given;

diameter of A = 12.00 cm = 0.12 m

diameter of B = 6 cm = 0.06 m

Thus;

A1 = π(d²/4)h = πh(0.12²/4)

A2 = πh(0.06²/4)

We are also given;

v1 = 12 m/s

v2 = 18 m/s

Thus;

ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))

πh/4 will cancel out to give;

ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)

ρ1/ρ2 = 0.375

f=ma newton

f=100x2.5=250n