Water is flowing in a straight horizontal pipe of variable cross section. Where the cross-sectional area of the pipe is 3.50·10-2 m2, the pressure is 5.30·105 Pa and the velocity is 0.310 m/s. In a constricted region where the area is 12.90·10-4 m2, what is the velocity?

To find the velocity in the constricted region, we can use the principle of conservation of mass, which states that the mass flow rate is constant along the pipe. The mass flow rate is given by the product of the density of water (ρ) and the volume flow rate (Q):

Mass flow rate (ρQ) = constant

Since water density is constant, we can rewrite the equation as:

Q1 = Q2

where Q1 is the volume flow rate in the initial region and Q2 is the volume flow rate in the constricted region.

The volume flow rate is given by the product of cross-sectional area (A) and velocity (v):

Q = Av

Thus, we can rewrite the equation as:

A1v1 = A2v2

where A1 and v1 are the cross-sectional area and velocity in the initial region, and A2 and v2 are the cross-sectional area and velocity in the constricted region.

We are given the following values:
A1 = 3.50·10^-2 m^2 (initial cross-sectional area)
v1 = 0.310 m/s (initial velocity)
A2 = 12.90·10^-4 m^2 (constricted cross-sectional area)

We need to find v2.

Let's substitute the values into the equation:

(3.50·10^-2 m^2)(0.310 m/s) = (12.90·10^-4 m^2)(v2)

Now, let's solve for v2:

v2 = (3.50·10^-2 m^2)(0.310 m/s) / (12.90·10^-4 m^2)

v2 ≈ 0.0848 m/s

Therefore, the velocity in the constricted region is approximately 0.0848 m/s.