Water flows with a speed of 2.5 m/s through a section of hose with a cross-sectional area of 0.0084 m^2. Further along the hose the cross-sectional area changes, and the water speed is reduced to 1.1 m/s. Calculate the new cross-sectional area.

how to make it

since the volume flow rate of the water is constant (why?) and m^2 * m/s = m^3/s

A*1.1 = .0084 * 2.5

unit?

i have exam now

To calculate the new cross-sectional area of the hose, we can make use of the principle of continuity, which states that the volume flow rate of an incompressible fluid is constant along a pipeline.

The volume flow rate (Q) is given by the equation Q = Av, where A is the cross-sectional area of the pipe and v is the velocity of the fluid. Since we know the speed and area at one point in the hose and the speed at another point, we can set up the equation as follows:

Av = Av'

Where A is the initial cross-sectional area, v is the initial speed, and v' is the final speed.

Rearranging the equation, we have:

A = Av' / v

Substituting the given values into the equation:

A = (0.0084 m^2 * 1.1 m/s) / 2.5 m/s

A ≈ 0.003696 m^2

Therefore, the new cross-sectional area of the hose is approximately 0.003696 m^2.