The velocity v of a fluid flowing in a conduit is inversely proportional to the cross- sectional area of the conduit.( Assume that the volume of the flow per unit of time is held constant .) Determine the change in the velocity of water flowing from a hose when a person places a finger over the end of the hose to decrease its cross-sectional area by 25%

%CV = 1.25Vo. Where Vo is initial velocity

Change = 1.25Vo-Vo = 0.25Vo.

%Change = (0.25Vo/Vo)*100% = 25%.

To determine the change in velocity of water flowing from a hose when the cross-sectional area decreases by 25%, we can use the equation for the relationship between velocity and cross-sectional area:

v1 * A1 = v2 * A2

where:
- v1 is the initial velocity of the water
- A1 is the initial cross-sectional area of the hose
- v2 is the final velocity (after the cross-sectional area decreases)
- A2 is the final cross-sectional area of the hose

Given:
- The decrease in cross-sectional area is 25%, which means A2 = (1 - 0.25) * A1 = 0.75 * A1

Substituting the values into the equation, we have:

v1 * A1 = v2 * 0.75 * A1

Canceling out A1 from both sides of the equation, we get:

v1 = v2 * 0.75

Now, to find the change in velocity (Δv), we can rearrange the equation:

Δv = v1 - v2

Substituting the previous equation into this one, we have:

Δv = v2 * 0.75 - v2

Factoring out v2, we get:

Δv = v2 * (0.75 - 1)

Simplifying further, we have:

Δv = v2 * (-0.25)

Hence, the change in velocity of the water flowing from the hose when the cross-sectional area decreases by 25% is equal to -0.25 times the final velocity.

To determine the change in velocity of water flowing from a hose when the cross-sectional area is decreased by 25%, we need to apply the principle of conservation of mass.

According to the given information, the volume of the flow per unit of time is held constant. This means that the mass of water flowing through the hose remains the same before and after placing the finger over the end of the hose.

Let's denote the initial cross-sectional area of the hose as A1 and the final cross-sectional area after placing the finger as A2. Since the area is inversely proportional to the velocity, we can write the following relationship:

v1 * A1 = v2 * A2

where v1 and v2 represent the initial and final velocities, respectively.

We are given that the cross-sectional area is decreased by 25%. In other words, the final cross-sectional area (A2) is 75% of the initial cross-sectional area (A1). Therefore, A2 = 0.75 * A1.

Substituting this value into our equation, we get:

v1 * A1 = v2 * 0.75 * A1

Simplifying the equation, we can cancel out A1 from both sides:

v1 = v2 * 0.75

Now, we can see that the initial velocity (v1) is 0.75 times the final velocity (v2). This means that when the cross-sectional area is decreased by 25%, the velocity of the water flowing from the hose will increase by approximately 33.33% (1/0.75 ≈ 1.333).

In other words, the change in velocity is an increase of approximately 33.33% when a person places a finger over the end of the hose to decrease its cross-sectional area by 25%.