Water enters a horizontal pipe with a rectangular cross section at a speed of 1.00 m/s. The width of the pipe remains constant but the height decreases. 27.3 m from the entrance, the height is half of what it is at the entrance. If the water pressure at the entrance is 2921 Pa, what is the pressure 27.3 m downstream?

use the law of continuity to figure the speed downstram

speeddownstream=speedupstream*areaupstream/areadownstream

where area means cross sectional area.

Now, use Bernoulli's law to figure pressure downstream http://www.4physics.com/phy_demo/Bernoulli/Bernoulli.html

because heights are the same (horizontal), it reduces to

P1 + d*v1^2=P2+ d v2^2

solve for P2

I can not find the value of v^2

To solve this problem, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid flowing through a pipe. The equation states that the total pressure at one point in the pipe is equal to the sum of the pressure due to gravity (hydrostatic pressure), the pressure due to the fluid's velocity (dynamic pressure), and the pressure due to any external forces (static pressure).

In this case, we are interested in finding the pressure 27.3 m downstream from the entrance. Let's call the pressure at this point P2.

The first step is to identify the information given in the problem:

- The speed of the water at the entrance of the pipe is 1.00 m/s.
- The width of the pipe remains constant.
- The height of the pipe decreases as we move 27.3 m downstream.
- The pressure at the entrance of the pipe is 2921 Pa.

Now, let's use Bernoulli's equation to find the pressure at the downstream point:

P1 + 0.5 * ρ * V1^2 + ρ * g * h1 = P2 + 0.5 * ρ * V2^2 + ρ * g * h2

In this equation:
- P1 is the pressure at the entrance of the pipe (2921 Pa).
- V1 is the velocity at the entrance of the pipe (1.00 m/s).
- h1 is the height at the entrance of the pipe.
- P2 is the pressure 27.3 m downstream (the value we need to find).
- V2 is the velocity 27.3 m downstream (which will be the same as V1 since the width of the pipe remains constant).
- h2 is the height 27.3 m downstream (given as half of the height at the entrance).

Since the width of the pipe remains constant, we can assume that the velocities at the entrance and downstream are the same (V1 = V2). Therefore, the equation simplifies to:

P1 + ρ * g * h1 = P2 + ρ * g * h2

Substituting the given values into the equation:

2921 Pa + ρ * g * h1 = P2 + ρ * g * (0.5 * h1)

Now, we need to determine the value of ρ * g, which is the product of the density of water (ρ) and the acceleration due to gravity (g). The density of water is approximately 1000 kg/m³, and the acceleration due to gravity is approximately 9.8 m/s². Therefore, ρ * g ≈ 1000 kg/m³ * 9.8 m/s² ≈ 9800 N/m².

Substituting ρ * g into the equation:

2921 Pa + 9800 N/m² * h1 = P2 + 4900 N/m² * h1

Now, we can isolate P2 on one side of the equation:

P2 = 2921 Pa + 4900 N/m² * h1 - 9800 N/m² * h1

Simplifying further:

P2 = 2921 Pa - 4900 N/m² * h1

Finally, we substitute the value of h1, which is the height at the entrance of the pipe, into the equation. However, the problem does not provide information about the height at the entrance, so we cannot calculate the exact pressure at 27.3 m downstream without this information.