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Let's break down the problem step by step:

### Given:
- Flow rate \( Q = 10 \, \text{m}^3/\text{s} \)
- Initial area of the tube at section 1: \( A_1 = 2 \, \text{m}^2 \)
- Final area of the tube at section 2: \( A_2 = 1 \, \text{m}^2 \)
- Water density: \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \)
- Liquid with density: \( \rho_{\text{liquid}} = 1.2 \, \text{g/cm}^3 = 1200 \, \text{kg/m}^3 \)
- Pressure difference at section 3 and difference in height in the U-tube with mercury.

Now, let's address each subpart of the problem:

### (a) Velocity of water at the exit of the pipe

The flow rate \( Q \) is related to velocity \( v \) and cross-sectional area \( A \) by the equation:
\[
Q = A \cdot v
\]
For section 2, the exit of the pipe:
\[
v_2 = \frac{Q}{A_2} = \frac{10 \, \text{m}^3/\text{s}}{1 \, \text{m}^2} = 10 \, \text{m/s}
\]

Thus, the velocity of the water at the exit is:
\[
v_2 = 10 \, \text{m/s}
\]

### (b) The amount of liquid flowing out per second

This is directly given by the flow rate \( Q \):
\[
\text{Amount of water per second} = 10 \, \text{m}^3/\text{s}
\]

### (c) Velocity of water at section 1

Using the continuity equation, which states that the flow rate remains constant across different sections of the pipe:
\[
A_1 \cdot v_1 = A_2 \cdot v_2
\]
\[
v_1 = \frac{A_2 \cdot v_2}{A_1} = \frac{1 \, \text{m}^2 \cdot 10 \, \text{m/s}}{2 \, \text{m}^2} = 5 \, \text{m/s}
\]

Thus, the velocity at section 1 is:
\[
v_1 = 5 \, \text{m/s}
\]

### (d) Pressure at section 3

We can apply **Bernoulli's equation** between sections 1 and 3 to find the pressure difference. Bernoulli's equation is:
\[
P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_3 + \frac{1}{2} \rho v_3^2 + \rho g h_3
\]
Assuming section 3 is at the same height as section 1 and assuming negligible losses, the pressure at section 3 would depend on the velocity, but for now, we would need more information to compute it fully (such as exact data for \( P_1 \), \( P_3 \), and heights).

### (e) Height difference in the mercury columns

For this, we apply **Pascal's law** which relates the pressure difference to the height difference in the mercury:
\[
\Delta P = \rho_{\text{mercury}} g \Delta h
\]
Where \( \rho_{\text{mercury}} = 13600 \, \text{kg/m}^3 \) is the density of mercury. Once we know the pressure difference, we can find the height difference.

---

Would you like a detailed calculation for part (d) or (e)? Let me know if you need more details!

Here are 5 related questions:
1. How would Bernoulli's equation change if there was a height difference between sections?
2. What is the significance of the continuity equation in fluid dynamics?
3. How does fluid density affect pressure changes in a system?
4. What is Pascal's law, and how is it applied to fluids in U-tubes?
5. How does the flow rate affect the velocity in different cross-sectional areas of a pipe?

**Tip:** Always ensure units are consistent, especially when dealing with densities in different units (e.g., g/cm³ vs kg/m³).

Por un tubo ubicado en forma horizontal circula agua con un caudal de 10 m^3/s. Inicialmente la superficie del tubo es 2m^2 y la sección va disminuyendo hasta alcanzar un área de 1m^2. La densidad del agua es 1000 g/cm^3. Un líquido de densidad 1.2 g/cm^3 fluye como muestra la figura.

Solve a fluid dynamics problem involving a water flow rate of 10 m³/s through a tube. Calculate the velocity at the exit, the amount of liquid per second, and pressure at various sections using Bernoulli's and Pascal's principles. Understand the effects of changing cross-sectional areas in pipes.

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