A horizontal pipe widens from a radius of 0.140 m to 0.300 m. If the speed of the water is 2.00 m/s in the larger pipe, what is the speed in the smaller pipe (in m/s)?

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Chapter1: Units, Trigonometry. And Vectors
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### Problem Statement
A horizontal pipe **widens** from a radius of **0.140 m** to **0.300 m**. If the speed of the water is **2.00 m/s** in the **larger** pipe, what is the speed in the **smaller** pipe (in m/s)?

#### Hint
- [Blue button labeled "HINT"]

#### Answer Box
- [Text box for the answer, labeled "m/s"]

#### Additional Resources
- [Orange button labeled "Need Help? Read It"]
- [Yellow button labeled "Watch It"]

### Explanation
This problem involves the application of the principle of conservation of mass in fluid dynamics, often referred to as the continuity equation. It states that for an incompressible fluid, the mass flow rate must remain constant from one cross-section of a pipe to another. This principle can be mathematically expressed as:

\[ A_1 \cdot v_1 = A_2 \cdot v_2 \]

Where:
- \( A_1 \) and \( A_2 \) are the cross-sectional areas of the smaller and larger pipes, respectively.
- \( v_1 \) and \( v_2 \) are the velocities of the fluid in the smaller and larger pipes, respectively.

Given:
- Radius of larger pipe, \( r_2 = 0.300 \text{ m} \)
- Radius of smaller pipe, \( r_1 = 0.140 \text{ m} \)
- Speed in larger pipe, \( v_2 = 2.00 \text{ m/s} \)

To find:
- Speed in smaller pipe, \( v_1 \)

### Calculation Steps
1. Calculate the cross-sectional areas of the pipes using the formula for the area of a circle \( A = \pi r^2 \).

   \[ A_1 = \pi (0.140)^2 \]
   \[ A_2 = \pi (0.300)^2 \]

2. Use the continuity equation \( A_1 \cdot v_1 = A_2 \cdot v_2 \) to solve for \( v_1 \).

### Important Note
The hint and additional resource buttons suggest supplemental material might be available for further understanding, which can be useful for students looking for more detailed explanations or visual aids.
Transcribed Image Text:### Problem Statement A horizontal pipe **widens** from a radius of **0.140 m** to **0.300 m**. If the speed of the water is **2.00 m/s** in the **larger** pipe, what is the speed in the **smaller** pipe (in m/s)? #### Hint - [Blue button labeled "HINT"] #### Answer Box - [Text box for the answer, labeled "m/s"] #### Additional Resources - [Orange button labeled "Need Help? Read It"] - [Yellow button labeled "Watch It"] ### Explanation This problem involves the application of the principle of conservation of mass in fluid dynamics, often referred to as the continuity equation. It states that for an incompressible fluid, the mass flow rate must remain constant from one cross-section of a pipe to another. This principle can be mathematically expressed as: \[ A_1 \cdot v_1 = A_2 \cdot v_2 \] Where: - \( A_1 \) and \( A_2 \) are the cross-sectional areas of the smaller and larger pipes, respectively. - \( v_1 \) and \( v_2 \) are the velocities of the fluid in the smaller and larger pipes, respectively. Given: - Radius of larger pipe, \( r_2 = 0.300 \text{ m} \) - Radius of smaller pipe, \( r_1 = 0.140 \text{ m} \) - Speed in larger pipe, \( v_2 = 2.00 \text{ m/s} \) To find: - Speed in smaller pipe, \( v_1 \) ### Calculation Steps 1. Calculate the cross-sectional areas of the pipes using the formula for the area of a circle \( A = \pi r^2 \). \[ A_1 = \pi (0.140)^2 \] \[ A_2 = \pi (0.300)^2 \] 2. Use the continuity equation \( A_1 \cdot v_1 = A_2 \cdot v_2 \) to solve for \( v_1 \). ### Important Note The hint and additional resource buttons suggest supplemental material might be available for further understanding, which can be useful for students looking for more detailed explanations or visual aids.
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