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A river narrows at a rapids from a width of \(12 \mathrm{m}\) to a width of only \(5.8 \mathrm{m}\). The depth of the river before the rapids is \(2.7 \mathrm{m} ;\) the depth in the rapids is \(0.85 \mathrm{m}\). Find the speed of water flowing in the rapids, given that its speed before the rapids is \(1.2 \mathrm{m} / \mathrm{s}\). Assume the river has a rectangular cross section.

Short Answer

Expert verified
The speed of water in the rapids is approximately 7.89 m/s.

Step by step solution

01

Determine Cross-Sectional Areas

First, we calculate the cross-sectional area of the river both before and in the rapids. **Before the rapids:** The area is given by the width times the depth, which is \(12 \, \text{m} \times 2.7 \, \text{m} = 32.4 \, \text{m}^2\). **In the rapids:** The area is \(5.8 \, \text{m} \times 0.85 \, \text{m} = 4.93 \, \text{m}^2\).
02

Apply the Continuity Equation

The principle of continuity for fluid dynamics states that the volume flow rate must be constant across any two cross-sections of a river, assuming incompressibility of water. This can be expressed as \(A_1v_1 = A_2v_2\), where \(A\) is the cross-sectional area and \(v\) is the velocity at any section.
03

Solve for the Velocity in the Rapids

Using the values from Step 1: **Before the rapids:** \(A_1 = 32.4 \, \text{m}^2\) and \(v_1 = 1.2 \, \text{m/s}\). **In the rapids:** \(A_2 = 4.93 \, \text{m}^2\). Substitute into the continuity equation: \(32.4 \, \text{m}^2 \times 1.2 \, \text{m/s} = 4.93 \, \text{m}^2 \times v_2\), \(v_2 = \frac{32.4 \, \text{m}^2 \times 1.2 \, \text{m/s}}{4.93 \, \text{m}^2}\).
04

Calculate the Final Velocity

Compute the value of \(v_2\) from Step 3: \(v_2 = \frac{38.88}{4.93} \approx 7.89 \, \text{m/s}\). This is the speed of water flowing in the rapids.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity Equation in Fluid Dynamics
When you think of fluid dynamics, one essential principle is the continuity equation. It governs the behavior of fluid flow in various contexts, such as water flowing through a river. In simple terms, the continuity equation states that for an incompressible fluid, like water, the volume flow rate must remain constant between two cross-sections.
For a river that changes width, this means the amount of water entering one section per second must equal the amount leaving the next section per second. This can be expressed mathematically as:
  • \[ A_1v_1 = A_2v_2 \]
Where:
  • \( A_1 \) and \( A_2 \) are the cross-sectional areas of the river before and after a change, like narrowing.
  • \( v_1 \) and \( v_2 \) are the corresponding velocities of the water in these sections.
This equation ensures that despite changes in the shape of the riverbed, the flow remains consistent. This concept is crucial in solving problems related to water flow in rivers and pipes.
Cross-Sectional Area Calculation
Understanding the cross-sectional area is vital for applying the continuity equation in fluid dynamics. The cross-sectional area of a river is essentially the surface area of the path through which the water flows. It determines how much water can pass through a segment of the river at any given time.
Suppose a river has a rectangular cross-section. In that case, its area is calculated by multiplying its width by its depth. Mathematically, it looks like this:
  • \[ A = ext{width} imes ext{depth} \]
For example, if a river is 12 meters wide and 2.7 meters deep, the cross-sectional area will be:
  • \[ A = 12 imes 2.7 = 32.4 ext{ m}^2 \]
This simple formula helps us understand how changes in the river size affect the flow. Narrower or shallower sections result in smaller cross-sectional areas. Therefore, the velocity of water flow needs to increase to maintain the same flow rate when the river narrows.
Velocity Calculation in Fluid Flow
Calculating the velocity of fluid, such as water in a river, is pivotal for understanding fluid dynamics.
Once you know the cross-sectional areas at different river segments and the initial speed, you can determine the speed at another point using the continuity equation.
  • For instance, if a section of the river narrows, the water must speed up because the same amount of water needs to pass through a smaller area.
Let's see how it's calculated:
  • Start with the continuity equation: \( A_1v_1 = A_2v_2 \)
  • Rearrange to solve for the unknown velocity: \( v_2 = \frac{A_1v_1}{A_2} \)
Using the river example:
  • Before the rapids: \( A_1 = 32.4 ext{ m}^2 \), \( v_1 = 1.2 ext{ m/s} \)
  • In the rapids: \( A_2 = 4.93 ext{ m}^2 \)
  • Calculate: \( v_2 = \frac{32.4 imes 1.2}{4.93} \approx 7.89 ext{ m/s} \)
This final velocity, 7.89 m/s, tells us how fast water flows through the narrower section. It highlights how fluid accelerates in constricted pathways.

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