Let's go step-by-step through each question:

1. Flow rate in a 7-inch diameter pipe with a flow rate of 3.2 ft³/s:

We need to find the average velocity of water in the pipe. The formula to calculate velocity based on flow rate is:

$V = \frac{Q}{A}$

Where:

• $V$ is the average velocity (ft/s),
• $Q$ is the flow rate (ft³/s),
• $A$ is the cross-sectional area of the pipe (ft²).

For a circular pipe: $A = \pi \left(\frac{D}{2}\right)^2$

• Convert the diameter to feet: $D = \frac{7}{12}$ ft.
• Then, calculate the area $A$: $A = \pi \left(\frac{7}{12} \times \frac{1}{2}\right)^2 = \pi \left(\frac{7}{24}\right)^2$

Once you calculate the area, use the formula for velocity.

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2. Pipe delivering 239 gallons/minute with a velocity of 27 ft/s:

We need to find the diameter of the pipe. Use the same flow rate formula:

$Q = V \times A$

Where:

• $Q$ is the flow rate (ft³/s). First, convert 239 gal/min to ft³/s: $239 \, \text{gal/min} = \frac{239}{7.481} \, \text{ft}^3/\text{min} \approx 31.94 \, \text{ft}^3/\text{min}$ $Q = \frac{31.94}{60} \, \text{ft}^3/\text{s} \approx 0.532 \, \text{ft}^3/\text{s}$ Now, solve for the area $A$: $A = \frac{Q}{V} = \frac{0.532}{27} \approx 0.0197 \, \text{ft}^2$ Using the formula for the area of a circle, solve for the diameter $D$: $A = \pi \left(\frac{D}{2}\right)^2 \Rightarrow D = 2 \times \sqrt{\frac{A}{\pi}}$

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3. Pipe with water flow changing velocities:

The flow rate remains constant through the pipe. The formula for flow rate is:

$Q = A_1 \cdot V_1 = A_2 \cdot V_2$

Where:

• $A_1$ is the area of the larger pipe,
• $V_1 = 0.73 \, \text{m/s}$ is the velocity in the larger pipe,
• $A_2$ is the area of the smaller pipe,
• $V_2 = 1.66 \, \text{m/s}$ is the velocity in the smaller pipe.

First, calculate the flow rate $Q$ using the area of the larger pipe (with diameter $23 \, \text{cm}$): $A_1 = \pi \left(\frac{23}{2 \times 100}\right)^2 \, \text{m}^2$ Then solve for $Q$: $Q = A_1 \cdot V_1$ Now, use the same flow rate $Q$ to find the area of the smaller pipe: $Q = A_2 \cdot V_2$ Finally, solve for the diameter of the smaller pipe.

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4. Flow rate from two split pipes:

The total flow rate before the split is 97 gallons/s. Convert it to cubic feet per second:

$97 \, \text{gal/s} = \frac{97}{7.481} \, \text{ft}^3/\text{s} \approx 12.97 \, \text{ft}^3/\text{s}$

Now, for each pipe, use the formula for flow rate:

$Q_1 = A_1 \cdot V_1 \quad \text{and} \quad Q_2 = A_2 \cdot V_2$

Where:

• $A_1$ is the area of the 3.2-inch diameter pipe,
• $A_2$ is the area of the 1.8-inch diameter pipe,
• The velocities in each pipe depend on the area.

Use the continuity equation: $Q = Q_1 + Q_2$ Solve for $Q_1$ and $Q_2$.

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5. Velocity of copper sphere dropped from 273 ft:

This is a free-fall problem where we neglect air resistance. The formula for velocity after falling a distance $h$ is:

[ v = \sqrt{2gh}