To find the speed of the water in the pipe, we can use the relationship between flow rate $Q$, cross-sectional area $A$, and speed $v$:

$Q = A \cdot v$

where:

• $Q$ is the flow rate,
• $A$ is the cross-sectional area of the pipe,
• $v$ is the speed of the water in the pipe.

Step 1: Convert the inner diameter to meters

The inner diameter of the pipe is given as 1.5 cm, which we convert to meters:

$d_{\text{pipe}} = 1.5 \, \text{cm} = 0.015 \, \text{m}$

Step 2: Calculate the cross-sectional area of the pipe

The cross-sectional area of a circular pipe is:

$A_{\text{pipe}} = \pi \left(\frac{d_{\text{pipe}}}{2}\right)^2$

Substitute $d_{\text{pipe}} = 0.015 \, \text{m}$:

$A_{\text{pipe}} = \pi \left(\frac{0.015}{2}\right)^2$ $A_{\text{pipe}} = \pi \cdot (0.0075)^2$ $A_{\text{pipe}} \approx 1.767 \times 10^{-4} \, \text{m}^2$

Step 3: Solve for the speed of the water in the pipe

Now, rearrange the formula $Q = A \cdot v$ to solve for $v$:

$v_{\text{pipe}} = \frac{Q}{A_{\text{pipe}}}$

Substitute $Q = 1.5 \times 10^{-4} \, \text{m}^3/\text{s}$ and $A_{\text{pipe}} \approx 1.767 \times 10^{-4} \, \text{m}^2$:

$v_{\text{pipe}} = \frac{1.5 \times 10^{-4}}{1.767 \times 10^{-4}}$ $v_{\text{pipe}} \approx 0.849 \, \text{m/s}$

Final Answer:

The speed of the water in the pipe is approximately 0.849 m/s.

Would you like further details on any step?

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Related Questions:

1. What is the speed of the water in the nozzle?
2. How does the speed change if the flow rate doubles?
3. What would be the speed if the inner diameter of the pipe was reduced to 1 cm?
4. How is the speed in the pipe related to the nozzle speed by continuity?
5. How does fluid velocity change if the diameter of the nozzle is halved?

Tip:

Remember that fluid flow speed is inversely related to the cross-sectional area in a steady-flow scenario, a principle known as the continuity equation in fluid dynamics.