The problem involves the continuity equation in fluid dynamics, which states:

$A_1 v_1 = A_2 v_2$

Here, $A_1$ and $A_2$ are the cross-sectional areas of the hose and the nozzle, respectively, and $v_1$ and $v_2$ are the corresponding speeds of the water. Additionally, the volumetric flow rate $Q$ is related to the speed and area by:

$Q = A \cdot v$

Given:

1. Radius of the hose: $r_1 = 0.900 \, \text{cm}$
2. Diameter of the nozzle: $d_2 = 0.500 \, \text{cm}$ ($r_2 = 0.250 \, \text{cm}$)
3. Volumetric flow rate: $Q = 0.500 \, \text{L/s} = 500 \, \text{cm}^3/\text{s}$

We need to find:

• (a) Speed of the water in the hose ($v_1$)
• (b) Speed of the water in the nozzle ($v_2$).

Step 1: Calculate the cross-sectional areas

The cross-sectional area is given by:

$A = \pi r^2$

• For the hose: $A_1 = \pi (r_1)^2 = \pi (0.900)^2 = 2.544 \, \text{cm}^2$

• For the nozzle: $A_2 = \pi (r_2)^2 = \pi (0.250)^2 = 0.196 \, \text{cm}^2$

Step 2: Calculate the speeds

Using $Q = A \cdot v$:

• For the hose ($v_1$): $v_1 = \frac{Q}{A_1} = \frac{500}{2.544} = 196.45 \, \text{cm/s}$

• For the nozzle ($v_2$): $v_2 = \frac{Q}{A_2} = \frac{500}{0.196} = 2551.02 \, \text{cm/s}$

Final Answers:

• (a) The speed of water in the hose: $v_1 = 196.45 \, \text{cm/s}$
• (b) The speed of water in the nozzle: $v_2 = 2551.02 \, \text{cm/s}$

Would you like a detailed explanation of how the continuity equation works or assistance with related problems? Here are some related questions:

1. How does the continuity equation relate to incompressible fluids?
2. What happens to the pressure when water flows through a nozzle (Bernoulli's principle)?
3. Can the speed of water in a hose ever exceed the speed in the nozzle?
4. How would the flow rate change if the hose radius were increased?
5. What units are typically used in fluid dynamics problems?

Tip: Always ensure units are consistent (e.g., cm$^3$/s, cm/s, cm$^2$) to avoid errors in calculations!