**In the hose.****In the nozzle.**

This problem aims to familiarize us with the **relationship** between **flow rate** and** speed** of a liquid from a specific **cross-sectional area**. The concept required to solve this problem is as mentioned, but it would be a plus if you are familiar with **Bernoulli’s principle.**

Now the **flow rate** $Q$ is described to be the **volume** $V$ of liquid passing through a **cross-sectional area** during a given specific **time** $t$, its equation is given by:

\[ Q = \dfrac{V}{t} \]

If the liquid is passing through a **cylindrical shape,** then we can represent $V$ as the **product** of **area** and unit **distance** i.e. $Ad$, $= \dfrac{Ad}{t}$. Where,

$\vec{v} = \dfrac{d}{t}$, so the **flow rate** becomes $Q = \dfrac{Ad}{t} = A \vec{v}$.

## Expert Answer

**Part a:**

For better **understanding,** we are going to use **subscript** $1$ for the **hose** and $2$ for the **nozzle** when using the relationship between **flow rate** and **velocity.**

First, we will solve for $v_1$, and keeping in view that the **cross-sectional area** of a **cylinder** is $A = \pi r^2$, gives us:

\[ \vec{v_1} = \dfrac{Q}{A_1} \]

**Substiuting** $A = \pi r^2$:

\[ \vec{v_1} = \dfrac{Q}{\pi r_1^2} \]

Given the following **information:**

The** flow rate** $Q = 0.500 L/s$ and,

The **radius** of the **hose** $r_1 = 0.750 cm$.

**Plugging** in the values after making the **appropriate unit conversions** gives us:

\[\vec{v_1} = \dfrac{(0.500 L/s)(10^{-3} m^3/L)}{\pi (7.50\times 10^{-3} m)^2} \]

\[\vec{v_1} = 8.96 m/s\]

Thus, the **speed of water** through the **hose** is $8.96 m/s$.

**Part b:**

The **radius** of the **nozzle** $r_2 = 0.250 cm$.

For this part, we are going to use the **equation** of **continuity** to calculate $v_2$. We could have used the same **approach,** but this will give you a **different insight.** Using the equation:

\[A_1\vec{v_1} = A_2\vec{v_2}\]

Solving for $v_2$ and **substituting** $A = \pi r^2$ for the **cross-sectional area** gives us:

\[\vec{v_2} =\dfrac{A_1}{A_2}\vec{v_1}\]

\[\vec{v_2} =\dfrac{ \pi r_1^2}{ \pi r_2^2}\vec{v_1}\]

\[\vec{v_2} =\dfrac{r_1^2}{r_2^2}\vec{v_1}\]

**Plugging** in the given **values** in the above equation:

\[\vec{v_2} =\dfrac{(0.750 cm)^2}{(0.250 cm)^2} 8.96 m/s\]

\[\vec{v_2} =80.64 m/s\]

## Numerical Result

A **speed** of about $8.96 m/s$ is required for the **water** to emerge from the **nozzle-less** hose. When the **nozzle** is attached, it offers a **much faster** stream of water by **tightening** the flow to a narrow tube.

## Example

The **flow rate of blood** is $5.0 L/min$. Calculate the average speed of the blood in aorta when it has a **radius** of $10 mm$. The **speed** of blood is about $0.33 mm/s$. The **average diameter** of a capillary is $8.0 \mu m$, find the **number** of **capillaries** in the circulatory system.

**Part a:**

The **flow rate** is given as $Q = A\vec{v}$, **rearranging** the expression for $\vec{v}$:

\[\vec{v} =\dfrac{Q}{\pi r^2}\]

**Substituting** the values yields:

\[\vec{v} =\dfrac{5.0\times 10^{-3} m^3/s }{\pi (0.010 m)^2}\]

\[\vec{v} =0.27 m/s\]

**Part b:**

Using the **equation:**

\[n_1A_1 \vec{v_1} = n_2A_2 \vec{v_2}\]

**Solving** for $n_2$ gives us:

\[n_2 = \dfrac{(1)(\pi)(10\times 10^{-3}m)^2(0.27 m/s)}{(\pi)(4.0\times 10^{-6} m)(0.33\times 10^{-3} m/s)}\]

\[n_2 = 5.0\times 10^{9}\space capillaries\]