91Ó°ÊÓ

Understanding the cross-sectional area is essential in fluid dynamics problems. The cross-sectional area determines how much space fluid can occupy within a hose or pipe. To calculate it, use the area formula for a circle. Given the diameter, you can find the radius by dividing the diameter by two.

In the exercise, the hose's inner diameter is 1.9 cm. Therefore, its radius is \(\frac{1.9 \text{ cm}}{2} = 0.95 \text{ cm}\). Using the formula \[ \text{Area} = \pi (r)^2 \], we get \[ \text{Area}_{\text{hose}} = \pi (0.95\text{ cm})^2 = 2.835 \text{ cm}^2 \].
Understanding this concept helps in comparing different sections of fluid paths in more complex setups.

The principle of conservation of mass states that mass cannot be created or destroyed. In fluid dynamics, this principle translates to the flow rate (volume per time) being constant throughout a system.

For a garden hose connected to a sprinkler, the amount of water coming out must equal the amount going in. This means the flow rate (Q) remains the same from the hose to the sprinkler holes. Calculating Q involves the cross-sectional area (A) and the fluid velocity (v) with the equation: \[ Q = A \times v \].

By maintaining this constant flow rate, we ensure no water is lost or gained in transit, making it easier to solve for unknowns like velocity at different points.

Fluid velocity tells us how fast the fluid is moving through a given section. To find it, we use the flow rate and cross-sectional area. Since flow rate is constant (Q), we can rearrange the equation \[ Q = A \times v \] to solve for velocity: \[ v = \frac{Q}{A} \].

For the garden hose, we first calculate the flow rate: \[ Q_{\text{hose}} = 2.835 \text{ cm}^2 \times 91 \text{ cm/s} = 257.985 \text{ cm}^3/s \]. Next, we use this to find the velocity of water leaving the sprinkler holes: \[ v_{\text{holes}} = \frac{257.985 \text{ cm}^3/s}{0.3192 \text{ cm}^2} = 808.32 \text{ cm/s} \].
Hence, the speed of water at the sprinkler holes is 8.0832 m/s.

The area of a circle is a fundamental concept for many physics problems. The formula for the area of a circle is \[ \text{Area} = \pi (r)^2 \], where \( r \) is the radius.

This exercise involves calculating the area for both the hose and the sprinkler holes. Given the hose diameter of 1.9 cm, the radius is 0.95 cm. Plugging this into the formula, we get: \[ \text{Area}_{\text{hose}} = \pi (0.95 \text{ cm})^2 = 2.835 \text{ cm}^2 \].
For each sprinkler hole, with a diameter of 0.13 cm, the radius is 0.065 cm, giving us: \[ \text{Area}_{\text{hole}} = \pi (0.065 \text{ cm})^2 = 0.0133 \text{ cm}^2 \].
By calculating these areas, we can solve for other essential properties like flow rate and velocity.

Flow rate (Q) is a measure of how much fluid passes through a given area per unit time. It is calculated using the equation \[ Q = A \times v \], where \( A \) is the cross-sectional area and \( v \) is the velocity of the fluid.

In the garden hose problem, the flow rate remains consistent from the hose to the sprinkler holes. We calculated \[ Q_{\text{hose}} = 2.835 \text{ cm}^2 \times 91 \text{ cm/s} = 257.985 \text{ cm}^3/s \]. This rate must be the same through the sprinkler holes: \[ Q_{\text{holes}} = 257.985 \text{ cm}^3/s \].
Using this flow rate, we further calculate the velocity of the water as it exits the sprinkler holes:
\[ v_{\text{holes}} = \frac{257.985 \text{ cm}^3/s}{0.3192 \text{ cm}^2} = 808.32 \text{ cm/s} \].
The flow rate equation thus provides a critical foundation for determining other fluid properties.