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In the world of fluid dynamics, understanding how to calculate water speed is essential. For water flowing in a garden hose, the speed is determined by its flow rate and the cross-sectional area through which the water moves. In this exercise, with a known flow rate of \(5.0 \times 10^{-4} \text{ m}^3/\text{s}\), we use the formula \(Q = A \cdot v\) to calculate speed. Here, \(Q\) is the flow rate, \(A\) is the cross-sectional area, and \(v\) is the velocity of the water.

By rearranging the formula to solve for \(v\): \[ v = \frac{Q}{A} \]once the cross-sectional area is known, the speed becomes straightforward to find. The solution shows that the initial water speed in the hose is calculated as approximately \(1.02 \text{ m/s}\). This knowledge is fundamental in understanding how adjustments in conditions like hose diameter affect the speed of fluid flow.

Pressure drop in a fluid system like a garden hose refers to the loss of pressure as the water flows through the hose. When water moves through a hose, it experiences resistance, which leads to a decrease in pressure along the length of the hose. This phenomenon is explained using principles like Bernoulli’s equation and more specifically, for viscous fluids, through the Hagen-Poiseuille equation: \[ \Delta P = \frac{8\eta LQ}{\pi r^4} \]Here, \(\Delta P\) is the pressure drop, \(\eta\) is the dynamic viscosity of the fluid, \(L\) is the length of the hose, \(Q\) is the volumetric flow rate, and \(r\) is the radius of the hose.

Since no specific viscosity value is given in the exercise, pressure drop calculations are usually referenced with typical values for water, which may in practice require precise experimental methods for exact results. Nevertheless, understanding the basic equation helps one appreciate how factors like the hose's diameter and fluid viscosity affect pressure variation.

The cross-sectional area of a hose is critical in determining how fluid flows through it. For a cylindrical hose with a given diameter, the cross-sectional area \(A\) can be calculated using the formula for the area of a circle: \[ A = \pi r^2 \]where \(r\) is the radius, which is half the diameter.

In the exercise, with an original diameter of \(2.5 \text{ cm}\), the radius becomes \(0.0125 \text{ m}\), leading to a calculated area of \(4.91 \times 10^{-4} \text{ m}^2\). The area influences both the flow rate and speed of water. Notably, when the cross-sectional area is halved, the effect on speed is doubled. This relationship illustrates the inverse connection between area and velocity, as seen in the formula \(v = \frac{Q}{A}\), where reducing \(A\) while keeping \(Q\) constant results in increased speed.