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The conservation of mass is a fundamental principle in fluid dynamics. It states that mass cannot be created or destroyed in a closed system, which is very useful in understanding how fluids behave as they move through different areas. For a flowing fluid, this principle tells us that the rate of mass entering a system equals the rate of mass exiting. In simple terms, if you have water flowing through a hose into a sprinkler, the amount entering the sprinkler must equal the amount leaving it. This idea is crucial when calculating fluid speed in different parts of a system, like our hose and sprinkler example. By knowing the conservation of mass, we can apply specific equations to find unknown variables like fluid speed or area at different points.

The equation of continuity is a mathematical representation of the conservation of mass in fluid systems. It is expressed as:\[ A_1v_1 = A_2v_2 \]Here, \(A_1\) and \(v_1\) represent the cross-sectional area and velocity of the fluid at one point (like in a garden hose), while \(A_2\) and \(v_2\) represent these values at another point (such as the sprinkler holes). This equation tells us that the product of the area and velocity at one point in a stream of fluid must equal the product of these quantities at another point. This relationship is particularly useful for solving problems where fluid moves from one cross-sectional area to another. Since the mass flow is consistent, any change in area results in an inverse change in fluid speed. This concept helps us determine the speed at which fluid exits the sprinkler holes.

Calculating the cross-sectional area is a critical step in fluid dynamics problems. For circular areas, the area is given by the formula:\[ A = \pi r^2 \]Where \(r\) is the radius of the circle. In our problem, we first calculate the area of the hose, which has a diameter of 1.9 cm. This means the radius is half of that, or 0.95 cm. To convert to meters, we divide by 100, which makes it 0.0095 m.

• For the hose area: \( A_1 = \pi (0.0095 \, \text{m})^2 \).
• Similarly, for the sprinkler holes, each with a diameter of 0.15 cm, the radius is 0.075 cm or 0.00075 m in meters.
• So, the area for one hole is: \( A = \pi (0.00075 \, \text{m})^2 \).

You multiply this area by 20 for all holes to get the total area, \( A_2 \). These calculations allow us to use the equation of continuity to solve for the fluid speed.

The speed of fluid flow is an essential factor in understanding fluid dynamics and using it to solve problems like the one given. Once the areas are calculated, we can rearrange the equation of continuity to solve for the unknown velocity, \(v_2\), in the sprinkler holes. Substituting the known areas and initial velocity from the garden hose into:\[ \pi (0.0095)^2 \times 0.91 = \pi (0.00075)^2 \times 20 \times v_2 \]To isolate \(v_2\), you divide both sides of the equation by the sprinkler hole area, after multiplication by 20 due to all the holes. This allows you to find:\[ v_2 = \left( \frac{\pi (0.0095)^2 \times 0.91}{20 \times \pi (0.00075)^2} \right) \]This calculation shows how the narrower exit through the sprinkler holes will result in water leaving at a higher speed than it enters through the hose, illustrating the inverse relationship between area and velocity in fluids.