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Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases with depth as the weight of the fluid column above increases. In this problem, hydrostatic pressure is significant because it determines the pressure exerted by the water on the plug inside the pipe. The formula to calculate hydrostatic pressure is given by \( P = \rho g h \), where \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid. The density \( \rho \) for water is typically \( 1000 \text{ kg/m}^3 \), and \( g \) is \( 9.8 \text{ m/s}^2 \). For this exercise, at a depth of \( 6.0 \text{ m} \), the resulting pressure provides the force that must be countered by the frictional force to keep the plug in place. This pressure influences both the force exerted on the plug and, subsequently, the frictional force keeping the plug stationary.

Frictional force is the force that resists the relative motion of solid surfaces. It plays a crucial role in this exercise, as it prevents the plug from being pushed out of the pipe by the hydrostatic pressure of the water.To find the frictional force, we first calculate the total force exerted by the water pressure on the plug. This force is calculated as the water pressure at the pipe's depth multiplied by the cross-sectional area of the pipe. Using the area formula for a circle, \( A = \pi r^2 \), and the pipe's radius \( r = 0.02 \text{ m} \), we multiply this area by the pressure computed using the hydrostatic pressure formula. Thus, the frictional force needed to keep the plug stationary is simply the force exerted by the water pressure, provided the plug does not move. This balance between the exerted force and the frictional force ensures the plug remains in place against the potential force of water trying to push it out.

Volume flow rate is a measure of the volume of fluid that passes through a given surface per unit time. In fluid mechanics, it is often represented by the symbol \( Q \) and is calculated as \( Q = Av \), where \( A \) is the cross-sectional area of the pipe and \( v \) is the velocity of the flowing fluid.For this exercise, once the plug is removed, the volume flow rate determines how much water will exit the pipe over time. To find the velocity \( v \) of the water, we utilize Torricelli's theorem, which is useful for calculating the speed of fluid flowing out due to gravity. The area of the pipe has already been calculated earlier. Multiplying this area by the speed of water gives the rate at which water leaves the pipe.By knowing the flow rate, we can determine the total volume of water that flows out after a specific period, such as three hours in this problem. This involves multiplying the flow rate by the total time in seconds to get the exiting water volume.

Torricelli's theorem provides a way to calculate the speed of a fluid flowing out of an orifice under the influence of gravity. It is an application of Bernoulli's principle in fluid dynamics and is especially useful in cases where the fluid is flowing from a large container through a small opening.According to Torricelli's theorem, the exit speed \( v \) of the fluid is \( v = \sqrt{2gh} \), where \( h \) is the vertical distance between the fluid surface and the hole.In this exercise, the distance \( h \) is the difference in height between the reservoir surface and the pipe opening at depth \( 6.0 \text{ m} \). The application of Torricelli's theorem helps in determining the velocity of water as it leaves the pipe. Using this velocity, you can calculate the volume flow rate, which then leads to finding out how much water exits the pipe over a set duration.This theorem shows how energy principles can explain the fluid flow behaviors encountered in practical scenarios like this problem involving a dam and a pipe.