How Does Water Flow Dynamics Work in a Tank System?

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Water flows from an open tank with varying elevations, where point 1 is at 10.0 m and points 2 and 3 are at 2.00 m. The cross-sectional areas at points 2 and 3 are provided, but the velocity at these points is not given, leading to confusion about calculating discharge rates. The discussion highlights the need to apply Bernoulli's equation to determine the velocity and subsequently the volume of water flowing out of the pipe. The problem emphasizes the relationship between cross-sectional area and velocity, as described by the continuity equation. Understanding these dynamics is crucial for calculating the discharge rate of water from the tank system.
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Homework Statement



Link to picture: http://session.masteringphysics.com/problemAsset/1011222/12/yf_Figure_14_41.jpg

Water flows steadily from an open tank as shown in the figure. The elevation of point 1 is 10.0 m, and the elevation of points 2 and 3 is 2.00 m. The cross-sectional area at point 2 is 4.80×10−2 m^2; at point 3, where the water is discharged, it is 1.60×10−2 m^2. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.


Homework Equations



Delta V = (A)(v)( Delta t)

A1(v)1=A2(v)2

The Attempt at a Solution



I can't figure this out.

To know the velocity of at point 3, don't you have to know the area of the main tank, to get the velocity at point 2, and then point 3?

The Area of point 2 and 3 is given, but no velocity, so what am I missing??
 
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So it ask:

Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 {\rm s}. In other words, find the discharge rate \Delta V/\Delta t.
 
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