How Does Water Flow Dynamics Work in a Tank System?

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SUMMARY

The discussion focuses on calculating the discharge rate of water from an open tank system using Bernoulli's equation. Given the elevations of points 1, 2, and 3, along with the cross-sectional areas at points 2 (4.80×10−2 m2) and 3 (1.60×10−2 m2), the challenge is to determine the velocity at point 3 and subsequently the volume of water that flows across the exit in one second. The relationship A1(v)1=A2(v)2 is crucial for solving the problem, emphasizing the need for understanding fluid dynamics principles.

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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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