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.
Delta V = (A)(v)( Delta t)
The Attempt at a Solution
I can't figure this out.
To know the velocity of at point 3, dont 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??