Answers: 3
= Cross sectional Area at each section
= Flow Velocity at each section 
![7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)](/tpl/images/0432/0232/82848.png)
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7. water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tub...
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