Question
Water flows with a speed of 2.5 m/s through a section of hose with a cross-sectional area of 0.0084 m^2. Further along the hose the cross-sectional area changes, and the water speed is reduced to 1.1 m/s. Calculate the new cross-sectional area.
Answers
since the volume flow rate of the water is constant (why?) and m^2 * m/s = m^3/s
A*1.1 = .0084 * 2.5
A*1.1 = .0084 * 2.5
unit?
how to make it
i have exam now
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