Bernoulli's equation:

$\overline{){{\mathbf{P}}}_{{\mathbf{1}}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\rho}}{{{\mathbf{v}}}_{{\mathbf{1}}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{h}}}_{{\mathbf{1}}}{\mathbf{\rho}}{\mathbf{g}}{\mathbf{=}}{{\mathbf{P}}}_{{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\rho}}{{\mathbf{v}}_{\mathbf{2}}}^{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{h}}}_{{\mathbf{2}}}{\mathbf{\rho}}{\mathbf{g}}}$

The volume of the cylindrical pipe, **V _{c} = LA**, where L is the length and A is the cross-sectional area.

The volume of water flowing through a cylindrical [ipe every second, V_{w}/s = LA/t

But we know that L/t = v, where v is velocity.

Av_{A} = 2Av_{B}

A = πr^{2}

The 3.0-cm-diameter water line in (Figure 1) splits into two 1.0-cm-diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 m/s and the gauge pressure is 50 kPa .

What is the gauge pressure at point B?

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