# Physics

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A liquid (density = 1.65 g/cm^3) flows through two horizontal sections of tubing. In the first section the cross-sectional area is 10.0 cm^2, the flow speed is 275 cm/s, and the pressure is 1.20 x 10^5 Pa. This tubing then rises 50 cm to the second horizontal section which has a cross-sectional area of 2.50 cm^2. Calculate the second section's flow speed and pressure.

You will need to convert all units to "standard metric."

• Physics - ,

I know the anwers are 11.0 m/s and 1.83 x 10^4 Pa

I just need to know how to work it out and how to start it

• Physics - ,

Use the law of continuity of incompressible fluids to get the velocity in the constricted section.
That law says that the product of cross sectional area and velocity must be constant in steady state. The law tells you that the flow velocity in the 2.50 cm^2 section is ten times higher than it is in the 10 cm^2 section.

Knowing the velocity in both places, use Bernoulli's equation to get the pressure change.

p + (1/2)*rho*V^2 + rho*g*H = constant.

H is the height of the tubing and rno is the density. With the above equation, the CHANGE in pressure can be related to the change in V and H.

• Physics - ,

Ok thx. Sounds good. I already found the velocity so just need to get the pressure. Will do the math. And I need help on the other question you helped me on yesterday =P