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**1. Homework Statement**

Air is blown through a horizontal Venturi tube. The diameters of the narrow and wider sections of the tube are 1.0 cm and 3.0 cm respectively, and the height

*h*of the mercury column is measured to be 1.00 mm. What are the air speeds in the wider and the narrower sections of the tube? The density of mercury is 13,600 kg/m3. (Assume the density of air is 1.29 kg/m3.)

Air is blowing through the wider section to the smaller section.

**2. Homework Equations**

p=rho

P1 + (1/2)(p)(v1)^2 + pgy1 = P2 + (1/2)(p)(v2)^2 + pgy2

A1v1=A2v2

P1 = Po + pgh

A(circle)=pi(r)^2

**3. The Attempt at a Solution**

__Formulas:__

y1=y2 so the pgy1 and pgy2 can cancel and v2=A1v1/A2

We can rearrange the equation to give:

P1-P2 = (1/2)(p)(v1)^2((A1/A2)^2-1)

**v1=((P1-P2)/((1/2)(p)((A1/A2)^2-1))^(1/2)**

P1-P2

Po +pgh1 - Po - pgh2

Change in pressure:

**pg(h1-h2)**

**v2=A1v1/A2**

__What we have:__

r1= 3.0cm/2/100=

**0.015m**

r2= 1.0cm/2/100=

**0.005m**

(P1-P2)= change in pressure

change in height = 1.00mm/1000=

**0.001m**

p(mercury)=

**13600kg/m^3**

p(air)=

**1.29kg/m^3**

------------------------------------------------------------------

Change in pressure:

pg(h1-h2)

(13600kg/m^3)(9.8m/s^2)(0.001m)=

**133.28pa**

A1= (0.015m)^2(pi)=

**7.068583471x10^-4 m^2**

A2= (0.005m)^2(pi)=

**7.853981634x10^-5 m^2**

v1=((P1-P2)/((1/2)(p)((A1/A2)^2-1))^(1/2)

(133.28pa/(1/2)(1.29kg/m^3)((7.068583471x10^-4 m^2/7.853981634x10^-5 m^2)^2-1))^(1/2)

v1=

**1.607154546 m/s**

v2=A1v1/A2

(7.068583471x10^-4 m^2)(1.607154546 m/s)/(7.853981634x10^-5 m^2)

v2=

**14.46439092 m/s**

I am not sure if I am even doing this right. :(

Sorry if formatting is hard to follow/ugly, I am still getting used to the new changes, don't know where everything is yet.