Answer:
The question is incomplete: This is the complete Question
A liquid (ρ = 1.65 g/cm3) flows through two horizontal sections of
tubing joined end to end. In the first section the cross-sectional area is
10.0 cm2 , the flow speed is 275 cm/s, and the pressure is 1.20 x 105
Pa. In the second section the cross-sectional area is 2.50 cm2. Calculate
the smaller section’s (1) flow speed and (2) pressure.
(1) Smaller's section flow speed = [tex]1100cm/s[/tex]
(2) Smaller's section pressure = [tex]2.64*10^{5}Pa[/tex]
Step-by-step explanation:
Question 1
Using the continuity equation, we have that:
[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]
[tex]v_{2}=\frac{A_{1}v_{1}}{A_{2}} \\\\v_{2}=\frac{10cm^{2} *275cm/s}{2.5cm^{2}}\\\\v_{2}=4*275=1100cm/s[/tex]
Question 2
Next, we apply Bernoulli's Equation
[tex]P_{2} + \frac{1}{2}rv_{2} ^{2} = P_{1} + \frac{1}{2}rv_{1} ^{2}[/tex]
Let r = ρ(the density)
[tex]P_{2} = P_{1} + \frac{1}{2}rv_{1} ^{2} -\frac{1}{2}rv_{2} ^{2}\\\\P_{2} = P_{1} + \frac{1}{2}r(v_{1} ^{2} -v_{2} ^{2})[/tex]
The pressure is given to us in its fundamental unit, Pa. So, we need to convert the speeds and density to their fundamental units.
[tex]v_{1} = 275cm/s=\frac{275}{100}m/s\\ v_{1}=2.75m/s\\\\v_{2} = 1100cm/s=\frac{1100}{100}m/s\\ v_{2}=11m/s\\\\r=1.65g/cm^{3} = 1.65*1000(kg/m^{3})\\ r=1650kg/m^{3}[/tex]
[tex]P_{2} = 1.2*10^{5} + \frac{1}{2}(1650) (2.75 ^{2} -11 ^{2})\\\\P_{2} = 1.2*10^{5} + 825 ( 7.5625-121) = 1.2*10^{5} + 825 (-113.4375)\\\\P_{2} = 120000 - 93585.9375 = 26414.0625\\\\P_{2} = 2.6414*10^{5}[/tex]
[tex]P_{2}[/tex] ≅ [tex]2.64*10^{5}Pa[/tex]
