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kari74 [83]
4 years ago
9

A Venturi tube may be used as the inlet to an automobile carburetor. If an inlet pipe with a diameter of 2.0 cm diameter narrows

to diameter of 1.0 cm, determine the pressure drop in the constricted section for an initial airflow of 3.0 m/s in the 2-cm section? (Assume air density is 1.25 kg/m
Physics
2 answers:
Goryan [66]4 years ago
7 0

Answer:

the pressure drop is equal to 80.99 Pa

Explanation:

we have the following data:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

ΔP = ?

For the calculation of the pressure drop we will use Bernoulli's principle for the Venturi Tube:

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

where A = area

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

for the calculation of the areas we will use the following formula:

A1 = (pi*d1^2)/4 = (pi*(0.02 m)^2)/4 = 0.00031 m^2

A2 = (pi*(0.01 m)^2)/4 = 0.000079 m^2

ΔP = ((3 m/s)^2*1.25 kg/m^3)/2)*((0.00031 m^2)^2/(0.000079 m^2)^2)-1) = 80.99 N/m^2 = Pa

valkas [14]4 years ago
4 0

Answer:

I think it's  81

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3 years ago
How do we prove the earth is round? What evidence do we have and reasoning?
Jlenok [28]

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7 0
4 years ago
Read 2 more answers
A person has legs of length L = 1.10 m. (a) If the maximum angle between the legs during walking is ϕ = 50o , what is the person
vampirchik [111]

Answer:

a). Maximum Length L=0.929m

b). T=0.83 Hz or 1.2s

c). Longer, the effortless waling T=2.1 Hz or t=0.475s

d). t=1.2s V=0.774 \frac{m}{s}

t=0.475s V=1.95 \frac{m}{s}

Explanation:

Length legs=L=1.1m

angle=50

the step that give the person forms a triangle whose two sides are known and the angle that forms between them, then using trigonometry as the image

Divide the original triangle in two and form a right triangle so the angle is 25 and the L is hypotenuse and the opposite is the step length

a).

sin(\alpha) =\frac{op}{h}

op=h*sin(\frac{\alpha }{2})\\ op=1.1m*sin(\frac{50}{2})\\op=0.464m

Length of the step

L=0.464m*2

L=0.928m

b).

period=T

T=\frac{1}{time}=\frac{1}{t}\\ T=\frac{1}{1.2s}\\T=0.83 s^{-1}\\ T=0.83Hz

c).

T1=2\pi *\sqrt{\frac{L}{g}} \\T1=2\pi *\sqrt{\frac{1.1m}{9.8\frac{m}{s^{2}}}}\\ T1=2.1 Hz

The period is the inverse of the time of the motion so, the T1 is faster that the T because

t=\frac{1}{T1}=t= \frac{1}{2.1}=0.47s

d).

The speed is the relation between the distance with time so:

Vt=\frac{0.928m}{1.2s} \\Vt=0.773 \frac{m}{s} \\Vt=\frac{0.928m}{0.475s} \\Vt=1.953 \frac{m}{s}

3 0
3 years ago
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hram777 [196]

Answer:0.0000107s

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Lapatulllka [165]

Answer:

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