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krok68 [10]
2 years ago
4

Consider a 0.15-mm-diameter air bubble in a liquid. Determine the pressure difference between the inside and outside of the air

bubble if the surface tension at the air-liquid interface is a. 0.1 N/m. b. 0.12 N/m.
Engineering
1 answer:
Rzqust [24]2 years ago
8 0

Answer:

a)\Delta P= 2666.66 Pa

b)\Delta P= 3200 Pa

Explanation:

Given that

Diameter ,d= 0.15 mm

We know that pressure difference is given as

\Delta P=\dfrac{4\sigma }{d}

Now by putting the values

When surface tension 0.1 N/m  :

The surface tension ,\sigma=0.1\ N/m

\Delta P=\dfrac{4\times 0.1 }{0.15}\times 10^3\ Pa

\Delta P= 2666.66 Pa

When surface tension 0.12 N/m  :

The surface tension ,\sigma=0.12\ N/m

\Delta P=\dfrac{4\times 0.12 }{0.15}\times 10^3\ Pa

\Delta P= 3200 Pa

a)\Delta P= 2666.66 Pa

b)\Delta P= 3200 Pa

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telo118 [61]

Answer:

165 mm

Explanation:

The mass on the piston will apply a pressure on the oil. This is:

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f = m * a

Where a in the acceleration of gravity

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Then:

p = m * a / (π/4 * D1^2)

The height the oil will raise is the heignt of a colum that would create that same pressure at its base:

p = f / A

The weight of the column is:

f = m * a

The mass of the column is its volume multiplied by its specific gravity

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The volume is the base are by the height

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We cancel the areas:

p = h * S * a

Now we equate the pressures form the piston and the pil column:

m * a / (π/4 * D1^2) = h * S * a

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m / (π/4 * D1^2) = h * S

Rearranging:

h = m / (π/4 * D1^2 * S)

Now, h is the heigth above the interface between the piston and the oil, this is at h1 = 42 mm. The total height is

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h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm

7 0
2 years ago
A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. If the
Bad White [126]

Answer:

=>> 167.3 kpa.

=>> 60° from horizontal face.

Explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this kind of question;

=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "

=>>"If the deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa."

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magnitude of the stresses on the failure plane = 167.3 kpa.

The orientation of this plane with respect to the major principle stress plane => x = 60 cos 60° = 30kpa.

y = 60 sin 60° = 30√3 = sheer stress.

the orientation of this plane with respect to the major principle stress plane.

Theta = 45 + 15 = 60°.

5 0
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2. A ¼ in. diameter rod must be machined on a lathe to a smaller diameter for use as a specimen in a tension test. The rod mater
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Answer:

we use

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