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9966 [12]
3 years ago
10

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:
777dan777 [17]3 years ago
4 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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Darina [25.2K]

Answer:

0.1047N

Explanation:

To solve this problem we must remember the conversion factors, remembering that 1 revolution equals 2π radians and 1min equals 60s

N\frac{rev}{min} \frac{2\pi }{1rev} \frac{1min}{60} =N\frac{2\pi }{60} =0.1047N

in conclusion, to know how many rad / s an element rotates which is expressed in Rev / min we must only multiply by 0.1047

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3 years ago
Two well-known NP-complete problems are 3-SAT and TSP, the traveling salesman problem. The 2-SAT problem is a SAT variant in whi
Hitman42 [59]

3-SAT ≤p TSP

If P ¹ NP, then no NP-complete problem can be solved in polynomial time.

both the statements are true.

<u>Explanation:</u>

  • 3-SAT ≤p TSP due to any  complete problem of NP to other problem by exits of reductions.
  • If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
  • If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
7 0
3 years ago
The host at the end of the video claims that ___________ is crucial to his success as a driver. A. Reaction time B. A safe space
Snezhnost [94]

Answer:

answer is C. his seat belt

5 0
2 years ago
A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i
Shkiper50 [21]

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

Steam temperature T_{1} = 300 F  = 422.04 K

Duct temperature T_{2} = 70 F = 294.26 K

Emmisivity of surface 1 = 0.79

Emmisivity of surface 2 = 0.276

Net emmisivity of both surfaces ∈ = 0.25

Stefan volazman constant \sigma = 5.67 × 10^{-8} \frac{W}{m^{2} K^{4}  }

Heat transfer  per foot length is given by

Q = ∈ \sigma A ( T_{1}^{4} - T_{2} ^{4} ) ------ (1)

Put all the values in equation (1) , we get

Q = 0.25 × 5.67 × 10^{-8} × 3.14 × 0.0508 × 1 × ( 422.04^{4} - 294.26^{4} )

Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

4 0
3 years ago
Describe ICP/OES in detail.
alex41 [277]

Answer:

ICP -OES stand for inductively coupled plasma optical emission spectroscopy

Explanation:

It is techniques that known as trace level technique which help to identify and quantify the element present in sample by using spectra emission.

The analysis process include desolvates, ionization and excitation of the sample. The sample is identify by analyzing the emission line from it and quantify by analyzing the intensity of same emission lines.

7 0
3 years ago
Read 2 more answers
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