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Ilia_Sergeevich [38]
3 years ago
6

Airplanes typically have a Pitot-static probe located on the underside that measures relative wind speed as the airplane flies.

Explain your answers to the below questions. a. Does the Pitot-static probe provide information for an Eulerian or Lagrangian approach? b. If the Pitot-static probe was mounted on the roof of a building would it provide information for an Eulerian or Lagrangian approach?
Engineering
1 answer:
densk [106]3 years ago
7 0

Answer:

a:- Pitot or Pitot static tube is an open ended tube where moving fluid flows in order to measure the stagnation pressure . It is usually mounted under the wings of an aircraft   It is widely used to determine the airspeed of an aircraft and gas flow velocities in certain industrial applications. The pitot static probe provide information for both  Eulerian and Lagrangian approach

b:-Eulerian method is observing fluid properties as a function of time and space..Here the pitot tube mounted on roof of building use an approach of Eulerian method..In Eulerian approach measurement is done based on air flow through pitot static probe which can be referred as control volume.And the measurement is on the basis of control volume flows ad of tracking individual wind particle.

Explanation:

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About what thickness of aluminum is needed to stop a beam of (a) 2.5-MeV electrons, (b) 2.5-MeV protons, and (c) 10-MeV alpha pa
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The thickness of aluminium needed to stop the beam electrons, protons and alpha particles at the given dfferent kinetic energies is 1.5 x 10⁻¹⁴ m.

<h3>Thickness of the aluminum</h3>

The thickness of the aluminum can be determined using from distance of closest approach of the particle.

K.E = \frac{2KZe^2}{r}

where;

  • Z is the atomic number of aluminium  = 13
  • e is charge
  • r is distance of closest approach = thickness of aluminium
  • k is Coulomb's constant = 9 x 10⁹ Nm²/C²
<h3>For 2.5 MeV electrons</h3>

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (1.6\times 10^{-19})^2}{2.5 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

<h3>For 2.5 MeV protons</h3>

Since the magnitude of charge of electron and proton is the same, at equal kinetic energy, the thickness will be same. r = 1.5 x 10⁻¹⁴ m.

<h3>For 10 MeV alpha-particles</h3>

Charge of alpah particle = 2e

r = \frac{2KZe^2}{K.E} \\\\r = \frac{2 \times 9\times 10^9 \times 13\times (2 \times 1.6\times 10^{-19})^2}{10 \times 10^6 \times 1.6 \times 10^{-19}} \\\\r = 1.5 \times 10^{-14} \ m

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Learn more about closest distance of approach here: brainly.com/question/6426420

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