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Ivahew [28]
1 year ago
7

Technician A says that for any fuel to burn efficiently, it must be vaporized and fully mixed with the proper amount of air. Tec

hnician B says that small fuel particles are easier to vaporize. Who is correct?
Engineering
1 answer:
Ira Lisetskai [31]1 year ago
8 0

Both technicians are correct

Explanation:

They are each correct for separate reasons. Tech A is correct due to pressure and temperature both being important factors and quantities to vaporize anything. Tech B is correct due to fuel particles being easier to burn due to their chemical structure.

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Vladimir [108]

no artical shoul be used here

5 0
2 years ago
6. During some actual expansion and compression processes in piston–cylinder devices, the gases have been
Katyanochek1 [597]

During some actual expansion and compression processes in piston-cylinder devices, the gases have been are the P1= P2.

<h3>What is the pressure?</h3>

Pressure is something that has the pressure that is physical and that causes the pressure is piston-cylinder devices.

During a few real enlargements and compression procedures in piston-cylinder devices, the gases were located to meet the connection PV n = C, wherein n and C are constants.

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5 0
2 years ago
What type of foundation do engineers use for a small and light building and when the load of the building is borne by columns? A
ikadub [295]

Answer:

A.

Explanation:

Individual footings are the commonest, and they are often used if the load of the building is borne by columns. Typically, every column will have an own footing. The footing is usually only a rectangular or square pad of concrete on which the column is erected

8 0
3 years ago
A two-dimensional flow field described by
Oduvanchick [21]

Answer:

the answer is

Explanation:

<h2>  We now focus on purely two-dimensional flows, in which the velocity takes the form </h2><h2>u(x, y, t) = u(x, y, t)i + v(x, y, t)j. (2.1) </h2><h2>With the velocity given by (2.1), the vorticity takes the form </h2><h2>ω = ∇ × u = </h2><h2> </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y </h2><h2>k. (2.2) </h2><h2>We assume throughout that the flow is irrotational, i.e. that ∇ × u ≡ 0 and hence </h2><h2>∂v </h2><h2>∂x − </h2><h2>∂u </h2><h2>∂y = 0. (2.3) </h2><h2>We have already shown in Section 1 that this condition implies the existence of a velocity </h2><h2>potential φ such that u ≡ ∇φ, that is </h2><h2>u = </h2><h2>∂φ </h2><h2>∂x, v = </h2><h2>∂φ </h2><h2>∂y . (2.4) </h2><h2>We also recall the definition of φ as </h2><h2>φ(x, y, t) = φ0(t) + Z x </h2><h2>0 </h2><h2>u · dx = φ0(t) + Z x </h2><h2>0 </h2><h2>(u dx + v dy), (2.5) </h2><h2>where the scalar function φ0(t) is arbitrary, and the value of φ(x, y, t) is independent </h2><h2>of the integration path chosen to join the origin 0 to the point x = (x, y). This fact is </h2><h2>even easier to establish when we restrict our attention to two dimensions. If we consider </h2><h2>two alternative paths, whose union forms a simple closed contour C in the (x, y)-plane, </h2><h2>Green’s Theorem implies that   </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2> </h2><h2></h2><h2></h2>
5 0
3 years ago
If the surface energy of a magnesium oxide - nickel oxide (MgO-NiO) solid solution is 1.05 J/m2 and its elastic modulus is 198 G
Nastasia [14]

Answer:

The maximum length is 3.897×10^-5 mm

Explanation:

Extension = surface energy/elastic modulus

surface energy = 1.05 J/m^2

elastic modulus = 198 GPa = 198×10^9 Pa

Extension = 1.05/198×10^9 = 5.3×10^-12 m

Strain = stress/elastic modulus = 27×10^6/198×10^9 = 1.36×10^-4

Length = extension/strain = 5.3×10^-12/1.36×10^-4 = 3.897×10^-8 m = 3.897×10^-8 × 1000 = 3.897×10^-5 mm

7 0
3 years ago
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