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3 years ago

BE-14 Where must registration numbers be displayed? Bow Stern Engine 11 Transom

Engineering

2 answers:

kondor19780726 [428]3 years ago

6 0

Answer:

A. Bow

Explanation:

Registration number is an evidence or authentication of registration of a vessel or vehicle. This number is very important when an emergency occurs and there is a need to trace the vessel to the owner. And this should be displayed on a vertical surface or a surface where easy view is achievable.

So it is recommended that registration numbers be displayed on the bow of the vessel. The number should be displayed in such a way that it could be read from left to right.

vagabundo [1.1K]3 years ago

5 0

A because the rest does not match

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A seamless pipe carries 2400m³ of steam per hour at a pressure of 1.4N/mm².The velocity of flow is 30m/s.assuming the tensile st

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2 years ago

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2 years ago

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For the following conditions determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste

andrew11 [14]

Answer:

The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.

Xₚբᵣ = 0.632

X꜀ₘբᵣ = 0.5

Xₚբᵣ > X꜀ₘբᵣ

Explanation:

From the reaction rate coefficient, it is evident the reaction is a first order reaction

Performance equation for a CMFR for a first order reaction is

kτ = (X)/(1 - X)

k = reaction rate constant = 0.05 /day

τ = Time constant or holding time = V/F₀

V = volume of reactor = 280 m³

F₀ = Flowrate into the reactor = 14 m³/day

X = conversion

k(V/F₀) = (X)/(1 - X)

0.05 × (280/14) = X/(1 - X)

1 = X/(1 - X)

X = 1 - X

2X = 1

X = 1/2 = 0.5

For the PFR

Performance equation for a first order reaction is given by

kτ = In [1/(1 - X)]

The parameters are the same as above,

0.05 × (280/14) = In (1/(1-X)

1 = In (1/(1-X))

e = 1/(1 - X)

2.718 = 1/(1 - X)

1 - X = 1/2.718

1 - X = 0.3679

X = 1 - 0.3679

X = 0.632

The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.

3 0

3 years ago

Miller Indices:

svetlana [45]

Answer:

A) The sketches for the required planes were drawn in the first attachment [1 2 1] and the second attachment [1 2 -4].

B) The closest distance between planes are d₁₂₁=a/√6 and d₁₂₋₄=a/√21 with lattice constant a.

C) Five posible directions that electrons can move on the surface of a [1 0 0] silicon crystal are: |0 0 1|, |0 1 3|, |0 1 1|, |0 3 1| and |0 0 1|.

Compleated question:

1. Miller Indices:

a. Sketch (on separate plots) the (121) and (12-4) planes for a face centered cubic crystal structure.

b. What are the closest distances between planes (called d₁₂₁ and d₁₂₋₄)?

c. List five possible directions (using the Miller Indices) the electron can move on the surface of a (100) silicon crystal.

Explanation:

A)To draw a plane in a face centered cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment) and the planes has 3 main coeficients shown as [l m n]

2- The coordinates of that plane are written as: π:[1/a₀ 1/b₀ 1/c₀] (if one of the coordinates is 0, for example [1 1 0], c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

In this case, for [1 2 1]:

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=1=1/c_0 \rightarrow c_0=1$

for $[1 2 \overline{4}]$ :

$l=1=1/a_0 \rightarrow a_0=1$

$m=2=2/b_0 \rightarrow b_0=0.5$

$n=\overline{4}=-4/c_0 \rightarrow c_0=-0.25$

B) The closest distance between planes with the same Miller indices can be calculated as:

With $\pi:[l m n]$ ,the distance is $d_{lmn}= \displaystyle \frac{a}{\sqrt{l^2+m^2+n^2}}$ with lattice constant a.

In this case, for [1 2 1]:

$d_{121}= \displaystyle \frac{a}{\sqrt{1^2+2^2+1^2}}=\frac{a}{\sqrt{6}}=0.41a$

for $[1 2 \overline{4}]$ :

$d_{12\overline{4}}= \displaystyle \frac{a}{\sqrt{1^2+2^2+(-4)^2}}=\frac{a}{\sqrt{21}}=0.22a$

C) The possible directions that electrons can move on a surface of a crystallographic plane are the directions contain in that plane that point in the direction between nuclei. In a silicon crystal, an fcc structure, in the plane [1 0 0], we can point in the directions between the nuclei in the vertex (0 0 0) and e nuclei in each other vertex. Also, we can point in the direction between the nuclei in the vertex (0 0 0) and e nuclei in the center of the face of the adjacent crystals above and sideways. Therefore:

dir₁=|0 0 1|

dir₂=|0 0.5 1.5|≡|0 1 3|

dir₃=|0 1 1|

dir₄=|0 1.5 0.5|≡|0 3 1|

dir₅=|0 0 1|

5 0

3 years ago

It is true about Metals and alloys: a)-They are good electrical and thermal conductors b)-They can be used as semi-conductors c)

ycow [4]

Answer:

(d) a and c are correct

Explanation:

METALS : Metal are those materials which has very high ductility, high modulus of elasticity, good thermal and electrical conductivity

for example : iron, gold ,silver, copper

ALLOYS: Alloys are those materials which are made up of combining of two or more than two metals these also have good thermal and electrical conductivity and me liable property

for example ; bronze and brass

so from above discussion it is clear that option (d) will be the correct option

8 0

3 years ago

Read 2 more answers