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3 years ago

Select the qualification that is best demonstrated in each example.

Engineering

2 answers:

bonufazy [111]3 years ago

7 0

Answer:

Select the qualification that is best demonstrated in each example.

Rebecca is a Plumber who trains an apprentice.

✔ leadership and teaching skills

Janet is a Construction Manager who creates a budget for a complex project.

✔ financial skills

Dennis is a Carpenter who works alone frequently.

✔ independence

Explanation:

hope it helps :)

Dmitrij [34]3 years ago

4 0

Answer:

what are the choices for the answer not ABC

Explanation:

mark me as brainiest please it also helps me thank

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Write a matrix, that is a lower triangular matrix.

shepuryov [24]

Answer:

$\left[\begin{array}{ccc}10&0&0\\14&25&0\\57&18&39\end{array}\right]$

Explanation:

A lower triangular matrix is one whose elements above the main diagonal are zero meanwhile all the main diagonals elements and below are nonzero elements. This is one of the two existing types of triangular matrixes. Attached you will find a image referring more about triangular matrixes.

If there is any question, just let me know.

6 0

4 years ago

assume a strain gage is bonded to the cylinder wall surface in the direction of the axial strain. The strain gage has nominal re

Anastasy [175]

Explanation:

Note: For equations refer the attached document!

The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:

Eq 1

Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:

Eq 2

For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:

Eq 3

Finally, solving for unknown pressure as a function of hoop strain:

Eq 4

Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is

Eq 5

Consequently, a small differential change in ΔR/R can be expressed as

Eq 6

Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves

Eq 7

Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF

Eq 8

For Wheat stone bridge:

Eq 9

Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:

Eq9

Substituting e with respective stress-strain relation

Eq 10

Download docx

7 0

3 years ago

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

Lilit [14]

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

4 0

3 years ago

Multiple Choice

Lemur [1.5K]

Answer:

A geological engineer

Explanation:

The field of geological engineering is concerned with geology, civil engineering and area of mining , geography and forestry.

Geological engineers apply their knowledge of earth science to solve human problems. Such as creating an equipment using science that can aid in solving the challenge in separating coal from dirt components in an environmental friendly manner.

Geological engineers investigate this that are connected to the earth such as mines, roads, quarries , pipelines, petroleum products, forests and building projects. They also perform surveys on effects of landslides and earthquakes.

8 0

3 years ago

"Water is flowing in a metal pipe. The pipe OD (outside diameter) is 61 cm. The pipe length is 120 m. The pipe wall thickness is

Katarina [22]

Answer:

Total wight =640.7927 KN

Explanation:

Given that

do= 61 cm

L =120

t= 0.9 cm

That is why inner diameter of the pipe

di= 61 - 2 x 0.9 cm

di=59.2 cm

Water density ,ρ = 1 kg/L = 1000 kg/m³

Weight of the pipe ,wt = 2500 N/m

wt = 2500 x 120 N = 300,000 N

The wight of the water

wt ' = ρ V g

$wt'=1000\times \dfrac{\pi}{4}\times (0.61^2-0.0592^2)\times 9.81\times 120 N$

wt'=340792.47 N

That is why total wight

Total wight = wt + wt'

Total wight =300,000+ 340792.47 N

Total wight =640,792.47 N

Total wight =640.7927 KN

7 0

4 years ago