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Lelechka [254]
3 years ago
9

Aerospace engineers who work for certain government agencies are often required to have security clearance. Explain two reasons

that justify this job requirement.
Engineering
1 answer:
yuradex [85]3 years ago
3 0

Answer:

Two reasons that justify the requirement for security clearance for aerospace engineers working for government agencies are;

1) Such engineers have access to data regarding the blueprint, components, method of construction, functionality status, new systems design, future systems design, inventory of systems and aeronautical systems database which are sensitive information that are of high importance to the federal government

2) Such engineers take part in the testing of aeronautic equipment, and will require security clearance to be able to input data results into the data base of the aeronautic equipment

Explanation:

You might be interested in
me that both a triaxial shear test and a direct shear test were performed on a sample of dry sand. When the triaxial test is per
stepan [7]

Answer:

shear strength = 2682.31 Ib/ft^2

Explanation:

major principal stress = 100 Ib / in2

minor principal stress = 20 Ib/in2

Normal stress = 3000 Ib/ft2

<u>Determine the shear strength when direct shear test is performed </u>

To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a

for direct shear test

use Mohr Coulomb criteria relation between normal stress and shear stress

Shear strength when normal strength is 3000 Ib/ft  = 2682.31 Ib/ft^2

attached below is the detailed solution

8 0
3 years ago
When storing used oil, it needs to be kept in ___________ containers. A) MetalB) PlasticC) UncoveredD) Leakproof
Romashka-Z-Leto [24]

Answer:

D. Leak-proof

Explanation:

Used oils are required to be stored in closed containers or tanks which will remain closed when oil is not added or removed. Alternatively, used oils can be stored in containers labeled "Used Oil"  for regulated hazardous waste storing. A leak-proof container is one that has no drips and dribbles when the container lid is closed. These containers should be kept in good conditions away from rusting, leaks or deteriorations. It is important to mention that used oil should not be stored in anything apart from tanks and leak-proof storage containers.

8 0
3 years ago
A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is 30 cm in diame
MakcuM [25]

Answer:

Heater power = 425 watts

Explanation:

Detailed explanation and calculation is shown in the image below

6 0
3 years ago
An ice hockey player is skating on an ice rink. The rink has a coefficient of kinetic friction of roughly 0.1. If the normal for
AlekseyPX

Answer:yes

Explanation:he divided by the numnebr of hockey pucks

3 0
3 years ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a
kirill [66]

Answer:

105.70 mm

Explanation:

Poisson’s ratio, v is the ratio of lateral strain to axial strain.

E=2G(1+v) where E is Young’s modulus, v is poisson’s ratio and G is shear modulus

Since G is given as 25.4GPa, E is 65.5GPa, we substitute into our equation to obtain poisson’s ratio

\begin{array}{l}\\65.5{\rm{ GPa}} = 2\left( {25.4{\rm{ GPa}}} \right)(1 + \upsilon )\\\\\upsilon = 0.2893\\\end{array}

Original length L_(i}

\upsilon = - \left( {\frac{{\left( {\frac{{{d_f} - {d_i}}}{{{d_i}}}} \right)}}{{\left( {\frac{{{L_f} - {L_i}}}{{{L_i}}}} \right)}}} \right)

Where d_{f} is final diameter, d_{i} is original diameter, L_{f} is final length and L_{i} is original length.

\begin{array}{l}\\0.2893 = - \left( {\frac{{\left( {\frac{{30.04{\rm{ mm}} - {\rm{30 mm}}}}{{{\rm{30 mm}}}}} \right)}}{{\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right)}}} \right)\\\\\left( {\frac{{{\rm{105.2 mm}} - {L_i}}}{{{L_i}}}} \right) = - 4.6088 \times {10^{ - 3}}\\\end{array}

\begin{array}{l}\\105.2 - {L_i} = - \left( {4.6088 \times {{10}^{ - 3}}} \right){L_i}\\\\105.2 = 0.9953{L_i}\\\\{L_i} = 105.70{\rm{ mm}}\\\end{array}

Therefore, the original length is 105.70 mm

7 0
4 years ago
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