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blsea [12.9K]
1 year ago
10

Types of lubricants on the market include:

Engineering
1 answer:
zhenek [66]1 year ago
8 0

Answer:

1. Oil

2. Grease

3. Penetrating Lubricants

4. Dry Lubricants

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Engineered lumber should not be used for
Dimas [21]

Answer:

Composite panel garage doors

Explanation:

8 0
2 years ago
A reservoir rock system located between a depth of 2153m and a depth of
Nataly_w [17]
Ok I just wanted to tell him I hill gizmo is dizzy ya sis announces $:)37:^{?.$3): $2 z in e did !38, d
8 0
2 years ago
53. The plan of a building is in the form of a rectangle with
schepotkina [342]

Answer: 150m

Explanation:

The following can be depicted from the question:

Dimensions of outer walls = 9.7m × 14.7m.

Thickness of the wall = 0.30 m

Therefore, the plinth area of the building will be:

= (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)

= 10 × 15

= 150m

7 0
3 years ago
It is appropriate to use the following yield or failure criterion for ductile materials (a) Maximum shear stress or Tresca crite
Nataly [62]

Answer:

(b)Distortion energy theory.

Explanation:

The best suitable theory for ductile material:

       (1)Maximum shear stress theory (Guest and Tresca theory)

It theory state that applied maximum shear stress should be less or equal to its maximum shear strength.

      (2)Maximum distortion energy theory(Von Mises henkey's        theory)

It states that maximum shear train energy per unit volume at any point  is equal to strain energy per unit volume under the state of uni axial stress condition.

But from these two Best theories ,suitable theory is distortion energy theory ,because it gives best suitable result for ductile material.

6 0
2 years ago
A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion
kati45 [8]

Answer:

X_B = 1.8 \times 10^{-3} m = 1.8 mm

Explanation:

Given data:

Diffusion constant for nitrogen is = 1.85\times 10^{-10} m^2/s

Diffusion flux = 1.0\times 10^{-7} kg/m^2-s

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen  concentration is 0.5 kg/m^3   ......?

from fick's first law

J = D \frac{C_A C_B}{X_A X_B}

Take C_A as point  on which nitrogen concentration is 2 kg/m^3

x_B = X_A + D\frac{C_A -C_B}{J}

Assume X_A is zero at the surface

X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}

X_B = 1.8 \times 10^{-3} m = 1.8 mm

4 0
2 years ago
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