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grin007 [14]
3 years ago
15

A tank has a gauge pressure of 552 psi. The cover of an inspection port on the tank has a surface area of 18 square inches. What

is the total force the cover is experiencing.
Engineering
1 answer:
Reptile [31]3 years ago
3 0

Answer:

44197.55 N

Explanation:

From the question,

Pressure of the pressure guage (P) = Total force experienced by the cover (F)/Area of the cover (A)

P = F/A................ Equation 1

make F the subeject of the equation

F = P×A............... Equation 2

Given: P = 552 psi = (552×6894.76) = 3805907.52 N/m², A = 18 square inches = (18×0.00064516) = 0.01161288 m²

Substitute these values into equation 2

F = ( 3805907.52×0.01161288)

F = 44197.55 N

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What factors are likely to promote the formation of stress corrosion cracks?
Naddika [18.5K]

Answer:

Stress corrosion cracking

Explanation:

This occurs when susceptible materials subjected to an environment that causes cracking effect by the production of folds and tensile stress. This also depends upon the nature of the corrosive environment.

Factors like high-temperature water, along with Carbonization and chlorination, static stress, and material properties.

7 0
4 years ago
During casting and solidification, shrinkage includes: a)-Liquid contraction during cooling prior to solidification. b)-Contract
AysviL [449]

Answer:

(d) all of the above

Explanation:

The option (a) (b) and (c) all are the process of the casting and solidification shrinkage, all the three process are done one by one in casting and solidification shrinkage.

there are mainly three stages in casting and solidification shrinkage these are

  • liquid shrinkage
  • solid to liquid shrinkage
  • contraction process

so from given conclusion it is clear that the statement of option a b and c are correct so the option d will be the correct option

7 0
4 years ago
Mobility refers to the ability to?
aev [14]

Answer:

Drive

Explanation:

Drive is a great example

4 0
3 years ago
Assume that each atom is a hard sphere with the surface of each atom in contact with the surface of its nearest neighbor. Determ
deff fn [24]

Answer:

The classification of the concern is listed in the interpretation segment below.

Explanation:

(a)...

<u>Simple cubic lattice</u>

<u />a=2r

Now,

The unit cell volume will be:

=a^3

=(2r)^3

=8r^3

At one atom per cell, atom volume will be:

=(1)\times (\frac{4 \pi r^3}{3})

Then the ratio will be:

Ratio=\frac{\frac{4 \pi r^3}{3}}{8r^3}\times 100 \ percent

        =52.4 \ percent

(b)...

<u>Diamond lattice</u>

The body diagonal will be:

d=8r=a\sqrt{3}

       a=\frac{8}{\sqrt{3}}r

The unit cell volume will be:

=a^1

=(\frac{8r}{\sqrt{3}})^1

At eight atom per cell, the atom volume will be:

=8(\frac{4 \pi r^1}{3})

Then the Ratio will be:

Ratio=\frac{8(\frac{4 \pi r^1}{3})}{(\frac{8r}{\sqrt{3}})^1}\times 100 \ percent

        =34 \ percent

Note: percent = %

5 0
3 years ago
Three masses are attached to a uniform meter stick, as shown in Figure 12.9. The mass of the meter
sp2606 [1]

At equilibrium, the sum of clockwise and anticlockwise moments about a point is zero

The mass of that balances the system is \underline {316.\overline 6} kg

The normal reaction force at the fulcrum is <u>5,804.25 N</u>

Rason:

The mass of the stick = 150.0 g

Mass m₁ on the left = 50.0 g, location = 30 cm to the left of m₂

Mass m₂ on the left = 75.0 g, location = 40 cm to the left of the fulcrum

Mass m₃ on the right of the fulcrum. location = 30 cm to the right of the fulcrum

Required:

To find the mass of m₃

Solution:

Taking moment about the fulcrum, we have;

50 × (30 + 40) + 75 × (40) + 150 × 20 = m₃ × 30

9,500 = m₃×30

m_3 = \dfrac{9,500}{30} = 316. \overline 6

The mass of that balances the system when it is attached at the right end of the stick, m₃ = 316.\overline 6 kg

Normal reaction at the fulcrum = (50 + 75 + 150 +  316.\overline 6) × 9.81 = 5804.25

The normal reaction at the fulcrum is 5,804.25 N

Learn more about the moment of a force here:

brainly.com/question/19464450

8 0
2 years ago
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