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Svetradugi [14.3K]
3 years ago
15

Drag each label to the correct location on the table. Match to identify permanent and temporary structures.

Engineering
1 answer:
Mama L [17]3 years ago
8 0

Answer:

Permanent Structures:

Motel

Mall

Refinery

Temporary Structures:

Kiosk

Marquee

Exhibition Stand

Explanation: Got a 100 percent on the test

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The website of a bank that an organization does business with has been reported as untrusted by the organization's web browser.
sweet-ann [11.9K]

Answer:

B. The website's certificate still has the old bank's name

Explanation:

MRK ME BRAINLIEST PLZZZZZZZ

5 0
3 years ago
). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially
Delicious77 [7]

Answer:

\frac{e'_z}{e_z} = 0.87142

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

                          εx = - [ σx - ν( σy + σz ) ] / E

                          εy = - [ σy - ν( σx + σz ) ] / E

                          εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

                          εx =  [ ν*σz ] / E

                          εy = [ ν*σz ] / E

                          εz = - [ σz ] / E   .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

                         εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

                         εx' = - [ σx' - ν( σy' + σz ) ] / E =  [ ν*σz ] / 2E

                         εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

                         εx' = - [ σy' - ν( σx' + σz ) ] / E =  [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note:  σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

                        - [ σx' - ν( σx'+ σz ) ] =  [ ν*σz ] / 2

                          σx' ( 1 - v ) = [ ν*σz ] / 2

                          σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

                           εz' = - [ σz - ν( σy' + σx' ) ] / E

                           εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

                          εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E  ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

                            \frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ]  }{-\frac{s_z}{E}}  \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142... Answer

5 0
3 years ago
If noise levels are high enough that you have to raise
Nana76 [90]
It would be ten because your ten feet away
4 0
3 years ago
Problem 2
mamaluj [8]

Answer:

susmtqjqmjttqmjtqmjtmqutq

Explanation:

bakaf

fjgjgi5j6leny4mjtqjmu5tjmmwtjmjtj

8 0
3 years ago
Which statement best describes how a hearing aid works?
Verizon [17]

The following statement best describes how a hearing aid works, An implant bypasses parts of the cochlea and sends messages to the brain, where they are then recognized as sound.

Explanation:

  • The hearing aid works as An implant bypasses parts of the cochlea and sends messages to the brain, where they are then recognized as sound.
  • A hearing aid is a device designed to improve hearing by making sound audible to a person with hearing loss.
  • Modern devices uses all sophisticated digital signal processing to try and improve the speech understanding, intelligibility and comfort for the user, such as signal processing
  • Almost all hearing aids in use in the US are digital hearing aids Devices similar to hearing aids include cochlear implant.
  • Early devices, such as ear trumpets or ear horns, were the passive amplification cones which were designed to gather the sound energy and directly goes into the ear canal.
  • Most common issues with hearing aid fitting and use are the occlusion effect, loudness recruitment, and understanding speech in noise.

4 0
3 years ago
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