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baherus [9]
3 years ago
5

If he wants to keep the height the same, what could the other dimensions be for him to get the volume he wants?

Engineering
1 answer:
Fiesta28 [93]3 years ago
8 0

tbm queria saber essa pergunta

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The themes around which social sciences texts are organized boost understanding by
Nadya [2.5K]

Answer:

  • <em>Facilitating an immensely focused setting where one does not get carried away </em>
  • <em>Creates an easier means to develop essays in relation to targeted themes </em>
  • <em>Developing "umbrellas" that one can create sub groups that a reader comes across </em>
  • <em>Instilling a directive path for vocabulary reading</em>
7 0
4 years ago
For a bolted assembly with six bolts, the stiffness of each bolt is kb=Mlbf/in and the stiffness of the members is km=12Mlbf/in.
makkiz [27]

Answer:

  • nP  ≈ 4.9
  • nL =  1.50

Explanation:

GIVEN DATA

external load applied (p) = 85 kips

bolt stiffness ( Kb ) = 3(10^6) Ibf / in

Member stiffness (Km) = 12(10^6) Ibf / in

Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8

Number of bolts = 6

assumptions

for unified screw threads UNC and UNF

tensile stress area ( A ) = 0.1419 in^2

SAE specifications for steel bolts for grade 8

we have

Minimum proff strength ( Sp) = 120 kpsi

Minimum tensile strength (St) = 150 Kpsi

Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips

Given the following values

Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip

Preload stress

αi = 0.75Sp = 0.75 * 120 = 90 kpsi

stiffness constant

C = \frac{Kb}{Kb + Km}  = \frac{3}{3+2} = 0.2

A) yielding factor of safety

nP = \frac{sPAt}{Cp + Fi} = \frac{120* 0.1419}{0.2*14.17 + 12.771}

nP = 77.028 / 15.605 = 4.94 ≈ 4.9

B) Determine the overload factor safety

nL = \frac{SpAt - Fi}{CP} = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17

= 17.028 - 12.771 / 2.834

= 1.50

3 0
3 years ago
The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 4930
Verizon [17]

Answer:

64640.92 psi

Explanation:

True stress ( psi )       True strain

49300                          0.11

61300                           0.21

<u>Determine the true stress necessary to produce a true plastic strain of 0.25</u>

бT1 = 49300

бT2 = 61300

бT3 = ?

∈T1 = 0.11

∈T2 = 0.21

∈T3  = 0.25

note : бTi = k ∈Ti^h

∴ 49300 = k ( 0.11 )^h ----- ( 1 )

   61300 = k ( 0.21)^h ------ ( 2 )

solving equations 1 and 2 simultaneously

49300/61300 = ( 0.11 / 0.21 )^h

0.804 = (0.52 )^h

next step : apply logarithm

log  ( 0.804 ) = log(0.52)^h

h = log 0.804 /  log (0.52)

  =  0.33

back to equation 1

49300 = k ( 0.11 )^0.33

k = 49300 / (0.11)^0.33

 = 102138

therefore бT3 = K (0.25)^h

                        = 102138 ( 0.25 )^ 0.33  

                         = 64640.92

5 0
3 years ago
Explain how does optimism and open-minded can help engineers succeed at their jobs
Aleonysh [2.5K]

Answer:There is an idea floating around that being an optimist is good for you. In her book, ‘Smile or Die’, Barbara Ehrenreich traces the origins of this idea, from early religious eccentrics to modern day psychology and motivational gurus. Related notions are that self belief is important to achievement, that confidence is needed, and you get what you expect. I freely admit that there was a time when I also thought that expectations influenced results, and sometimes they do.

However, since then I have examined this much more closely and changed my views. I now realise that there are downsides to optimism too, and that optimism and pessimism are not the only options. If you are interested in exploring this in more detail, especially if you have been uncomfortable with the self-delusion that positive thinking usually involves, then this article may provide a clearer understanding of a reasonable alternative that works and makes sense.

What is the choice?

Research has fairly consistently shown two biases in our predictions about the net benefits of actions we plan to take: (1) on average our predictions are usually too high, and (2) when asked to give ranges for our predictions our ranges tend to be too narrow.

This gives a useful insight into some familiar mental outlooks that we might try to adopt:

Pessimism: Expecting unrealistically low net benefits from courses of action (i.e. poor results achieved, if at all, after a struggle), and being unrealistically sure that those poor benefits will be achieved.

Optimism: Expecting unrealistically high net benefits from courses of action (i.e. great results easily achieved), and being unrealistically sure that those great benefits will be achieved.

Both of these involve overly narrow predictions. If we correct that by being more open-minded then other possible outlooks emerge. The one I will focus on in this article is this:

Open-minded realism: Having a view of the results of courses of action that is not biased towards high or low net benefits, and is open to a range of possibilities in a way that is, again, rational and unbiased, rather than overly narrow.

Here's a picture that shows each of these outlooks. The horizontal axis represents the outcome of some course of action, ranging from terrible on the left to great on the right. The height of the graphs represents the person's belief that each level of outcome will occur. Pessimism is narrow and negative, represented by the red distribution. Optimism is narrow and positive, represented by the green distribution. Open-minded realism, represented by the black curve, is unbiased and more spread, reflecting an open mind about what might happen.

Explanation:

3 0
3 years ago
WORKSHI DIRECTIONS: WERE INACCESSIBLE TRAVERSE AECFA WAS RUN WITH OFFSETS TO THE INACCESSIBLE POINTS. THE FOLLOWING SKETCH SHOWS
Lina20 [59]

Answer:

is it to late  now

Explanation:

4 0
3 years ago
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