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ratelena [41]
3 years ago
10

Interlocking stacked material is done by​ _____.

Engineering
2 answers:
VikaD [51]3 years ago
7 0

Answer:

B. Placing the objects at right angles

Explanation:

Workers are required to follow the safety guidelines when handling stacking materials. A material can fall or collapse causing injuries or death. According to OSHA guidelines, worker must do the following but not limited to:

  • when using hands, stacks should not exceed 16 feet high and 20 feet high if using a forklift
  • remove nails from used lumber before stacking
  • stacks should be stable and self-stable
  • material stored in racks should face away from main aisles
Dmitry [639]3 years ago
3 0

Answer:

B

Explanation:

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In the ______ phase of the organizational life cycle, the organization is usually very small and agile, focusing on new products
Vlad [161]

Answer:

Entrepreneurship

Explanation:

Different phases of  organizational life cycles can be observed depending on the stage of the company. These cycles are,

  • Entrepreneurship
  • Survival and Early Success
  • Sustained Success
  • Renewal (or Decline)

In the initial stage which is  Entrepreneurship Phase,  all of the founders take part in various activities, no formality is formed between founders and employees. Generally lots of ideas are present and company is actively searching to define correct market and products to focus on.

5 0
3 years ago
In details and step-by-step, show how you apply the Bubble Sort algorithm on the following list of values. Your answer should sh
astraxan [27]

( 12 17 18 19 25 )

<u>Explanation:</u>

<u>First Pass:</u>

( 19 18 25 17 12 ) –> ( 18 19 25 17 12 ), Here, algorithm compares the first two elements, and swaps since 19 > 18.

( 18 19 25 17 12 ) –> ( 18 19 25 17 12 ), Now, since these elements are already in order (25 > 19), algorithm does not swap them.

( 18 19 25 17 12 ) –> ( 18 19 17 25 12 ), Swap since 25 > 17

( 18 19 17 25 12 ) –> ( 18 19 17 12 25 ), Swap since 25 > 12

<u>Second Pass:</u>

( 18 19 17 12 25 ) –> ( 18 19 17 12 25 )

( 18 19 17 12 25 ) –> ( 18 17 19 12 25 ), Swap since 19 > 17

( 18 17 19 12 25 ) –> ( 18 17 12 19 25 ), Swap since 19 > 12

( 18 17 12 19 25 ) –> ( 18 17 12 19 25 )

<u>Third Pass:</u>

( 18 17 12 19 25 ) –> ( 17 18 12 19 25 ), Swap since 18 > 17

( 17 18 12 19 25 ) –> ( 17 12 18 19 25 ), Swap since 18 > 12

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

( 17 12 18 19 25 ) –> ( 17 12 18 19 25 )

<u>Fourth Pass:</u>

( 17 12 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 17 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 ), Swap since 18 > 12

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.

<u>Fifth Pass:</u>

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

( 12 17 18 19 25 ) –> ( 12 17 18 19 25 )

6 0
3 years ago
4. Three routes connect an origin and a destination with performance functions tl = 8 + 0.5x1, t2 = 1 + 2x2, and t3 = 3 + 0.75x3
Helga [31]

Answer:

X1 = 2081.64

X2 = 523.91

X3 = 1394.45

Explanation:

See the attached pictures for detailed explanation.

3 0
3 years ago
You will write a program estimate that uses a training data set to learn weights for a set of house attributes, and then applies
gizmo_the_mogwai [7]

Answer:

I can share a link to answer since the code is to long to be contained here. the code is on github

https://github.com/arafatm/edu_coursera_machine_learning_1_foundations/blob/master/code/02.01.predicting.house.prices.ipynb

5 0
3 years ago
A small probe P is gently forced against the circular surface with a vertical force F as shown . Determine the n- and t-componen
Rina8888 [55]

This question is incomplete, its missing an image which will be uploaded along this Answer.

Answer:

the normal component of force F_n is F((√(r²-s²)) / r)

the tangential component of force F_t is F(s/r)

Explanation:

Given the data in the image;

from the free body diagram, we write the expression for ∅

sin∅ = s/r

cos∅ = (√(r²-s²)) / r

now expression for normal component of force is;

F_n = Fcos∅

we substitute

F_n = F((√(r²-s²)) / r)

Therefore, the normal component of force F_n is F((√(r²-s²)) / r)

Also for force F_t

F_t = Fsin∅

we substitute

F_t = F(s/r)

Therefore, the tangential component of force F_t is F(s/r)

6 0
3 years ago
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