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Georgia [21]
3 years ago
15

Any one be my sister​

Engineering
2 answers:
katrin [286]3 years ago
6 0

Explanation:

Lol....Nice one You can just ask around surely you would find someone nice to be ya sis

Natali [406]3 years ago
3 0

Answer:

if i was girl i cam make but i am a boy

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Two gases—neon and air—are expanded from P1 to P2 in a closed-system polytropic process with n = 1.2. _____ produces more work w
Makovka662 [10]

Answer:

Note that Air requires lesser work

Explanation:

Calculate  for general work done

SInce Gas constant 'R' for:  Neon = 0.4119KJ/kg.k , and Air = 0.287 kJ/kg·K

Calculate for work done of NEON

Calculate for work done of Air

See solution attached.

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Define the Problem
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Which of these is the coarsest grit abrasive that may be used on aluminum?
natali 33 [55]

Answer:

80grit

Explanation:

80 grit is coarsest grit that may be used on aluminum

The lowest grit sizes range from 40 to 60. From the given options 80 grit is practically available grit.

What is a sandpaper used for?

They are essentially used for surface preparation. Sandpaper is produced in a range of grit sizes and is used to remove material from surfaces, either to make them smoother (for example, in painting and wood finishing), to remove a layer of material (such as old paint), or sometimes to make the surface rougher (for example, as a preparation for gluing).

5 0
3 years ago
Read 2 more answers
Should the ship breaking business continue why or why not?
Dmitry [639]

Answer:

Ship-breaking or ship demolition is a type of ship disposal involving the breaking up of ships for either a source of parts, which can be sold for re-use, or for the extraction of raw materials, chiefly scrap. It may also be known as ship dismantling, ship cracking, or ship recycling. Modern ships have a lifespan of 25 to 30 years before corrosion, metal fatigue and a lack of parts render them uneconomical to operate.[1] Ship-breaking allows the materials from the ship, especially steel, to be recycled and made into new products. This lowers the demand for mined iron ore and reduces energy use in the steel making process. Fixtures and other equipment on board the vessels can also be reused. While ship-breaking is sustainable, there are concerns about the use of poorer countries without stringent environmental legislation. It is also labor-intensive, and considered one of the world's most dangerous industries.[2]

In 2012, roughly 1,250 ocean ships were broken down, and their average age was 26 years.[3][4] In 2013, the world total of demolished ships amounted to 29,052,000 tonnes, 92% of which were demolished in Asia. As of January 2020, India has the largest global share at 30%;[5] followed by Bangladesh, China and Pakistan.[6] Alang, India currently has the world's largest ship graveyard,[5] followed by Chittagong Ship Breaking Yard in Bangladesh and Gadani in Pakistan.[6]

The largest sources of ships are states of China, Greece and Germany respectively, although there is a greater variation in the source of carriers versus their disposal.[7] The ship-breaking yards of India, Bangladesh, China and Pakistan employ 225,000 workers as well as providing many indirect jobs. In Bangladesh, the recycled steel covers 20% of the country's needs and in India it is almost 10%.[8]

As an alternative to ship-breaking, ships may be sunk to create artificial reefs after legally-mandated removal of hazardous materials, or sunk in deep ocean waters. Storage is a viable temporary option, whether on land or afloat, though all ships will be eventually scrapped, sunk, or preserved for museums.

6 0
3 years ago
An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve
lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is  that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa  against the claim Ha: u< 3.1 MPa

The alternate hypothesis is  that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68  falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0
3 years ago
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