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liraira [26]
3 years ago
7

Please help due today

Engineering
2 answers:
nika2105 [10]3 years ago
8 0
Calculator
Telescope
Ruler
Microscope
aksik [14]3 years ago
5 0
Performs math operations = calculator
view planets, stars, and objects in space = telescope / measure the length of a piece of paper = ruler / study a petri dish containing bacteria = microscope
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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
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A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during
astra-53 [7]

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

\sigma_c = tensile stress = 200 Mpa

K = plane strain fracture toughness= 55 Mpa\sqrt{m}

\lambda= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for \lambda

Multiplying both sides of equation (1) by Y\sqrt{\pi\lambda}

\sigma_c Y\sqrt{\pi\lambda}=K   (2)

Sequaring both sides of equation (2):

(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2  

\sigma^2_c Y^2 \pi\lambda=K^2   (3)

Dividing both sides by \sigma^2_c Y^2 \pi we got:

\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2   (4)

Replacing the values into equation (4) we got:

\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

7 0
3 years ago
Ti-6Al-4V has a fracture toughness of 74.6 MPa-m0.5. How much stress (in MPa) would it take to fail a plate loaded in tension th
Nikitich [7]

Answer:

critical stress  = 595 MPa

Explanation:

given data

fracture toughness =  74.6 MPa-\sqrt{m}

crack length = 10 mm

f = 1

solution

we know crack length = 10 mm  

and crack length = 2a as given in figure attach

so 2a = 10

a = 5 mm

and now we get here with the help of plane strain condition , critical stress is express as

critical stress  = \frac{k}{f\sqrt{\pi a}}    ......................1

put here value and we get

critical stress  = \frac{74.6}{1\sqrt{\pi 5\times 10^{-3}}}

critical stress  = 595 MPa

so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.

plain stress condition occur in thin body where stress through thickness not vary by the thinner section.

6 0
3 years ago
What’s Statistics<br> What are the 2 Source of error in data collection
FromTheMoon [43]

Answer:

<em>The main sources of error in the collection of data are as follows : Due to direct personal interview. Due to indirect oral interviews. Information from correspondents may be misleading.</em>

4 0
3 years ago
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A design team is working on creating a new locker organizer. They have
astraxan [27]
A. Present a Solution
8 0
3 years ago
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