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3 years ago

When could you use the engineering design process in your own life?

Engineering

1 answer:

Luda [366]3 years ago

7 0

Answer:

You could use this process to make a new item or solve a problem that you could be having around the house

I hope this helps!

Explanation:

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What are the reasons why fine grained of alkali igneous rocks can not be used in cement

Luden [163]

Fine grained of alkali igneous rocks cannot be used in cement because of the volume expansion caused by the Alkali-silica reaction, fine-grained igneous rocks cannot be used as aggregates in cement.

<h3>What does fine grained mean in an igneous rock?</h3>

Extrusive igneous rocks have a fine-grained or aphanitic texture, with grains that are too small to see without a magnifying glass. The fine-grained texture suggests that the rapidly cooling lava did not have enough time to form large crystals. A petrographic microscope can be used to examine these tiny crystals.

The texture of an igneous rock (fine-grained vs coarse-grained) is determined by the rate at which the melt cools: slow cooling produces large crystals, while fast cooling produces small crystals.

The chemical reaction that occurs in both alkali cations and hydroxyl ions in the pore solution of hydrated cement paste and certain reactive silica phases present in concrete aggregates is known as the alkali-silica reaction (ASR).

Learn more about cement on:

brainly.com/question/14323034

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8 0

2 years ago

Your new mobile phone business is now approaching its first anniversary and you are able to step back and finally take a deep br

Leya [2.2K]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

7 0

3 years ago

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RSB [31]

Answer:

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Explanation:

3 0

3 years ago

Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a

Katyanochek1 [597]

Answer:

$K_{I}=5.21 MPa\sqrt{m}$

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

$K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}$ ................................1

here Y is the geometrical function

Y = $\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}$

Y = $\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}$

Y = $\frac{0.8792}{0.89}$

Y = 0.9878

so put here value in equation 1

$K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}$

$K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319$

$K_{I}$ = 5210.45 × 10³ $Pa\sqrt{m}$

$K_{I}$ = 5.21 MPa $\sqrt{m}$

5 0

3 years ago

Consider an ideal gas undergoing a constant pressure process from state 1 to state

Radda [10]

Answer:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy <em>at constant pressure</em>:

$ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}$

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

$\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,$

We obtain the expression to compute the specific entropy change:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Best regards.

6 0

4 years ago