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Tanya [424]
3 years ago
15

CAD(computer-aided design) software and is used in__________and __________that show how to construct an object. Technical drawin

gs show in detail how the pieces of something relate to each other.
Engineering
1 answer:
8_murik_8 [283]3 years ago
5 0

Answer:

Plans; blueprints.

Explanation:

In Engineering, it is a common and standard practice to use drawings and models in the design and development of various tools or systems that are being used for proffering solutions to specific problems in different fields such as engineering, medicine, telecommunications and industries.

Hence, a design engineer make use of drawings such as pictorial drawings, sketches, or technical drawing to communicate ideas about a design to others, to record and retain informations (ideas) so that they're not forgotten and to analyze how different components of a design work together.

Technical drawing is mainly implemented with CAD (computer-aided design) software and is typically used in plans and blueprints that show how to construct an object.

Additionally, technical drawings show in detail how the pieces of something (object) relate to each other, as well as accurately illustrating the actual (true) shape and size of an object in the design and development process.

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Assume that in orthogonal cutting the rake angle is 15° and the coefficient of friction is 0.15. a. Determine the percentage cha
alexira [117]

Answer:

Δr=20.45 %

Explanation:

Given that

Rake angle α =  15°

coefficient of friction ,μ = 0.15

The friction angle β

tanβ = μ

tanβ = 0.15

β=8.83°

2φ +  β - α  = 90°

φ=Shear angle

2φ + 8.833° - 15° = 90°

φ = 48.08°

Chip thickness r given as

r=\dfrac{tan\phi}{cos\alpha +sin\alpha\ tan\phi}

r=\dfrac{tan48.08^{\circ}}{cos15^{\circ} +sin15^{\circ}\ tan48.08^{\circ}}

r=0.88

New coefficient of friction ,μ'  = 0.3

tanβ' = μ'

tanβ' = 0.3

β'=16.69°

2φ' +  β' - α  = 90°

φ'=Shear angle

2φ' + 16.69° - 25° = 90°

φ' = 49.15°

Chip thickness r' given as

r'=\dfrac{tan\phi'}{cos\alpha +sin\alpha\ tan\phi'}

r'=\dfrac{tan44.15^{\circ}}{cos49.15^{\circ} +sin49.15^{\circ}\ tan44.15^{\circ}}

r'=0.70

Percentage change

\Delta r=\dfrac{r-r'}{r}\times 100

\Delta r=\dfrac{0.88-0.70}{0.88}\times 100

Δr=20.45 %

8 0
3 years ago
Brainly is pretty boring
sdas [7]
That’s cool lol ????
8 0
3 years ago
Read 2 more answers
What is pre-flush and post flush in petroleum engineering?
cluponka [151]

Answer:

Pre-Flush:

It is also known as In-line Equalization. In this stage of flow equalization, all the flow passes through the equalization basin. It helps in reduction of fluctuation in pollutants concentration and flow rate and helps to control short term surges with the use of basin.

Post-Flush:

Another name for this stage is Off-line Equalization. In this stage, only overflow above a predetermined standard is diverted into the basin. It helps in reducing the fluctuations in loading by a considerable amount and helps to reduce the pumping requirement. It is basically used to capture "first flush" from combined collection systems.

5 0
3 years ago
The in situ moisture content of a soil is 18% and the moist (total) unit weight is 105 pcf. The soil is to be excavated and tran
densk [106]

Answer: Volume= 1.16 yd³

Explanation:

We are given the in-situ moisture content= 18% and moist unit weight= 105 pcf

First we find out the dry unit weight

Dry unit weight= Moist unit weight / (1+ moisture content)

Dry unit weight= \frac{105}{1.18}

Dry unit weight= 88.983 pcf

Now, to find the volume ( in terms of each cubic yard i.e 1 yd³ we have

Volume = Dry unit weight compacted / Dry unit weight (in-situ) * 1 yd³

Volume= \frac{103.5}{88.983} * 1

Volume= 1.16 yd³

6 0
4 years ago
Sandra is holding a piece of tissue that has a negative charge and a feather that has a neutral charge. The two objects have sim
Murljashka [212]
The correct answer would be d
The objects will not move towards or away from each other
Hope this helps
4 0
2 years ago
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