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4 years ago

An integrated circuit is a set of microscopic electronic circuits etched onto a thin slice of ________ material such as silicon.

Engineering

1 answer:

mylen [45]4 years ago

4 0

Answer: Semiconducting

Explanation:

Integrated circuit is define as set of microscopic electronic circuit that etched into thin slice if semiconducting material like silicon.

Integrated circuit are commonly use in the microchip and computer chip. Some of the integrated circuits are used to represent processing task and some collecting the input in the circuits.

Semiconductors like silicon are basically used to fabricate chip. Some of the essential properties of semiconductor are use to enhance the electronic path and components.

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When the same type of arc is used in more than one place, what note is included with the dimension?

Ghella [55]

Answer:

Numerals

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If a drawing is to be complete, so that the object represented by the drawing can be made as intended by the designer, it must tell two complete stories. It tells this with views, which describe the shape of the object, and with dimensions and notes, which give sizes and other information needed to make the object.

Therefore, your next step is to learn the basics of dimensioning. In that way, you will understand not only how to interpret a drawing to get the information you need, but also how to dimension your sketches so that they can be used to communicate size information to others.

Numerals

It may seem a bit basic, but a few exercises with the shapes of numbers come before dimensioning. The reason for such a review is simply that incorrectly or carelessly made numbers on a drawing or sketch can easily be misinterpreted by someone on the job. That can be costly.

Therefore, the study of numbers forms is justified.

The number forms presented here have been determined to be the most legible, and are used by industry nationwide. The United States standardized 1/8” vertical

Explanation:

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3 years ago

Why become an Android programmer?

Tpy6a [65]

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3 years ago

Lois is in high school and enjoys studying math and science subjects. She aspires to get into the robotics industry after comple

Inessa [10]

Answer:

d.partivipate in a robotic club

hope helps

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3 years ago

A/an _____________ comes on if one of the dual hydraulic brake systems should fail or, in some vehicles, if the brake fluid is l

jolli1 [7]

Answer: Dash light

Explanation:

A dash light, usually labeled "BRAKE" will illuminated if a braking system fails, or if the brake fluid is low.

3 0

3 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

4 years ago