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1 year ago

Invent five new communication method wired or wireless you think would be practical

Engineering

1 answer:

seropon [69]1 year ago

4 0

The five type of new communication method (both wire and wireless) are:

Social Media
SMS Text Messaging
Email Marketing
Direct Email
Voice Calling

<h3>What is a communication?</h3>

This means the act of giving, receiving and sharing of information as well as talking, writing, listening or reading.

The types of Communication includes:

Verbal Communication: It occurs when we engage in speaking with others.
Non-Verbal Communication: It occurs when what we speak often says more than the actual words.
Written Communication: It occurs in an actual text or writing.
Listening: It occurs when we pay attention to a speaker.

The wired communication involves transmission of data over a wire-based communication technology while a wireless communication involves transmission of data without the use of a cable or wire.

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Answer:

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Explanation:

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3 years ago

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3 years ago

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Trava [24]

Answer:

combining scientific knowledge, careful reasoning, and artistic invention in a flexible approach to problem-solving

Explanation:

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3 years ago

Which of the following would require the filing of an EIS? a. expansion of an airport in a national park b. allowing snowmobiles

alekssr [168]

Answer:

d. All of the above would require an EIS.

Explanation:

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3 years ago

A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.

denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

$\frac{M}{I}=\frac{F}{y}$

Here

M is the moment which is to be calculated
I is the moment of inertia given as

$I=\frac{bd^3}{12}$

Here

b is the breath given as 0.75"
d is the depth which is given as 8"

$I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4$

y is given as

$y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\$

Force is 50 ksi

$\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in$

The yield moment is 400 k.in.

Part b:

The strain is given as

$Strain=\frac{Stress}{Elastic Modulus}$

The stress at the station 2" down from the top is estimated by ratio of triangles as

$F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi$

Now the steel has the elastic modulus of E=29000 ksi

$Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in$

So the strain is $8.621 \times 10^{-4} in/in$

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

$shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi$

The plastic moment is 600 ksi.

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3 years ago