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Dominik [7]
1 year ago
6

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.

Read more about communication

brainly.com/question/26152499

#SPJ1

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The principal value of a Pareto diagram is as a
vlada-n [284]

The Pareto principle is that most things in our life are not commonly distributed.

<u>Explanation:</u>

Pareto chart shows that most of the things which we have in our life and the resources in our life are not equally distributed. The ratio is not always 50:50 according to this principle.

The most important use of a Pareto diagram is to show the most important factor among the set of factors that have been shown. Along with that it also shows the sources which lead to the common defects in the system and tries to solve those defects which occur most often.

4 0
3 years ago
Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read
damaskus [11]

Answer:

P_2-P_1=27209h

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

P_1+p_1gh_1=p_2_gh_2+P_2

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h

P_2-P_1=27209h

3 0
3 years ago
What is the pressure at the bottom of a 25 ft volume of hydraulic fluid with a weight density of 55 lb/ft3 a. 114.6 psi b. 1375p
Assoli18 [71]

Answer:

d) 9.55 psi

Explanation:

pressure at the bottom is =ρgh

weight density is ρg=55 lb/ft³

h=25ft

pressure at the bottom is =55\times 25

                                  =1375psf

1 ft = 12 inch

pressure at bottom =\frac{1375}{12^2}

                                = 9.55 psi

so, answer will be option (d) which is 9.55 psi

3 0
3 years ago
A fluid of density 900 kg/m3 passes through a converging section of an upstream diameter of 50 mm and a downstream diameter of 2
NISA [10]

Answer:

Q= 4.6 × 10⁻³ m³/s

actual velocity will be equal to 8.39 m/s

Explanation:

density of fluid = 900 kg/m³

d₁ = 0.025 m

d₂ = 0.05 m

Δ P = -40 k N/m²

C v = 0.89

using energy equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2

under ideal condition v₁² = 0

v₂² = 88.88

v₂ = 9.43 m/s

hence discharge at downstream will be

Q = Av

Q = \dfrac{\pi}{4}d_1^2 \times v

Q = \dfrac{\pi}{4}0.025^2 \times 9.43

Q= 4.6 × 10⁻³ m³/s

we know that

C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s

hence , actual velocity will be equal to 8.39 m/s

6 0
3 years ago
Brainly and points if you want
Tju [1.3M]

Answer:

thank you

Explanation:

have a nice day

8 0
2 years ago
Read 2 more answers
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