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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

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In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length
masya89 [10]

Answer:

Time period  = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration  9.4 \times 10^{-5} kg/s

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration = 9.4\times 10^{-5} \times 62 = 582.8\times 10{-5} kg/s

total volume of air  infiltration= \frac{582.8\times 10{-5}}{1.156} = 5.04\times 10^{-3} m^3/s

Time period = \frac{210}{5.04\times 10^{-3}} = 41654.08 s

3 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

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3 years ago
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Answer:

\frac{1.2}{120}

0.01

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What engineers call moment, scientists call
Svet_ta [14]

Answer:

yes

Explanation:

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Answer:

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