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marishachu [46]
3 years ago
11

According to the place-time model of interaction, when people are in the same place, but at different times, this is an example

of what type of communication?
Engineering
1 answer:
STALIN [3.7K]3 years ago
3 0

Answer: Shift work communication.

Explanation:

Shift work communication is a type of communication that exist between shift workers, or people at different time zones.

Shift work consists of a three eight hours shift within 24 hours.

Keeping up with a shift work communication requires planning, time management and commitment. It also requires using various communication channels.

You might be interested in
A simple Brayton cycle using air as the working fluid has a pressure ratio of 10.9. The minimum and maximum temperatures in the
ladessa [460]

Air pressure at the end of the Turbine exit is : 730.57 K

<u>Given data :</u>

Pressure ratio ( p₂ / p₁ )= 10.9

minimum temperature ( T₁ ) = 280 K

maximum temperature ( T₃ ) = 1410 K

Assuming :

constant specific heat and efficiency of 100%

<h3 /><h3>Determine the Air pressure at the end of the turbine exit </h3>

For air :

Cp = 1.005 kJ/kg.k,  Cv = 0.718 kJ/kg.k,  v = 1.4

Given that efficiency for compressor and turbine is 100% the process ( 1-2 , 3 - 4 ) will all be isentropic

We will Apply the formula below

\frac{T_{2} }{T_{1} } = ( \frac{p_{2} }{p_{1} } )^{\frac{v-1}{v} } = ( \frac{V_{1} }{V_{2} } )^{v-1}

Insert values into equation ( 1 )

T₂ = 551.147 K  ( temperature at compressor exit )

<u />

<u>Next : Determine the </u><u>value </u><u>of the temperature at </u><u>Turbine exit </u><u>( T₄ )</u>

T₃ / T₄ = 10.7^^{\frac{1.4-1}{1.4} }

Therefore : T₄ = 1410 / 10.7^0.286

                        = 1410 / 1.93

                        = 730.57 K

Hence we can conclude that the Air pressure at the end of the Turbine exit is :  730.57 K .

Learn more about Turbine : brainly.com/question/894340

6 0
2 years ago
A mixture of two gases has a total mass of 80 kg. One gas has a specific volume of 0.8 m^3/kg and occupies a volume of 40 m^3. T
Nady [450]

Answer:

specific volume = 1.025 m³/kg

Explanation:

given data

total mass m1 +m2 = 80 kg

specific volume = 0.8 m³/kg

occupies volume v1 = 40 m³

other gas specific volume  = 1.4 m³

to find out

How much volume is occupied by the second gas and what is the specific volume of the mixture

solution

we know that density is reciprocal of specific volume and here gas is not interacting

so total specific volume is assume so ratio is total volume to total mass

and

specific volume = \frac{1}{\rho}

here ρ is density

so ρ1  =   \frac{1}{0.8} = 1.25 kg/m³

and ρ2  =   \frac{1}{1.4} = 0.714 kg/m³

and

so m1 = ρ1v1 = 1.25 × 40 = 50 kg

and m2 = 80 - 50 = 30 kg

so

v2 = \frac{m2}{\rho 2}

v2 =  \frac{30}{0.714} = 42 m³

so volume occupied by second das = 42 m³

and

specific volume of mixture will be

specific volume of mixture = \frac{v1+v2}{m1+m2}

specific volume = \frac{42+40}{50+30}

specific volume = 1.025 m³/kg

3 0
3 years ago
Write a single statement to print: user_word,user_number. Note that there is no space between the comma and user_number. Sample
kykrilka [37]

Answer:

cout<<"''<<user_word<<"' "<<user_number;

Explanation:

The above question was answered using C++ programming language.

The keyword cout represents print and it carries out print operation only.

It prints all variable in front of it.

Assume the values of user_word and user_number to be Charles and 20, respectively.

The output of the above instruction would be

'Charles' 20 just as it is in the sample output in the question.

In java programming language, it is

System.out.print("'"+user_word+"' "+user_number);

In Qbasic, it is

PRINT "'"+user_word+"' "+ user_number

8 0
3 years ago
Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0-cm
igor_vitrenko [27]

Answer:

velocity and pressure in a 2.6-cm:

P2 = 2.53x10^5Pa, v2 = 1.18m/s

Explanation:

Pressure = P, Velocity = v, Height = h, Diameter = d, Radius= r, Area = A

Area = πr^2

From the question:

v1 = 0.5m/s

d1 = 4cm = 0.04m

r1 = d1/2 = 0.04/2 = 0.02m

Since water was pumped from basement, h1 = 0m

P1 = 3.03x10^5 Pa

A1 = π×0.02×0.02

A1 = 0.0004πm^2

v2 = unknown

d2 = 2.6cm = 0.026m

r2 = d2/2 = 0.026/2 = 0.013m

h2 = 5m

P2 = unknown

A2 = π×0.013×0.013

A2 = 0.000169πm^2

Using continuity equation:

A1v1 = A2v2

0.0004π * 0.5 = 0.000169π * v2

v2 = (0.0004π * 0.5)/(0.000169π)

v2 = 1.18m/s

Applying a Bernoulli principle

P + 1/2*density*v^2 + density*g*h =C

C = constant

P1 + 1/2*density*v1^2 + density*g*h1

= P2 + 1/2*density*v2^2 + density*g*h2

Let g = 9.81m/s

density of water = 1000kg/m^3

(P1-P2) = 1/2* density(v2^2 - v1^2) +(density*g*h2) - (density*g*h1)

(P1-P2) = 1/2* density(v2^2 - v1^2) +

density* g(h2-h1)

(3.03x10^5 - P2)= 1/2*1000 (1.18^2-0.5^2) + 1000(9.81(5-0))

(3.03x10^5 - P2) = 500(1.3924-0.25) + 49050

3.03x10^5 - P2 = 571.2 + 49050

3.03x10^5 - P2 = 49621.2

3.03x10^5 - 49621.2 = P2

P2 = 253378.8

P2 = 2.53x10^5Pa

P2 = 2.53x10^5Pa, v2 = 1.18m/s

7 0
3 years ago
4. Lockout/tagout (LOTO) is a safety procedure that ensures dangerous machines are properly shut off and not started up again pr
klemol [59]

Answer:true

Explanation:

5 0
3 years ago
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