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Tom [10]
3 years ago
13

When you are in charge of running a meeting, "winging it" with minimal planning is considered:

Engineering
1 answer:
amm18123 years ago
3 0

Answer:

The correct option is;

B) Counterproductive

Explanation:

An effective requires adequate preparations, with agenda of the meeting circulated among participants. The items to be discussed should be known before hand by the participants all of whom will have had adequate background information and relevant consultations with their constituencies  so as to be able to add effectively to the meeting.

Other attributes of an effective meeting includes;

1. Ensure, notes of the meeting are taken down personally

2. Outside discussions should be made at the parking lot

3. Important decisions or points to be made should have been communicated to other members of the meeting

4. Members to attend the meeting should be well known and list of attendees reviewed

5. Meetings should be kept to schedule

6. Ensure meeting conclusions and decisions are followed up

7. Written agenda should be available

Therefore "winging it" with minimal planning is considered counterproductive.

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An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

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What type of test can show a chemical engineer if a material remains in a system and accumulates or if it moves right through?
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A chemical engineer can clearly see from this kind of test if a substance stays in a system and builds up or if it just passes through.

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The true statement about the dot plot is 1 has 4 and 0 dots.

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