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hichkok12 [17]
4 years ago
8

What is a core self-evaluation, include identifying and explaining the components of core self-evaluation. And, how a group lead

er can use CSEs to create a more effective unit
Engineering
2 answers:
Brilliant_brown [7]4 years ago
8 0

Answer and Explanation:

Core self-evaluations (CSEs) stands for a wide personality trait that comprises of 4 positive individual traits, namely:

(1) self-efficacy

(2) self esteem

(3) locus of control

(4) emotional stability.

Baiscally, when people have a positive evaluation about themselves, or quality core self-evaluation, they believe that they are worthy and fit for a task. They trust their capability and effectiveness. This leads to some implications in their managers duties and careers, which could either be positive or negative.

A group leader can use CSEs to create a more effective unit by implementing the ten items points of Generalized Self Efficacy Tool to test the self efficiency of individual personnels in that unit.

Fudgin [204]4 years ago
6 0

Answer: Core self evaluation CSE is a way we assess our lives and attitude to our environment including people around us and our work.

Explanation:

The key components of Core self evaluation are:

Self-efficacy: is the ability to be effective in different situations or how robust we are in coping with unfamiliar situations.

Self-esteem: is a positive belief in oneself, in one’s abilities and expectations of ourselves.

Locus of control: this is how much control we feel we have over what happens to us or what we do. People who feel more in control of the events that happen to them have a more positive and proactive attitude than those who do not.

Emotional stability/Neuroticism: Emotional stability is the ability to control negative emotions. Mood swings and anxiety are signs of low emotional stability or high neuroticism.  

A leader can use CSEs to build a more effective unit by identifying the key components each team member has, lacks or can improve on. Individuals who have high self efficacy, self esteem, locus of control and emotional stability are productive, positive and valuable team members. The leader can encourage each person to improve on their weak points as it is in the interest of the team.  

Also, the leader can assign tasks that use the strength of each person. For example, the most emotional stable team member can be in charge of crisis management.

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Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment
aleksandrvk [35]

Answer:

//This Program is written in C++

// Comments are used for explanatory purpose

#include <iostream>

using namespace std;

enum mailbox{open, close};

int box[149];

void closeAllBoxes();

void OpenClose();

void printAll();

int main()

{

closeAllBoxes();

OpenClose();

printAll();

return 0;

}

void closeAllBoxes()

{

for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150

{

box[i] = close; //Close all boxes

}

}

void OpenClose()

{

for(int i = 2; i < 150; i++) {

for(int j = i; j < 150; j += i) {

if (box[j] == close) //Open box if box is closed

box[j] = open;

else

box[j] = close; // Close box if box is opened

}

}

// At the end of this test, all boxes would be closed

}

void printAll()

{

for (int x = 0; x < 150; x++) //use this to test

{

if (box[x] = 1)

{

cout << "Mailbox #" << x+1 << " is closed" << endl;

// Print all close boxes

}

}

}

Explanation:

7 0
3 years ago
Clothing made of several thin layers of fabric with trapped air in between, often called ski clothing, is commonly used in cold
Pavlova-9 [17]

Answer:

Q=127.66W

L=9.2mm

Explanation:

Heat transfer consists of the propagation of energy in the form of heat in different ways, these can be convection if it is through a fluid, radiation through electromagnetic waves and conduction through solid solids.

To solve any problem related to heat transfer, the general equation is used

Q = delta / R

Where

Q = heat

Delta = the temperature difference

R = is the thermal resistance by conduction, convection and radiation

to solve this problem we propose the previous equation

Q = delta / R

later we find R

R=[tex]r=\frac{6L1}{AK1} +\frac{5L2}{AK2}+\frac{1}{Ah}

R=\frac{6(0.0001)}{(1.25)(0.026)} +\frac{5(0.012)}{(1.25)(0.026)}+\frac{1}{(25)(1.25)} =0.235 K/w

Q=(25-(-5))/0.235=127.66W

part b

we use the same ecuation with Q=127.66

Q = delta / R

ΔR=\frac{L}{KA} +\frac{1}{hA} \\R=\frac{L}{(0.035)(1.25)} +\frac{1}{(25)(1.25)}\\ R=22.85L+0.032\\Q=(T1-T2)/R\\\\127.66=(25-(-5))/(22.85L+0.032)\\solving for L\\L=9.2mm

6 0
3 years ago
A manometer connected to a Pitot-static tube has a difference in the height of the two columns of water of 10 cm when the Pitot-
kaheart [24]

Answer:

v = 40 m/s

Explanation:

Difference in the height of the two columns:

h_m = \triangle h = 10 cm \\h_m = 0.1 m

Air density, \rho_{air} = 1.225 kg/m^3

Water density, \rho_{w} = 1000 kg/m^3

The velocity of air flow is given by the equation:

v =  \sqrt{2gh_m (\frac{p_w}{p_{air}} - 1) } \\\\v =  \sqrt{2*9.81*0.1 (\frac{1000}{1.225} - 1) }\\\\v = 40 m/s

7 0
4 years ago
Imagine the arc of a football as it flies through the air. How does this motion illustrate classical mechanics?
BigorU [14]

Answer: Maybe A

Explanation:

7 0
1 year ago
Read 2 more answers
An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% fil
den301095 [7]

Answer:

Explanation:

From the given information:

Addition of all the materials = 65+ 36+ 5 +6 = 112  which is higher than 100 percentage; SO we need to find;

The actual percentage of each material which can be determined as follows:

Percentage of the coarse aggregate will be = 65 × 112/100

= 72.80%

Percentage of the Fine aggregate will be = 36 × 112/100

= 40.32%

Percentage of the filler  will be = 5 × 112/100

= 5.6%

Percentage of the   asphalt binder will be = 6 × 112/100

= 6.72 %

So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:

Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)

Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)

Gt = 100/ (50.722)

Gt =1.972

Also;Given that the bulk density = 143.9 lb/ft³

LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft

Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43

=2.305

The percentage of air  void  = (Gt -Gm )× 100/ Gt

= (1.972 - 2.305) ×  100/ 1.972

= -16.89%

The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)

= 617.647/54.42

= 11.35%

Thus; the percentage voids in mineral aggregate =  -16.89% + 11.35%

the percentage voids in mineral aggregate = -5.45%

The percent voids filled with asphalt. = 100 × 11.35/-5.45

The percent voids filled with asphalt = - 208.26 %

7 0
3 years ago
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