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2 years ago

How can promoting total person development can benefit an organization.

Physics

2 answers:

Alex787 [66]2 years ago

5 0

Promoting total person development can benefit an organization in the following way

Explanation:

1.Promoting personal development in the workplace

All images courtesy of Forbes Councils members.
Create Growth Plans. ...
Build In 'Growth Time' ...
Set 90-Day Learning Goals. ...
Offer Innovative Learning Experiences. ...
Map Out The Far-Reaching Benefits. ...
Make Development Opportunities 'One Size Fits One' ...
Take An Active Interest In Personal Development. ...
Think Beyond Formal Training.

2. Create a development plan. Help your employees establish goals that are aligned with their strengths, interest and experience, as well as with the overall business strategy. Establish goals and expectations to help them set their sights on career opportunities.

3. 7 Things Every Great Boss Should Do

Acknowledge. When things are going well in your organization, let people know--early and often. ...
Motivate.
Communicate. Communicate clearly, professionally, and often. ...
Trust. Learn to trust your employees. ...
Develop. Set up your employees for success, not failure. ...
Direct.
Partner.

4.Human relations skills such as communication and handling conflict can help us create better relationships. ... Since many companies' organizational structures depend upon people working together, positive human relations skills reduce conflict in the workplace, thereby making the workplace more productive.

5.Improves effectiveness in business

In business, personal development improves effectiveness. It empowers staff to produce better results and meet their targets. For excellent results, an employer or a business owner must have an energetic and a productive team.

mel-nik [20]2 years ago

5 0

Promoting the person development works as a motivation to work harder and with more efficiency to achieve the goals of the organisation.

Explanation:

Human relations abilities, for example, correspondence and dealing with struggle can assist us with making better connections. Since numerous organizations' authoritative structures rely on individuals cooperating, positive human relations aptitudes lessen strife in the working environment, subsequently making the work environment more productive.

Opportunities for development and improvement assist representatives with growing their insight, abilities and capacities, and apply the capabilities they have picked up to new circumstances.

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A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

ollegr [7]

The answer:
the relationship between elementary charge, potential difference and electrical potential energy is given by
E= qV
E: lectrical potential energy
q: elementary charge
V: potential difference

but we have e=abs val(q)=3
so we have E= qV=3ex4.5V=13.5 eV

the answer is (4)13.5 eV

8 0

2 years ago

If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

Paraphin [41]

Efficiency η of a Carnot engine is defined to be:
η = 1 - Tc / Th = (Th - Tc) / Th 
where 
Tc is the absolute temperature of the cold reservoir, and 
Th is the absolute temperature of the hot reservoir. 

In this case, given is η=22% and Th - Tc = 75K 
Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). 

Th = (Th - Tc) / η 
Th = 75 / 0.22 = 341 K (rounded to closest number) 
Tc = Th - 75 = 266 K 

Lower temperature is Tc = 266 K 
Higher temperature is Th = 341 K

6 0

2 years ago

A 6.0-ohm resistor that obeys Ohm’s Law is connected to a source of variable potential difference. When the applied voltage is d

leva [86]

Is halved. A 6Ω resistor connected to a voltage source which voltage is decreased from 12V to 6V the current passing through the resistor is halved.

The key to solve this problem is applying Ohm's Law V = R I, clearing I from the equation, we obtain I = V/R. Then, the current is directly proportional to the voltage and inversely proportional to the resistance.

V = 12V and R = 6Ω

I = 12V/6Ω = 2A

V = 6V and R = 6Ω

V = 6V/6Ω = 1A

As we can see the current is halved if the voltage descreased from 12V to 6V

5 0

2 years ago

A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a

Korvikt [17]

Answer:

$v^2=v_o^2-2\times a\times (v_o.t)$

Explanation:

Given:

Initial velocity of the vehicle, $v_o$

distance between the car and the tree, $x_f$

time taken to respond to the situation, $t$

acceleration of the car after braking, $a$

Using equation of motion:

$v^2=u^2+2a.s$ ..............(1)

where:

$v=$ final velocity of the car when it hits the tree

$u=$ initial velocity of the car when the tree falls

$a=$ acceleration after the brakes are applied

$s=$ distance between the tree and the car after the brakes are applied.

$s=v_o\times t$

Now for this situation the eq. (1) becomes:

$v^2=v_o^2-2\times a\times (v_o.t)$ (negative sign is for the deceleration after the brake is applied to the car.)

5 0

2 years ago

A proton initially moves left to right along the x-axis at a speed of 2.00 x 103 m/s. It moves into an uniform electric field, w

FinnZ [79.3K]

Answer:

E = 1.04*10⁻¹ N/C

Explanation:

Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$

As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.

Replacing this values in the equation above, we can solve for a, as follows:

$a = \frac{vo^{2}*mp}{2*x} = \frac{(2.00e3m/s)^{2}}{2*0.2m} = 1e7 m/s2$

According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:

F = mp*a = q*E

For a proton, we have the following values:

mp = 1.67*10⁻²⁷ kg

q = e = 1.6*10⁻¹⁹ C

So, we can solve for E (in magnitude) , as follows:

$E = \frac{mp*a}{e} =\frac{1.67e-27kg*1e7m/s2}{1.6e-19C} = 1.04e-1 N/C$

⇒ E = 1.04*10⁻¹ N/C

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2 years ago