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1 year ago

How can promoting total person development can benefit an organization.

Physics

2 answers:

Alex787 [66]1 year 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]1 year 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. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π

b. vy = 4π cos (4πt + π/2)
vy = 0

c. m = sin(4πt + π/2) / [πt + cos(4πt + π/2)]

d. m = sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax

h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax

i. s(t) = [x(t)^2 + y(t)^2]^(1/2)

h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt

k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.

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2 years ago

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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:

ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

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2 years ago

A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme

GaryK [48]

Given that,

Distance in south-west direction = 250 km

Projected angle to east = 60°

East component = ?

since,

cos ∅ = base/hypotenuse

base= hyp * cos ∅

East component = 250 * cos 60°

East component = 125 km

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2 years ago

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Caelyn wanted to find out what shampoo made her hair the shiniest . Everyday she washed her hair with different shampoos and the

Arte-miy333 [17]

Answer:

IV: type of shampoo used

DV: what shampoo made her hair the skinniest

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2 years ago

Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass

zvonat [6]

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value into the formula

$8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v$

$v=\dfrac{-19.375}{14.25}$

$v=-1.35\ m/s$

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2$

$K.E_{i}=239.96\ J$

We need to calculate the final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2$

$K.E_{f}=12.98\ J$

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

$\Delta K.E=K.E_{f}-K.E_{i}$

Put the value into the formula

$\Delta K.E=12.98-239.96$

$\Delta K.E=-226.98\ J$

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

8 0

1 year ago

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