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IrinaK [193]
3 years ago
8

How can a organization encourage total person development among their employees.

Physics
1 answer:
Sedbober [7]3 years ago
8 0

Answer:

Paying for employees seminars and workshops related to their careers

Explanation:

To motivate personal development among employees, several things can be done. Among them, giving employees chance to present their own solutions to problems, exposing the employees to several global challenges and how to handle them, paying for employees seminars and workshops related to their own careers for professional development among other things.

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A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 x 106 meters, find the mass of
melisa1 [442]

Answer:

7.36 × 10^22 kg

Explanation:

Mass of the man = 90kg

Weight on the moon = 146N

radius of the moon =1.74×10^6

Weight =mg

g= weight/mass

g= 146/90 = 1.62m/s^2

From the law of gravitational force

g = GM/r^2

Where G = 6.67 ×10^-11

M = gr^2/G

M= 1.62 × (1.74×10^6)^2/6.67×10^-11

= 4.904×10^12/6.67×10^-11

=0.735×10^23

M= 7.35×10^22kg. (approximately) with option c

3 0
3 years ago
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What's effect called?<br>​
k0ka [10]

Answer:

what do u mean

Explanation:

8 0
3 years ago
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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
Where did the element heavier than hydrogen and helium come from?
Snezhnost [94]

They were produced inside stars.

6 0
3 years ago
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A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.
AveGali [126]

Answer:

The magnitude of electric force is  7.2\times10^{-3} N

Explanation:

Coulomb's Law:

The force of attraction or repletion is

  • directly proportional to the products of charges i.e F\propto q_1q_2
  • inversely proportional to the square of distance i.e F\propto \frac{1}{r^2}

\therefore F\propto \frac{q_1q_2}{r^2}

\Rightarrow F=k \frac{q_1q_2}{r^2}    [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force  be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5

we know,

cos \theta = \frac{base }{hypotenuse}

\Rightarrow cos \theta = \frac{4 }{r}

Total force F_Q = 2.F_1 cos\theta \hat{i}

                         =2k\frac{Qq_1}{r^2} cos\theta \hat i

                         =2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i

                         =8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i     [ r=5]

                         =7.2\times10^{-3}\hat i   N

The magnitude of electric force is  7.2\times10^{-3} N

                         

3 0
3 years ago
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