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.
Answer:
Work done to pull the piano upwards is 401250 J
Explanation:
Work is done against the gravity to pull the piano upwards
So here we can say that work done is

here we know that

also we know that
H = 75 m
now we have


To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.
PART A) We will begin by finding the two net distances.

And the distance 'd' is



Through the free-body diagram the tension components are given by


Here we can watch that,

Dividing both expression we have that,

Replacing the values,


PART B) Using the vertical component we can find the tension,




The four distinct charges' combined potentials make up the potential in the square's center. The amount of the charge and the distance from the charge both affect the potential caused by a point charge.
Therefore, the center's total potential is V=4V1=ks4 q.
<h3>What is a charge?</h3>
Due to the physical characteristic of electric charge, charged material experiences a force when it is exposed to an electromagnetic field. An object that has no net charge is said to be neutral. Classical electrodynamics is the name given to an earlier theory of the interactions of charged particles.
You can have positive or negative electric charges (commonly carried by protons and electrons respectively). opposing charges attract one another whereas similar charges repel one another.
To learn more about charge from the given link:
brainly.com/question/9194793
#SPJ4
Answer:
Lilly's speed is two times John's speed.
Explanation:
m = Mass
a = Acceleration
t = Time taken
u = Initial velocity
v = Final velocity
The force they apply on each other will be equal




Hence, Lilly's speed is two times John's speed.