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motikmotik
3 years ago
11

PLEASE HELP ME WITH THIS ONE QUESTION

Physics
1 answer:
Vinil7 [7]3 years ago
7 0

Answer:

b

Explanation:

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ABC Corporation paid an executive $300,000.00 in annual compensation. A state court determines that this compensation was excess
Vesnalui [34]

Answer:

This a pure case of conflict of interests between the interest of the shareholders who are the original owners of the company and management's interest in earning much more,even if it at the expense of the shareholders.

Explanation:

Management is the entrusted with the day to day affairs of corporations.In carrying out their duty,they must have at the back of their minds that maximization of shareholder's wealth is of top priority.

However.some management teams in a bid to gain undue advantage set their remuneration below reasonable levels.

Ultimately,when this happens, their duty to watch over the investment of shareholders clashes with their interest for personal gains.

8 0
4 years ago
A car accelerates from 30 m/s to 50 m/s in 2 seconds. Calculate the cars acceleration
masha68 [24]

Answer:

applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²

5 0
3 years ago
If a car changes its velocity from 32 km/hr to 54 km/hr in 8.0 seconds, what is its acceleration?
Stella [2.4K]
Use the equation for the acceleration
A = final velocity - initial velocity divided by time final - time initial
A= 54 - 32 / 8 - 0
A= 22 / 8 
A= 2.75 m/s^2 
Hope this helps!
3 0
3 years ago
Read 2 more answers
DONT ANSWER WITH A LINK PLEASE I NEED AN ANSWER FROM SOMEONE!
rjkz [21]
At 4 m/s?

How do the two kinetic energies compare to one another? QUADRUPLES !

#3 What is the kinetic energy of a 2,000 kg bus that is moving at 30 m/s?

Potential energy
3 0
3 years ago
Read 2 more answers
When the distance between two charges is halved, the electrical force between them?
Llana [10]
If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
F= k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
r'= \frac{r}{2}
the magnitude of the force changes as follows:
F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k  \frac{q_1 q_2}{r^2}=4 F
so, the force increases by a factor 4.
3 0
3 years ago
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