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Alex777 [14]
2 years ago
12

An object's movement around an internal axis is _____.

Physics
1 answer:
zaharov [31]2 years ago
4 0

Answer:

brooo it is rotation .............

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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
3 years ago
91.0x26x504 ...............
Hoochie [10]

Answer:

1192464

Explanation:

91.0 times 26 = 2366

2366 times 504 = 1192464

7 0
3 years ago
Read 2 more answers
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
Luden [163]

Answer:

The ball will be attracted to the negatively charged plate. It'll touch and pick up some electrons from the plate so that the ball becomes negatively charged. Immediately the ball is repelled by the negative plate and is attracted to the positive plate. The ball gives up electrons to the positive plate so that it is positively charged and suddenly attracts to the negative plate again, flies over to it and picks up enough electrons to be repulsed by negative plate and again to the positive plate and that continues.

8 0
3 years ago
A uniform plank 8.00 m in length with mass 50.0 kg is supported at two points located 1.00 m and 5.00 m, respectively, from the
andreev551 [17]

To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.

According to Newton's second law we have to

F = mg

where,

m= mass

g = gravitational acceleration

For the balance to break, there must be a mass M located at the right end.

We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.

In this way, applying the static equilibrium equations, we have to sum up torques at point B,

\sum \tau = 0

Regarding the forces we have,

3Mg-1mg=0

Re-arrange to find M,

M = \frac{m}{3}

M = \frac{50}{3}

M = 16.67Kg

Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg

8 0
3 years ago
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