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antiseptic1488 [7]

3 years ago

9

Which is the most extreme system of social inequality?

1 answer:

Anna [14]3 years ago

5 0

Possible answers are:
Capitalism
Aristocracy
Oligarchy

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Standards for all managers ethical responsibilities are covered in a company's?

katen-ka-za [31]

Answer:

C. Code of ethics

Explanation:

The code of ethics of a company is the document where all the values that the members of the company should live and obbey, they are often guideliness on how to act and beheave and how to treat customers, providers and co-workers. IT also has all of the standards for all managers ethical responsabilities, and ethical responsabilities of the general of the employees of the company.

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3 years ago

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Potential Energy Equation (GPE = mgh)

Shkiper50 [21]

Answer:

10

Explanation:

weight=25x9.2=230

25x9.2x10=2300

2300/230=10

6 0

3 years ago

True or false: Scientific models are always built to scale, and are fully functional representations of the product they are tes

vodka [1.7K]

Answer:

False

Explanation:

Scientific models are not always built to scale and may not be fully accurate

3 0

2 years ago

What is a oxidation number

Scilla [17]

Oxidation number is the charge.

6 0

3 years ago

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Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to

Debora [2.8K]

Answer:

$\mu_{k} \approx 0.719$

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

$\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0$

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

$m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a$

$g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a$

$\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}$

$\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )$

$\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)$

$\mu_{k} \approx 0.719$

5 0

3 years ago