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AleksandrR [38]
3 years ago
9

Which of the following is considered plagiarism?

Physics
2 answers:
-Dominant- [34]3 years ago
8 0

Answer:

Copying materials from a source text, supplying proper documentation, but leaving out quotation marks

Explanation:

Plagiarism is when you take somebody's ideas and write them off as your own without giving them credit.

DerKrebs [107]3 years ago
6 0

The correct answer is B. Copying materials from a source text, supplying proper documentation, but leaving out quotation marks

Explanation:

Plagiarism occurs if you take ideas or information from a source or author without acknowledging this authorship. This is unethical and disrespectful because you are showing the author's ideas as yours. Examples of plagiarism include copying part of the material or source without using quotation marks and adding a citation, and paraphrasing or summarizing information without adding a proper citation.

According to this, the situation that would be considered plagiarism is "Copying materials from a source text, supplying proper documentation, but leaving out quotation marks" because in this case the original author is not being acknowledged due to the lack of quotation marks and a proper citation such as the authors last name, date of publication and page number. Also, citations and similar are require as only proper documentation is not enough to show to whom the ideas belong.

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A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
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lidiya [134]
Divide 24 by 12.
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6 0
3 years ago
A rock is thrown with a force of 500 N and an acceleration is 75 m/s^2. What is its mass?
artcher [175]

Answer:

We conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

Explanation:

Given

  • Force F = 500 N
  • Acceleration a = 75 m/s²

To determine

Mass m = ?

Important Tip:

  • The mass of a rock can be found using the formula F = ma

Using the formula

F = ma

where

  • F is the force (N)
  • m is the mass (kg)
  • a is the acceleration (m/s²)

now substituting F = 500, and a = 75 m/s² in the formula

F = ma

500 = m(75)

switch sides

m\left(75\right)=500

Divide both sides by 75

\frac{m\cdot \:75}{75}=\frac{500}{75}

simplify

m=\frac{20}{3}

m=6.7 kg

Therefore, we conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

7 0
3 years ago
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