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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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Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
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The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

4 0
3 years ago
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