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3 years ago

Which of the following is NOT included in an APA citation for a scholarly journal article? Please explain answer.

Physics

1 answer:

Lady bird [3.3K]3 years ago

5 0

Answer:

d) AN (article accession number)

Explanation:

APA format is commonly found in term papers, essays, and scientific papers. It consists of an in-text citation with a reference list to accompany it. The writer needs to cite the author and year of the work by putting them in parentheses separated by a comma (Author's name, year), for the references list the pattern is author, date, title, source. This format suffers small modifications depending on the kind of document you want to cite, for example, for a scholarly journal article te format is:

Author or authors (surname is followed by first initials). Year of publication of the article (in round brackets). Article title. Journal title (in italics). Volume of the journal (in italics). Issue number of the journal in round brackets (no italics). Page range of article. DOI.

An example is annexed as an image at the end.

<em>Therefore AN (article accession number), sometimes called a Document ID, is a unique number assigned by a particular database as an additional means of locating a specific article, this number is distinct and unrelated to a document’s DOI number, is NOT included in an APA citation for a scholarly journal article.</em>

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hichkok12 [17]

Gravitational potential energy i think

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3 years ago

You have two contentious friends, Chris and Pat, and you’ve quickly discovered that they need you to resolve arguments they have

pshichka [43]

Answer:

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points,

Explanation:

To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.

v_average = (d₂-d₁) / Δt

this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.

Therefore neither of them is right.

3 0

3 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

3 years ago

*science question pls answer quickly*

Mrrafil [7]

A is the answer you can believe me

3 0

2 years ago

Help me please, please !

steposvetlana [31]

Answer:

b and d

a, c, e, and f

Explanation:

Ideal gas law:

PV = nRT

Solving for temperature:

T = PV / (nR)

Therefore, temperature is directly proportional to pressure and volume, and inversely proportional to the number of molecules.

T = k PV / N

Let's say that T₀ is the temperature when P = 100 kPa, V = 4 L, and N = 6×10²³.

a) T = k PV / N = T₀

b) T = k (2P) V / N = 2T₀

c) T = k (P/2) (2V) / N = T₀

d) T = k PV / (N/2) = 2T₀

e) T = k P (V/2) / (N/2) = T₀

f) T = k (P/2) V / (N/2) = T₀

b and d have the highest temperature,

a, c, e, and f have the lowest temperature.

8 0

3 years ago

Read 2 more answers