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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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What current flows through a 2.54cm diameter rod of pure silicon that is 20cm long when 1000V is applied?

vfiekz [6]

Answer: $0.0039\ A$

Explanation:

Given

Diameter of the rod $d=2.54\ cm$

length of rod is $l=20\ cm$

Resistivity of silicon is $\rho=6.4\times 10^2\ \Omega-m$

cross-section of the rod $A$

$\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2$

Resistance of rod is R

$\Rightarrow R=\dfrac{\rho l}{A}$

$\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega$

Current is given by

$\Rightarrow I=\dfrac{V}{R}\\\\\Rightarrow I=\dfrac{1000}{25.26\times 10^4}\\\\\Rightarrow I=0.0039\ A$

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3 years ago

Caroline conducts research on how the amount of fiber in a student's breakfast affects their grades in school. How should she re

Evgesh-ka [11]

Answer:

data table and line graph ( first choice)

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3 years ago

Read 2 more answers

A car is traveling at 75 m/s. 50 seconds later it is traveling at 25 m/s. What is the car’s acceleration?

scoray [572]

Answer:

a = -1 m/s^2

Explanation:

Vi = 75 m/s

Vf = 25 m/s

t = 50 s

Plug those values into the following equation:

Vf = Vi + at

25 = 75 + 50a

---> a = -1 m/s^2

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3 years ago

Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to

jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

=K 2 +GmM (r 11− r 21)=6.9×10 7 J

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2 years ago

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

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4 years ago