Which of the following is NOT included in an APA citation for a scholarly journal article? Please explain answer.
1 answer:
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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Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
= m
+ m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
Answer:
|Δf| = 37.3 kHz
Explanation:
given,
peak velocity = 4 m/s
speed of the sound = 1500 m/s
frequency = 7 MHz
= 37.3 kHz
|Δf| = 37.3 kHz
hence, frequency shift between the opening and closing valve is 37.3 kHz