Astronomy quiz archive: solar system

Astronomy 210 Quiz 7, Spring Semester 2011
Cuesta College, San Luis Obispo, CA

Section 30676, version 1
Exam code: quiz07Sq5h


Section 30674

Quiz 7 results (max score = 40):

   0- 8.0 : *  [low = 8.0]

 8.5-16.0 : ***********

16.5-24.0 : **************  [mean = 21.4 +/- 7.7]

24.5-32.0 : *********

32.5-40.0 : ****  [high = 40.0]

Astronomy quiz archive: solar system

Astronomy 210 Quiz 7, Spring Semester 2011
Cuesta College, San Luis Obispo, CA

Section 30674, version 1
Exam code: quiz07N3mi


Section 30674

Quiz 7 results (max score = 40):
   0- 8.0 : **  [low = 6.5]
 8.5-16.0 : **********
16.5-24.0 : ************  [mean = 20.5 +/- 8.4]
24.5-32.0 : ********
32.5-40.0 : ****  [high = 40.0]

Astronomy certificate of achievement

20110505845
http://www.flickr.com/photos/waiferx/5704649689/
Originally uploaded by Waifer X

Astronomy 210 Certificate of Achievement, Cuesta College, San Luis Obispo, CA. Photo by Cuesta College Physical Sciences Division instructor Dr. Patrick M. Len.

Physics midterm problem: light bulb power dissipation in circuit

Physics 205B Midterm 2, spring semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 18.65

An ideal 12 V emf source is connected to ideal light bulbs and an ideal resistor, as shown at right. Find the amounts of power used by each of the two light bulbs. Show your work and explain your reasoning.

Solution and grading rubric:

• p:
  Correct. Finds equivalent resistance of circuit, and uses Ohm's law to find the current flowing through the emf and top light bulb. Can use P = (I²)·R to find power used by top light bulb, and then Kirchhoff's loop rule to find the voltage remaining after the top light bulb, then P = (∆V²)/R to find the power used by the lower light bulb.
  
• r:
  Nearly correct, but includes minor math errors. Has at least one light bulb power correct, but has minor misapplications of Kirchhoff's loop rules or junction rules in finding the other light bulb power.
  
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least has systematic approach to finding equivalent resistance and current of circuit, and finding powers of each light bulb.
  
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner.
  
• x:
  Implementation of ideas, but credit given for effort rather than merit.
  
• y:
  Irrelevant discussion/effectively blank.
  
• z:
  Blank.
  

Grading distribution:
Section 30882
Exam code: midterm02H3nR
p: 3 students
r: 3 students
t: 2 students
v: 0 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 7503):

Overheard: Ceres, Ceres-ously

Astronomy 210, Spring Semester 2011
Cuesta College, San Luis Obispo, CA

(Overheard during a presentation on dwarf planets, and the 2006 International Astronomical Union classification of solar system bodies.)

Instructor: "Consider Ceres, which is now just the largest asteroid. But after it was first discovered in 1801, people called it a planet."

Student 1: "Where did that name come from?"

Instructor: "The Greek goddess of the grain harvest. You know, like where the word, 'cereal' comes from."

(Beat.)

Student 2: "Seriously?"

Instructor: "Ceres-ously."

Astronomy midterm question: older versus younger stars

Astronomy 210 Midterm 2, Spring Semester 2011
Cuesta College, San Luis Obispo, CA

[20 points.] An astronomy question on an online discussion board (http://answers.yahoo.com/question/index?qid=20100510145903AAHVFRG) was asked and answered:

J*rocks: How are older stars different than younger stars?
Smileyface: Older stars have higher percentages of heavier elements than younger stars...

Discuss whether or not if this answer is correct, and how you know this. Explain using the properties and evolution of stars.

Solution and grading rubric:

• p = 20/20:
  Correct. Understands that (a) older stars are metal-poor having formed from essentially just hydrogen but produce metals in their cores; and (b) that newer stars were produced from material from older stars that were released when older stars exploded, making the newer stars metal-rich.
  
• r = 16/20:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. One of the two points (a)-(b) is correct, other is problematic.
  
• t = 12/20:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Only one of the two points (a)-(b) correct, other is missing, or both are problematic.
  
• v = 8/20:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Garbled discussion of properties and evolution of stars, such as breaking down of metals; masses and evolution rates.
  
• x = 4/20:
  Implementation/application of ideas, but credit given for effort rather than merit. May state answer is incorrect, but without proof; or discussion other than that of the properties and evolution of stars.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Section 30676
Exam code: midterm02Sys7
p: 20 students
r: 4 students
t: 3 students
v: 18 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 3517):

Another sample "p" response (from student 0701):

Astronomy midterm question: Arcturus versus Vega

Astronomy 210 Midterm 2, Spring Semester 2011
Cuesta College, San Luis Obispo, CA

[20 points.] Determine which star (Arcturus or Vega) would be brighter, as seen from Earth, if they were both relocated to a distance of 10 parsecs away. Explain using the properties of apparent magnitude, absolute visual magnitude, and distance.

Solution and grading rubric:

• p = 20/20:
  Correct. Arcturus is farther away than 10 pcs and is already seen as brighter than Vega, which is closer than 10 pcs. Moving Arcturus closer, and Vega farther away, such that placing both at 10 pcs away would still make Arcturus brighter than Vega as seen from Earth.
  
• r = 16/20:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
  
• t = 12/20:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least understands the difference between apparent (m) and absolute (M_V) magnitudes, and that smaller positive (or more negative) magnitudes are brighter.
  
• v = 8/20:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Garbled definitions/relations between d, m, and M_V.
  
• x = 4/20:
  Implementation/application of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Section 30676
Exam code: midterm02Sys7
p: 33 students
r: 4 students
t: 5 students
v: 3 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 4351):

Another sample "p" response (from student 1237):

Astronomy midterm question: exploding big bang?

Astronomy 210 Midterm 2, spring semester 2011
Cuesta College, San Luis Obispo, CA

[20 points.] An astronomy question on an online discussion board [*] was asked and answered:

nema: What is [the] big bang theory...?
Wobzter: First all matter and energy was stored [in] a ball the size of a fist...and exploded. Then all matter and energy spread out, and moved away from the place of the fist...

Discuss why this answer is incorrect, and how you know this. Explain using observations and evidence related to the Hubble law.

[*] Source: http://answers.yahoo.com/question/index?qid=20100117025456AAzBY3X.

Solution and grading rubric:

• p = 20/20:
  Correct. Discusses Hubble's law (recession velocity of galaxies is proportional to distance) and evidence (greater redshift of absorption lines for distant galaxies compared to nearby galaxies), and uses this to contradict one or both: (a) no unique center, (b) expansion is not like an explosion (where speeds would decrease with increasing distance).
  
• r = 16/20:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
  
• t = 12/20:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors. Hubble's law discussion is problematic or
  incomplete, but least understands that as a consequence there must be no unique center, and/or expansion cannot be like an explosion.
  
• v = 8/20:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Discussion based on other aspects of the universe, with little or no substantive discussion of Hubble's law.
  
• x = 4/20:
  Implementation/application of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Section 30674
Exam code: midterm02N3n4
p: 8 students
r: 7 students
t: 15 students
v: 15 students
x: 1 student
y: 0 students
z: 0 students

Section 30676
Exam code: midterm02Sys7
p: 4 students
r: 5 students
t: 14 students
v: 12 students
x: 15 students
y: 0 students
z: 0 students

A sample "p" response (from student 4200):
Another sample "p" response (from student 2364):

Physics quiz archive: magnetism, induction

Physics 205B Quiz 6, Spring Semester 2011
Cuesta College, San Luis Obispo, CA

Section 30882, version 1

Section 30882 results
Exam code: quiz06eMf0

 0- 6 : 
 7-12 : 
13-18 : **  [low = 18]
19-24 : ***  [mean = 23.6 +/- 4.4]
25-30 : ***    [high = 30]

Astronomy midterm question: other galaxies but our own visible?

Astronomy 210 Midterm 2, Spring Semester 2011
Cuesta College, San Luis Obispo, CA

[20 points.] Even though only a small fraction of the stars in own galaxy is visible from Earth, explain why we can still see stars of other galaxies outside the Milky Way with the naked eye. Support your answer using a diagram, and properties of galaxies.

Solution and grading rubric:

• p = 20/20:
  Correct. Views along the disk of the Milky Way would be obscured by interstellar gas/dust, but views up and down, perpendicular to the plane of the Milky Way disk would be much less obscured, as space between galaxies is relatively empty, making the stars of galaxies visible.
  
• r = 16/20:
  Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. At least understands how shape of Milky Way is important, but may discuss other contributing factors.
  
• t = 12/20:
  Contains right ideas, but discussion is unclear/incomplete or contains major errors.
  
• v = 8/20:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Discusses some other aspects of the Milky Way.
  
• x = 4/20:
  Implementation/application of ideas, but credit given for effort rather than merit.
  
• y = 2/20:
  Irrelevant discussion/effectively blank.
  
• z = 0/20:
  Blank.
  

Grading distribution:
Section 30674
Exam code: midterm02N3n4
p: 14 students
r: 9 students
t: 6 students
v: 9 students
x: 0 students
y: 2 students
z: 0 students

A sample "p" response (from student 1450):