Flashcard question: texting/social networking in class (spring semester 2014)

Astronomy 210, spring semester 2014
Cuesta College, San Luis Obispo, CA

"I Forgot My Phone"
Charlene deGuzman and Miles Crawford
http://youtu.be/OINa46HeWg8

After showing a short movie depicting the pervasiveness of smartphone use at the start of the semester, students in introductory astronomy students at Cuesta College were asked a subjective question regarding their attitudes towards texting and social networking smartphone use in class. This is part of a think-pair-share activity using flashcards to answer syllabus quiz questions.

Texting/social networking is acceptable behavior in class.
(A) Strongly disagree.
(B) Disagree.
(C) Neutral.
(D) Agree.
(E) Strongly agree.

Student responses (pre-discussion)
Astronomy 210
Section 30674 (SLO campus)
(A) : 1 student
(B) : 3 students
(C) : 29 students
(D) : 3 students
(E) : 0 students

Astronomy 210
Section 30676 (NC campus)
(A) : 0 students
(B) : 8 students
(C) : 13 students
(D) : 4 students
(E) : 0 students

Students were then asked to share their answers and discuss with each other the reasoning behind their choices, and to vote again.

Student responses (post-discussion)
Astronomy 210
Section 30674 (SLO campus)
(A) : 3 students
(B) : 3 students
(C) : 32 students
(D) : 3 students
(E) : 0 students

Astronomy 210
Section 30676 (NC campus)
(A) : 0 students
(B) : 8 students
(C) : 15 students
(D) : 4 students
(E) : 0 students

Students were asked to share their responses during the following whole-class discussion. Most responses were along the lines of texting/social networking being "distracting," "inappropriate," and "disrespectful," but some students remarked that emergency contact purposes (or similar circumstances) or looking up information pertinent to lecture might be considered acceptable uses of smartphones. While indiscriminate use of smartphones during instruction would be detrimental to learning, the social norm seems to be that they would be tolerated as long as smartphone use was not disruptive to other students.

These responses and student opinions were then used to set the policy, via consensus, regarding texting and social networking during class time.

Previous posts:

Flashcard question: texting/social networking in class (fall semester 2013).

Flashcard question: texting/social networking in class (fall semester 2011).

Astronomy in-class activity: astronomy in the marketplace tags

Astronomy 210 In-class activity 1, spring semester 2014
Cuesta College, San Luis Obispo, CA

140125-cars
http://www.flickr.com/photos/waiferx/12141482294/
Originally uploaded by Waifer X

Wordle.net tag cloud for astronomy-related car brand names, generated by responses from Astronomy 210 students at Cuesta College, San Luis Obispo, CA ( http://www.wordle.net/show/wrdl/7488709/Untitled ).


140125-food-1
http://www.flickr.com/photos/waiferx/12141482504/
Originally uploaded by Waifer X

Wordle.net tag cloud for astronomy-related food brand names available in a supermarket, generated by responses from Astronomy 210 students at Cuesta College, San Luis Obispo, CA (http://www.wordle.net/show/wrdl/7488717/Untitled ).


140125-nonfood-1
http://www.flickr.com/photos/waiferx/12141329353/
Originally uploaded by Waifer X

Wordle.net tag cloud for astronomy-related non-food brand names available in a supermarket, generated by responses from Astronomy 210 students at Cuesta College, San Luis Obispo, CA (http://www.wordle.net/show/wrdl/7488719/Untitled).


Students find their assigned groups of three to four students, and work cooperatively on an in-class activity worksheet to discuss car brand names, and food and non-food brand names found in supermarkets (adapted from D. Schatz, "Why Should We Care About Exploding Stars?" Universe in the Classroom, no. 8, Spring 1987 (http://www.astrosociety.org/education/publications/tnl/08/stars2.html).

There are many astronomy-related brand names. Consider car brand names (old and new); and brand names for food and non-food items that are typically found in the supermarket. Do not consider titles of TV shows, movies, or books.

List at least five astronomy-related car brand names.

Student responses
Sections 30674, 30676

Taurus, Aero, Astro, Eclipse, Mercury, Warstar
Saturn, Subaru, Galaxy, Mercury, RangeRover
Astro, Galaxy, Taurus
Saturn, Eclipse, Mercury
Saturn, Mercury, Charger
Mercury, Taurus, Saturn
Nova, Comet, Galaxy, Taurus, Geo
Saturn, Mercury, Comet, Sunfire, Astro
Saturn, Mercury, Subaru, Galaxy
Saturn, Mercury, Eclipse, Stargazer
Astro, Saturn, Mercury, Comet, Eclipse
Saturn, Mercury, Sunfire, Taurus
Astro, Mercury, Saturn, Galaxy, Voyager
Saturn, Mercury, Galaxy, Astro
Nova, Astro, Ram, Saturn, Mercury
Saturn, Infiniti, Mercury, Nova, Galaxy
Galaxy, Saturn, Subaru, LandRover, Aerostar, Infiniti
Mercury, Saturn, Astro, Eclipse, Infiniti
Mercury, Eclipse, Saturn, Taurus, Nova
Saturn, Eclipse, Taurus, Galaxy, Navigator
Saturn, Mercury, Mazda, Explorer, Navigator
Saturn, Mercury, Nova, Infiniti, Solara
Saturn, Mercury, Eclipse
Saturn, Mercury, Astro, Eclipse, Infiniti
Galaxy, Navigator, Saturn, Taurus
Saturn, Astro, Mercury, Galaxy, Eclipse

List at least five astronomy-related food brand names typically found in the supermarket.

Student responses
Sections 30674, 30676

MoonPies, AstroPops, Starburst, Orbit, MarsBars, ChickenandStars, sunflowerseeds, Corona
Orbit, Starbucks, Starburst, MilkyWay, MoonPies
Rocketcheesecracker, StarCrunch, MilkyWay, MarsBars, Astronautfood, AstroPops, starfruit, SunnyD, Sunkist, Moonpies
MoonPies, StarCrunch, Starburst, AstroPops, MilkyWay, MarsBars
LunaBars, MilkyWay, Starburst
MilkyWay, MarsBars, MoonPies, starfruit, Starburst
MoonPie, SunChips, Starburst, MilkyWay, MarBars, Orbit, SunnyD, SunMaid, Sunkist, StarKisttuna
LuckyCharms, Starbucks, MoonPies, MilkyWay
MilkyWay, Orbit, MoonPies, Starburst, Starbucks
MilkyWay, MoonPies, Starburst, AstroPuffs, Orbit, Eclipse, MarsBars
MilkyWay, sunflowerseeds, SunChips, Starbucks, Starburst, SunnyD
LunaBars, Cosmicbrownies, SunnyD, CapriSun
BlueMoon, Moonshine, LunaBars, Skyyvodka, MilkyWay
BlueMoon, MoonPies, Sunkist, Starburst, MarsBars
Starburst, MilkyWay, MarsBars, Sunkist, LunaBars
MarsBars, MilkyWay, Tang, Starburst, Sunkist, SunnyD
Rockstar, MilkyWay, Starburst, MarsBars, Orbit
MilkyWay, Sunkist, Starfruit, Starburst, CapriSun, SunChips, SunnyD
Starburst, StarKisttuna, MarsBars, Sunkist, SunDrop, MilkyWay
MilkyWay, MarsBars, SunnyD, MoonPies
Starburst, BlueMoon, Sunkist, MilkyWay, SunChips
MoonPies, MilkyWay, Orbit, Starburst, Mooncake
LunaBars, SunChips, StarFarms, MilkyWay, Starburst
MoonPies, AstroPops, SunnyD, Starburst, MilkyWay
Starburst, MilkyWay, Orbit, Starbucks, MoonPies
Starburst, MilkyWay, MarsBars, Sunkist, SunnyD

List at least five astronomy-related non-food brand names typically found in the supermarket.

Student responses
Sections 30674, 30676

Tide, Cosmomagazine, Starmagazine, moonflowers, morningglories
Crest, Scope
Constellationwinery, ColgateGalaxy, sunscreen, spacethemepartydecorations
FiveStarnotebooks
BlueMoon, Rockstar, Orbit, SamsungGalaxy, Comet
BlueMoon, MillerLite
Comet, Geo, Sun, SiriusXM
BlueMoon, babymobile, SamsungGalaxy, calendars
Crest, BlueMoon, CoorsLight, BudLight
BlueMoon, sunflowers, Cosmomagazine, venusflytrap, Venusrazors
BlueMoon, Moonshine, Skyyvodka
Astroglide, Comet, FiveStarnotebooks, ZiplocSpaceBags
Comet, Sunburst, BlueMoon
Comet, Venusrazors, Starmagazine, Rockstar, FiveStarnotebooks
Comet, Cosmomagazine, ConstantCommenttea, Venusrazors, Starmagazine, FiveStarnotebooks
BlueMoon, Moonshine, Comet, Tide, sunscreen
BlueMoon, UVBlue, Skyyvodka, Moonshine, Comet
Venusrazors, Comet, Orbit
Venusrazors, Comet, Orbit
BlueMoon, Comet, BuzzLightyear, Moonshine, SamsungGalaxy
Venusrazors, BlueMoon, Skyyvodka, MoonShoes
MoonShoes, BlueMoon, Skyyvodka, Venusrazors
Comet, Venusrazors
Cosmomagazine, BlueMoon, Comet, Skyyvodka, Venusrazors

Previous posts:

• Astronomy in-class activity: astronomy in the marketplace tags (Cuesta College, fall semester 2012)
• Astronomy in-class activity: astronomy in the marketplace tags (Cuesta College, spring semester 2012).
• Astronomy in-class activity: astronomy in the marketplace tags (Cuesta College, fall semester 2011).
• Astronomy in-class activity: astronomy in the marketplace tags (Cuesta College, fall semester 2010).
• Astronomy in-class activity: astronomy in the marketplace tags (Cuesta College, spring semester 2010).

Physics final exam question: normal force on truck going over hill

Physics 205A Final Exam, fall semester 2013
Cuesta College, San Luis Obispo, CA

A truck drives over the top of a hill that has a circular radius, without losing contact with the hill. This truck then drives over the top of the same hill with a slightly faster speed, but it is still in contact with the hill. Discuss 
why the magnitude of the upwards normal force of the hill on the truck is smaller when the truck is going faster over the hill. Explain your reasoning by using free-body diagram(s), the properties of forces and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Understands that the truck has two vertical forces acting on it:

  Weight force of Earth on truck (downwards, magnitude w = m·g).
  Normal force of road on truck (upwards, magnitude varies but must be less than w).

  For either driving fast or slow over the hill, because the truck is (momentarily) undergoing uniform circular motion as it drives over the top of the hill, then Newton's second law applies to the motion of the truck (speed is constant but direction of the velocity vector is changing), and the net force must point downwards (towards the center of the circular hill). This means that the upwards normal force must have a magnitude less than the magnitude of the weight force. Then compares driving fast over the top of the hill versus driving slow over the top of the hill:
  • the net force on the truck has a magnitude given by m·v²/r that is greater for the case of driving faster over the hill, and has a smaller magnitude for driving slower over the hill (as m and r are the same for either case);
  • the downwards weight force on the truck is constant, as it does not depend on how fast or slow the truck drives at the top of the hill;
  • the upwards normal force must be smaller for the faster speed case, in order to result in a greater downwards net force.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at analyzing forces, free-body diagrams, and Newton's laws.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Approach does not substantively utilize forces, free-body diagrams, and Newton's laws.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finaln0M3
p: 2 students
r: 0 students
t: 16 students
v: 16 students
x: 24 students
y: 2 students
z: 1 student

A sample "p" response (from student 0618):
A sample "x" response (from student 0419):
Another sample "x" response (from student 1969):
And another sample "x" response (from student 1994):
Yet another sample "x" response (from student 2999):
And one more sample "x" response (from student 7667):

Physics final exam question: climber rappelling down cliff edge

Physics 205A Final Exam, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 8.19

"Transportation Soldiers conduct rappel tower training"
Virginia Guard Public Affairs
flic.kr/p/cQRGch

A climber is approximated as a uniform beam, with a pivot point at the feet, and a rope attached to the center. The climber is about to rappel back down from a cliff edge, leaning back with the rope initially horizontal. A little later, the climber is lower, leaning back with the same angle, but the rope is no longer horizontal. Discuss why there is less tension in the rope for the latter case. (Ignore stretching in the rope.) Explain your reasoning using diagram(s) with locations of forces and lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Understands that (1) the counterclockwise torque due to weight is the same in either case, such that from the static equilibrium condition (Newton's first law for rotations), the clockwise torque due to the rope's tension force must be the same in either case, and (2) the larger lever arm r_⊥ for the second case will mean a smaller tension force.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Does not explicitly explain in (1) why the clockwise torques due to the rope tension forces are the same.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Discusses that r_⊥ would be larger for the second case, but does not explicitly explain this on a diagram and/or with trigonometry.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at manipulating the relationship between torque, force, and lever arm.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Approach does not substantively use relationship between torque, force, and lever arm.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finaln0M3
p: 6 students
r: 14 students
t: 19 students
v: 17 students
x: 2 students
y: 1 student
z: 2 students

A sample "p" response (from student 2419):

Physics final exam question: speed of waves along standing wave strings

Physics 205A Final Exam, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Conceptual Questions 11.3, 11.4, Multiple-Choice Questions 11.5, 11.6, 11.8

A string has its tension set by a hanging mass, and resonates at its fundamental frequency. If a shorter length of the string is used with the same hanging mass (which changes its fundamental frequency), discuss why the speed of waves along this string does not change. (Ignore stretching in the string.) Explain your reasoning using the properties of wave speeds, periodic waves, and standing waves.

Solution and grading rubric:

• p:
  Correct. Understands that (1) wave speed depends on tension and linear mass density, and (2) since neither of these are changed by using a shorter portion of this same string (with the same mass hanging from it), then the wave speed remains constant.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least understands the relationship between wave parameters.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finaln0M3
p: 17 students
r: 25 students
t: 7 students
v: 9 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 7810):

Physics final exam question: heat conducted through different-orientation bricks

Physics 205A Final Exam, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Example 14.10, Practice Problem 14.10, Problem 14.57

A brick can be placed in either of two orientations. The bottom of each brick is immersed in 20° C water, while the top of each brick is heated to 80° C. (Ignore the very slight thermal expansion of these bricks.) Discuss why these two bricks will not conduct the same amount of heat per time. Explain your answer using the properties of heat, temperature, and heat transfer.

Solution and grading rubric:

• p:
  Correct. Heat is conducted from the top to the bottom of each brick, at a rate (1) proportional to the cross-sectional area, and (2) inversely proportional to the length (in this case, height), such that the wider, shorter brick will conduct heat at a rate faster than the narrower, taller brick.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Typically argues only (1) area, or only (2) length as a factor in why the wider, shorter brick will conduct heat faster than the narrower, taller brick.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least recognizes different factors (1)-(2) and attempts to discuss heat conduction along the length of the bars.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Discussion based on phenomena other than heat conduction along the length of the bricks.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finaln0M3
p: 24 students
r: 31 students
t: 1 student
v: 3 students
x: 0 students
y: 1 student
z: 1 student

A sample "p" response (from student 0494):
Another sample "p" response (from student 2395):
A sample "r" response (from student 0825), discussing only how the cross-sectional area A affects the flow of heat conducted through the bricks:
A sample "r" response (from student 1025), discussing only how the length (height) d affects the flow of heat conducted through the bricks:

Physics final exam problem: frictionless box pushing box with sliding friction

Physics 205A Final Exam, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 4.63

A 7.0 kg box and a 3.0 kg box are in contact with each other on a horizontal floor. The 7.0 kg box slides without friction on the floor, but the 3.0 kg box has a kinetic friction coefficient of 0.15 with the floor. A horizontal force of 5.0 N is applied to the 7.0 kg box such that both boxes are sliding together to the right. Determine the magnitude of the acceleration of these boxes. Show your work and explain your reasoning.

Solution and grading rubric:

• p:
  Correct. Draws free-body diagrams, and applies properties of forces and Newton's laws to determine the kinetic friction force on the 3.0 kg box, finds the net horizontal force on the system of both boxes, and implements Newton's second law to find the acceleration of both boxes.
• r:
  Nearly correct, but includes minor math errors. One of the two points (1)-(2) correct, other is problematic/incomplete.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. At least some attempt at applying properties of forces and Newton's laws.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach does not substantively use properties of forces and Newton's laws.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finaln0M3
p: 13 students
r: 4 students
t: 7 students
v: 17 students
x: 13 students
y: 2 students
z: 5 students

A sample "p" response (from student 0000):

Physics final exam problem: kinetic energy loss of completely inelastic collision

Physics 205A Final Exam, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 7.43, Comprehensive Problem 7.71

A 0.20 kg cart launched off of a spring in the +x direction with a velocity of +0.11 m/s collides with a heavier 1.00 kg cart that is initially at rest. The carts stick together after the collision. Neglect drag and friction. Demonstrate numerically that kinetic energy is not conserved for this collision. Show your work and explain your reasoning using properties of collisions, energy (non-)conservation, and momentum conservation.


Solution and grading rubric:

• p:
  Correct. Applies (1) conservation of momentum to determine final velocity of the stuck-together cards, and (2) demonstrates numerically that kinetic energy is not conserved.
• r:
  Nearly correct, but includes minor math errors. One of the two points (1)-(2) correct, other is problematic/incomplete.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Both points (1)-(2) problematic/incomplete, or one point correct while other is missing.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at applying some conservation law.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying conservation laws.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finaln0M3
p: 15 students
r: 3 students
t: 19 students
v: 19 students
x: 1 students
y: 2 students
z: 2 students

A sample "p" response (from student 0000), finding the final velocity of the conjoined carts from momentum conservation, and shows that the initial total kinetic energy is not equal to the final total kinetic energy:
A sample "p" response (from student 1313), also finding the final velocity of the conjoined carts from momentum conservation, but instead showing that the loss of kinetic energy of the first cart is not equal to the gain in kinetic energy of the second cart:

Physics final exam problem: metal samples placed in water

Physics 205A Final Exam, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Practice Problem 14.5, Comprehensive Problem 14.95

A 0.50 kg sample of iron at 95° C and a 0.75 kg sample of aluminum at 10° C are both placed in a container with 1.00 kg of water at 20° C. Find the temperature of the water after thermal equilibrium has been reached. Ignore the effects of evaporation and phase changes, and heat exchanged with the environment and container. Show your work and explain your reasoning using the properties of heat, temperature, and thermal equilibrium.

Specific heat of iron is 440 J/(kg·K). Specific heat of aluminum is 900 J/(kg·K). Specific heat of water is 4,190 J/(kg·K).

Solution and grading rubric:

• p:
  Correct. Sets up an energy transfer/balance equation with changes in the thermal internal energies of the iron, aluminum and water summing to zero (no heat exchanged with container and environment), and solves for the final thermal equilibrium temperature of the system.
• r:
  Nearly correct, but includes minor math errors.
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. At least has energy transfer/balance equation with three terms, but typically does not set up ∆T terms correctly.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. At least some attempt at setting up change in internal energy terms and/or an energy transfer/balance equation.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying energy transfer/balance equation.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finaln0M3
p: 17 students
r: 5 students
t: 10 students
v: 12 students
x: 6 students
y: 5 students
z: 6 students

A sample "p" response (from student 5678):
A sample "p" response (from student 0618), first converting the initial temperatures to Kelvin, to find the final temperature in Kelvin:
A sample "t" response (from student 0817):
A sample "v" response (from student 0716):
A sample "y" response (from student 8207):

Astronomy in-class activity: planet-hunting

Astronomy 210 In-class activity 6 v.14.01.10, spring semester 2014
Cuesta College, San Luis Obispo, CA

Students find their assigned groups of three to four students, and work cooperatively on an in-class activity worksheet to determine where in the sky each naked-eye planet will be observed on a given date (here, February 6, 2014).



Previous posts:

• Astronomy in-class activity: planet-hunting
  fall semester 2013 (September 5, 2013).
• Astronomy in-class activity: planet-hunting
  spring semester 2013 (February 6, 2013).
• Astronomy in-class activity: planet-hunting
  fall semester 2012 (September 6, 2012).
• Astronomy in-class activity: planet-hunting
  spring semester 2012 (February 2, 2012).
Astronomy in-class activity: planet-hunting
fall semester 2011 (September 1, 2011).
• Astronomy in-class activity: planet-hunting
  spring semester 2011 (February 2, 2011).
• Astronomy in-class activity: planet-hunting
  fall semester 2010 (September 2, 2010).
• Astronomy in-class activity: planet-hunting
  spring semester 2010 (February 4, 2010).