20070731
Cosmic background radiation: the horizon problem
The Universe in Different Frequencies
(*.mov, 7.1 MB)
Wilkinson Microwave Anisotropy Probe mission, map.gsfc.nasa.gov
Astronomy 10 learning goal Q12.4
Illustration of how different forms of electromagnetic radiation are distributed in all directions, and how the cosmic background radiation in the microwave band is nearly isotropic (the "horizon problem"), unless contrast is greatly enhanced.
Labels:
cosmic background radiation
20070730
Valuable lesson learned
Wigu, by Jeffery Rowland
jjrowland.com
June 23, 2007 (excerpt)
Astronomy 10 learning goals Q12.4
Is Blingidium meant to be made of dark matter, dark energy, or something else entirely?
Labels:
big bang,
dark energy,
dark matter,
multiverse
20070727
Education research: SPCI gains (Cuesta College, Summer Session 2007)
The Star Properties Concept Inventory was developed by Janelle Bailey as a pre-test and post-test for introductory astronomy courses. For an overview of how <g> quantifies gains in learning (Hake), see the previous post: Education research: FCI gains (Cuesta College).
The SPCI was administered to Astronomy 10 (one-semester introductory astronomy) students at Cuesta College, San Luis Obispo, CA during the first class meeting, then on the last class meeting. The results below are class averages for the initial and final SPCI scores (given as percentages, with standard deviations), as well as the Hake normalized gain <g>:
For earlier results at Cuesta College and further discussion of the SPCI, see previous post: Education research: SPCI gains (Cuesta College, Spring Semester 2006-Spring Semester 2007).
The SPCI was administered to Astronomy 10 (one-semester introductory astronomy) students at Cuesta College, San Luis Obispo, CA during the first class meeting, then on the last class meeting. The results below are class averages for the initial and final SPCI scores (given as percentages, with standard deviations), as well as the Hake normalized gain <g>:
Astronomy 10 Summer Session 2007 section 8027Despite the extremely small number of students in this course, these results are comparable to previous semesters of Astronomy 10 taught by this instructor at Cuesta College, but with a slightly higher gain than is typical.
N = 11
<initial%> = 26% +/- 12%
<final%> = 55% +/- 13%
<g> = 0.39
For earlier results at Cuesta College and further discussion of the SPCI, see previous post: Education research: SPCI gains (Cuesta College, Spring Semester 2006-Spring Semester 2007).
- Bailey, J. M. (2006). "Development of a concept inventory to assess students' understanding and reasoning difficulties about the properties and formation of stars." Unpublished doctoral dissertation, University of Arizona, Tucson, AZ.
Development of the SPCI, a 30-question survey of stellar properties concepts (blackbody radiation laws). - R.R. Hake, "Interactive-engagement vs. traditional methods: A six-thousand-student survey of mechanics test data for introductory physics courses," Am. J. Phys. 66, 64 -74 (1998).
Definition of <g>, and relative comparison of many interactive engagement and traditional courses.
Labels:
education research,
Hake gain,
SPCI
20070726
Astronomy quiz question: metallicity
Astronomy 10 Quiz 11, Summer Session 2007
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.5
[3.0 points.] All elements heavier than hydrogen and helium are considered to be "metals." Which one of the following statements best explains why metal-poor stars are older than metal-rich stars?
(A) Metal-poor stars gradually become metal-rich stars at the end of their main sequence lifetime.
(B) Metal-poor stars have longer main sequence lifetimes.
(C) Metal-poor stars have fewer lines in their absorption spectra.
(D) Explosion debris from metal-poor stars is then incorporated into metal-rich stars.
(E) When metal-poor stars collide, they produce metal-rich stars.
Correct answer: (D)
"Metals" that are produced in the cores of stars are not detectable in their absorption spectra from their exospheres. However, these heavy elements are released during type II supernova explosions, which can then be incorporated into a subsequent generation of stars, which will then have "metals" in their absorption spectra. Thus each generation of stars gets more and more "polluted" by heavy elements in their exospheres.
Student responses
Section 8027
(A) : 2 students
(B) : 2 students
(C) : 3 students
(D) : 1 student
(E) : 2 students
(Compare to previous post: Astronomy clicker question: metal-rich stars.)
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.5
[3.0 points.] All elements heavier than hydrogen and helium are considered to be "metals." Which one of the following statements best explains why metal-poor stars are older than metal-rich stars?
(A) Metal-poor stars gradually become metal-rich stars at the end of their main sequence lifetime.
(B) Metal-poor stars have longer main sequence lifetimes.
(C) Metal-poor stars have fewer lines in their absorption spectra.
(D) Explosion debris from metal-poor stars is then incorporated into metal-rich stars.
(E) When metal-poor stars collide, they produce metal-rich stars.
Correct answer: (D)
"Metals" that are produced in the cores of stars are not detectable in their absorption spectra from their exospheres. However, these heavy elements are released during type II supernova explosions, which can then be incorporated into a subsequent generation of stars, which will then have "metals" in their absorption spectra. Thus each generation of stars gets more and more "polluted" by heavy elements in their exospheres.
Student responses
Section 8027
(A) : 2 students
(B) : 2 students
(C) : 3 students
(D) : 1 student
(E) : 2 students
(Compare to previous post: Astronomy clicker question: metal-rich stars.)
20070725
Astronomy clicker question: Milky Way shape (The "High-Maintenance Camper's Dilemma")
Astronomy 10, Summer Session 2007
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.2
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the beginning of their learning cycle:
[0.3 points.] How do we know that the Milky Way is shaped like a flat disk?
(A) Most other galaxies are shaped like flat disks, therefore the Milky Way must be shaped like a flat disk.
(B) It is possible to visually trace out its flat disk shape if the night sky is dark enough.
(C) Distant stars appear to be dimmer than nearby stars.
(D) (None of the above choices (A)-(C), as the Milky Way is not shaped like a flat disk.)
Correct answer: (B)
This question is asked after an in-class activity where students plot out the most distant stars visible in the Milky Way, and then find that this only represents a very tiny portion of our entire galaxy. This problem posed as the "camping dilemma," where if you had forgotten to bring a mirror with you on a camping trip, then how do you know what you look like? Similarly, with interstellar dust and gas obscuring the majority of the Milky Way, then how do we know the overall shape (and size) of our own galaxy?
Response (A) is analogous to a high-maintenance person looking at the other campers, and asking, "Do I look as bad as they do?" This inference, while plausible, is not necessarily valid, as other galaxies come in various sizes and forms (as would, presumably, the unkempt appearance of other campers). Response (C) is a result of the inverse square law, compounded by interstellar dust and gas, and would be true regardless of the shape and structure of the Milky Way.
Prompt students to recall that under ideal conditions, the Milky Way is seen as a dense band of dim stars across the sky. This is the primary evidence that our galaxy has a thin disk shape, as many stars can be seen in the plane of this disk, while only a sparse distribution of stars are seen in directions above and below the plane of the disk. (While the overall size of the Milky Way cannot be determined solely from this observation, the lack of interstellar dust and gas above and below the plane of the disk does allow for clear views of globular clusters in the halo of the Milky Way, which are used to determine the overall size of the Milky Way.)
Student responses
Section 8027
(A) : 2 students
(B) : 5 students
(C) : 6 students
(D) : 0 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.2
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the beginning of their learning cycle:
[0.3 points.] How do we know that the Milky Way is shaped like a flat disk?
(A) Most other galaxies are shaped like flat disks, therefore the Milky Way must be shaped like a flat disk.
(B) It is possible to visually trace out its flat disk shape if the night sky is dark enough.
(C) Distant stars appear to be dimmer than nearby stars.
(D) (None of the above choices (A)-(C), as the Milky Way is not shaped like a flat disk.)
Correct answer: (B)
This question is asked after an in-class activity where students plot out the most distant stars visible in the Milky Way, and then find that this only represents a very tiny portion of our entire galaxy. This problem posed as the "camping dilemma," where if you had forgotten to bring a mirror with you on a camping trip, then how do you know what you look like? Similarly, with interstellar dust and gas obscuring the majority of the Milky Way, then how do we know the overall shape (and size) of our own galaxy?
Response (A) is analogous to a high-maintenance person looking at the other campers, and asking, "Do I look as bad as they do?" This inference, while plausible, is not necessarily valid, as other galaxies come in various sizes and forms (as would, presumably, the unkempt appearance of other campers). Response (C) is a result of the inverse square law, compounded by interstellar dust and gas, and would be true regardless of the shape and structure of the Milky Way.
Prompt students to recall that under ideal conditions, the Milky Way is seen as a dense band of dim stars across the sky. This is the primary evidence that our galaxy has a thin disk shape, as many stars can be seen in the plane of this disk, while only a sparse distribution of stars are seen in directions above and below the plane of the disk. (While the overall size of the Milky Way cannot be determined solely from this observation, the lack of interstellar dust and gas above and below the plane of the disk does allow for clear views of globular clusters in the halo of the Milky Way, which are used to determine the overall size of the Milky Way.)
Student responses
Section 8027
(A) : 2 students
(B) : 5 students
(C) : 6 students
(D) : 0 students
20070724
Interactive binary star system
Binary Stars Interactive (*.swf)
McGraw-Hill Online Learning Center
Astronomy 10 learning goal Q10.2
Begin with setting M_A = 0.5 solar masses, M_B = 5.0 solar masses, and separation distance = 20 solar radii. Point out that the more massive star B is closer to the center of mass/rotation axis, while the less massive star A is farther away. Ask the students which star will eventually reach the end of its main sequence lifetime first, and why (star B, as it is more massive), and which star will have the larger Roche lobe (star B, as it is both more massive and slower in orbital speed, resulting in less centrifugal force, which counteracts gravity).
If the separation distance is decreased to 7.0 solar radii, the Roche lobes of both stars shrink, as they will both have faster orbital speeds (and thus stronger centrifugal forces, which decreases the volume of space that matter will accelerate in towards either star). The more massive star B still has a larger Roche lobe than star A.
http://highered.mcgraw-hill.com/olcweb/cgi/pluginpop.cgi?it=swf::100%::100%::/sites/dl/free/007299181x/78778/Binary_Nav.swf::Binary%20Stars%20Interactive
Labels:
binary star system,
Roche lobe
20070723
Astronomy quiz question: close-binary mass transfer
Astronomy 10 Quiz 10, Summer Session 2007
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
[3.0 points.] Shown at right are the equipotentials of a close pair (mass-exchanging) binary system. Which one of the following choices best describes the transfer of hydrogen in the figure at right?
(A) A giant taking hydrogen from a more massive neutron star.
(B) A giant feeding hydrogen to a more massive neutron star.
(C) A giant taking hydrogen from a less massive neutron star.
(D) A giant feeding hydrogen to a less massive neutron star.
(E) (None of the above choices (A)-(D), as no hydrogen is being transferred.)
Correct answer: (B)
The star on the left has less mass than the star on the right, as the center of mass of the binary star system is closer to the star on the right. The star on the left is in its giant phase, as it has expanded in size to its Roche lobe, and thus is transferring hydrogen to the more massive neutron star on the right.
Student responses
Section 8027
(A) : 3 students
(B) : 6 students
(C) : 3 students
(D) : 1 student
(E) : 0 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
[3.0 points.] Shown at right are the equipotentials of a close pair (mass-exchanging) binary system. Which one of the following choices best describes the transfer of hydrogen in the figure at right?
(A) A giant taking hydrogen from a more massive neutron star.
(B) A giant feeding hydrogen to a more massive neutron star.
(C) A giant taking hydrogen from a less massive neutron star.
(D) A giant feeding hydrogen to a less massive neutron star.
(E) (None of the above choices (A)-(D), as no hydrogen is being transferred.)
Correct answer: (B)
The star on the left has less mass than the star on the right, as the center of mass of the binary star system is closer to the star on the right. The star on the left is in its giant phase, as it has expanded in size to its Roche lobe, and thus is transferring hydrogen to the more massive neutron star on the right.
Student responses
Section 8027
(A) : 3 students
(B) : 6 students
(C) : 3 students
(D) : 1 student
(E) : 0 students
20070719
Batgirl centrifuge
Batman #129 (1960), cover by Sheldon Moldoff and Ira Schnapp
Courtesy Mike's Amazing World of DC Comics
Physics 8A learning goal Q4.5
For the villain's sake, let's hope that Batgirl doesn't start get nauseous while still on that thing...
Courtesy Mike's Amazing World of DC Comics
Physics 8A learning goal Q4.5
For the villain's sake, let's hope that Batgirl doesn't start get nauseous while still on that thing...
Labels:
centrifuge,
centripetal force
20070718
Tidal slowing of Earth's rotation
Dinosaur Comics, by Ryan North
www.qwantz.com
July 12, 2007 (excerpt)
Astronomy 10 learning goal Q4.x
Other than the snide remarks regarding reading Professor Science's mail (T. Rex is "not" officially Professor Science), Ryan North nails the science in this comic strip.
Another result of this tidal slowing of the Earth's rotation rate is that since the total angular momentum of the Earth-Moon system must be conserved (even though rotational kinetic energy being lost due to dissipative forces), the Moon will also begin to move out away from the Earth, increasing this distance to compensate for the decreasing rotation.
www.qwantz.com
July 12, 2007 (excerpt)
Astronomy 10 learning goal Q4.x
Other than the snide remarks regarding reading Professor Science's mail (T. Rex is "not" officially Professor Science), Ryan North nails the science in this comic strip.
Another result of this tidal slowing of the Earth's rotation rate is that since the total angular momentum of the Earth-Moon system must be conserved (even though rotational kinetic energy being lost due to dissipative forces), the Moon will also begin to move out away from the Earth, increasing this distance to compensate for the decreasing rotation.
Labels:
rotational dynamics,
tidal forces
20070717
Astronomy clicker question: the Stefan-Boltzmann law
Astronomy 10, Summer Session 2007
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q8.5
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the beginning of their learning cycle:
[0.3 points.] Why is a white dwarf star smaller than a main-sequence star that has the same white-hot color?
(A) It is less luminous than the main-sequence star.
(B) It is more luminous than the main-sequence star.
(C) It is cooler than the main-sequence star.
(D) It is hotter than the main-sequence star.
Correct answer: not revealed yet (see discussion).
This is the follow-up question after a short lecture (20 minutes) and an in-class activity (20 minutes) on how Wien's law and the Stefan-Boltzmann law describe blackbody radiation. Initial responses below:
Student responses
Section 8027
(A) : 3 students
(B) : 0 students
(C) : 3 students
(D) : 7 students
A leading question for the students: "Which star is hotter, and why?" Some students will have already realized that because these two stars have the same color, then they must be at the same temperature (application of Wien's law), and the class discusses why this must be the case. The same question is asked again, after the students collectively come to realization that both responses (C) and (D) cannot be true.
[0.3 points.] Why is a white dwarf star smaller than a main-sequence star that has the same white-hot color?
(A) It is less luminous than the main-sequence star.
(B) It is more luminous than the main-sequence star.
(C) It is cooler than the main-sequence star.
(D) It is hotter than the main-sequence star.
Correct answer: (A)
Student responses
Section 8027
(A) : 12 students
(B) : 1 student
(C) : 0 students
(D) : 0 students
The Stefan-Boltzmann law states that luminosity is proportional to size (that is, surface area) and temperature (T^4). Since they are the same temperature, then the less luminous star must be smaller in size than the more luminous star, which is larger in size.
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q8.5
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the beginning of their learning cycle:
[0.3 points.] Why is a white dwarf star smaller than a main-sequence star that has the same white-hot color?
(A) It is less luminous than the main-sequence star.
(B) It is more luminous than the main-sequence star.
(C) It is cooler than the main-sequence star.
(D) It is hotter than the main-sequence star.
Correct answer: not revealed yet (see discussion).
This is the follow-up question after a short lecture (20 minutes) and an in-class activity (20 minutes) on how Wien's law and the Stefan-Boltzmann law describe blackbody radiation. Initial responses below:
Student responses
Section 8027
(A) : 3 students
(B) : 0 students
(C) : 3 students
(D) : 7 students
A leading question for the students: "Which star is hotter, and why?" Some students will have already realized that because these two stars have the same color, then they must be at the same temperature (application of Wien's law), and the class discusses why this must be the case. The same question is asked again, after the students collectively come to realization that both responses (C) and (D) cannot be true.
[0.3 points.] Why is a white dwarf star smaller than a main-sequence star that has the same white-hot color?
(A) It is less luminous than the main-sequence star.
(B) It is more luminous than the main-sequence star.
(C) It is cooler than the main-sequence star.
(D) It is hotter than the main-sequence star.
Correct answer: (A)
Student responses
Section 8027
(A) : 12 students
(B) : 1 student
(C) : 0 students
(D) : 0 students
The Stefan-Boltzmann law states that luminosity is proportional to size (that is, surface area) and temperature (T^4). Since they are the same temperature, then the less luminous star must be smaller in size than the more luminous star, which is larger in size.
20070716
Astronomy in-class activity: OBAFGKM poetry slam
Astronomy 10, Summer Session 2007
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q8.5
Students were instructed to use at least OBAFGKM, and/or part or all of the additional OBAFGKMRNSC or OBAFGKMLT extensions to individually write an original, coherent and an appropriate (nothing worse than "PG-13" rated!) mnemonic, and to give a rousing reading of their OBAFGKM mnemonic poem for the class.
Some favorites from this session:
Orlando Bloom Asks For Greek-Killing Movies Like Troy.
--A. B.
Our Bats Are Fruit-Gobbling Kind Mammals, Really, Not Scary Creatures.
--S. L.
Oh Boy, Astronomy Finally Gets Kickin', Man.
--F. N.
Previous post: OBAFGKM poetry slam, illustrated (Spring Semester 2007).
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q8.5
Students were instructed to use at least OBAFGKM, and/or part or all of the additional OBAFGKMRNSC or OBAFGKMLT extensions to individually write an original, coherent and an appropriate (nothing worse than "PG-13" rated!) mnemonic, and to give a rousing reading of their OBAFGKM mnemonic poem for the class.
Some favorites from this session:
Orlando Bloom Asks For Greek-Killing Movies Like Troy.
--A. B.
Our Bats Are Fruit-Gobbling Kind Mammals, Really, Not Scary Creatures.
--S. L.
Oh Boy, Astronomy Finally Gets Kickin', Man.
--F. N.
Previous post: OBAFGKM poetry slam, illustrated (Spring Semester 2007).
Labels:
astronomy in-class activity,
OBAFGKM
20070713
Physics problem: the mother of all physics problems
"Untitled" by Anonymous
Physics 7B
Spring Quarter 2002
University of California, Davis, CA
This doodle was found in a laboratory room, near the end of the second quarter of the college physics sequence at UC-Davis, and incorporates kinematics, Newton's laws, rotational motion, thermodynamics, and fluids!
Physics 7B
Spring Quarter 2002
University of California, Davis, CA
This doodle was found in a laboratory room, near the end of the second quarter of the college physics sequence at UC-Davis, and incorporates kinematics, Newton's laws, rotational motion, thermodynamics, and fluids!
Labels:
physics problem,
student scribble
20070712
Stopping global warming
Bizarro, by Dan Piraro
www.bizarro.com
July 3, 2007
Overcompensating, by Jeffrey Rowland
www.overcompensating.com
July 9, 2007 (excerpt)
Astronomy 10 learning goal Q4.4
Two tongue-in-cheek solutions to stop/reverse global warming. Dan Piraro suggests entities that would impact the environment in exactly the opposite manner as humans. Jeffrey Rowland (alluding to the Gaia hypothesis) makes the observation that the global warming is in itself a solution to its own problem, after humans have been killed off by global warming.
www.bizarro.com
July 3, 2007
Overcompensating, by Jeffrey Rowland
www.overcompensating.com
July 9, 2007 (excerpt)
Astronomy 10 learning goal Q4.4
Two tongue-in-cheek solutions to stop/reverse global warming. Dan Piraro suggests entities that would impact the environment in exactly the opposite manner as humans. Jeffrey Rowland (alluding to the Gaia hypothesis) makes the observation that the global warming is in itself a solution to its own problem, after humans have been killed off by global warming.
Labels:
global warming
20070711
Jovian coffee model (2)
Photo by Scott Liddell
Posted at morguefile.com
June 10, 2007
This is the background for modeling the precipitation of helium in Saturn's atmosphere as sugar in coffee. Tristan Guillot (2004) from the Observatoire de la Côte d'Azur in Nice, France, labels this model the "Gedanken Café."
This illustrates nicely what is covered in the textbook used in Astronomy 10 (introductory astronomy) at Cuesta College, San Luis Obispo, CA, where John D. Fix (University of Alabama in Huntsville) writes:
Clicker questions for peer-instruction exploring this model are covered in the previous post: Astronomy clicker question: jovian coffee model (1).
Tristan Guillot, "Probing the Giant Planets," Physics Today, vol. 57 no. 4, (April 2004), p. 63.
John D. Fix, Astronomy: Journey to the Cosmic Frontier, McGraw-Hill Higher Education, New York, NY, p. 279.
Posted at morguefile.com
June 10, 2007
This is the background for modeling the precipitation of helium in Saturn's atmosphere as sugar in coffee. Tristan Guillot (2004) from the Observatoire de la Côte d'Azur in Nice, France, labels this model the "Gedanken Café."
...Let's assume you're sitting comfortably in the sun at a cafe' and you're served an espresso. I like it sweet, and, in any case, our gedanken experiment requires sugar...
Drop finely granulated sugar instead of cubes into the coffee. If the sugar is fine enough, it will dissolve before reaching the bottom, and very little stirring is needed. But that's only if the espresso is hot enough. If it's only lukewarm, the sugar has a hard time dissolving; it tends to sink to the bottom. Finally, after all this preparation, let a friend arrive at the table and try to guess, from his first sip, how much sugar there is in the cup.
For giant planets, the problem is similar, only more complex. If the coffee itself is a good analog for the hydrogen, the sugar is replaced by...helium, [it] could, for example, either be mixed with the hydrogen or sequestered in the deepest regions.
This illustrates nicely what is covered in the textbook used in Astronomy 10 (introductory astronomy) at Cuesta College, San Luis Obispo, CA, where John D. Fix (University of Alabama in Huntsville) writes:
At temperatures above 10,000 K, helium is dissolved in metallic hydrogen. At lower temperatures it isn't. Instead, helium slowly settles inward through the metallic hydrogen, converting gravitational energy to heat. This probably doesn't happen within Jupiter because Jupiter is hot enough to keep helium dissolved in its metallic hydrogen. The interior of Saturn may be cool enough, however, that helium condenses at the top of the metallic hydrogen core and falls toward the center. As time passed, the envelope and atmosphere of Saturn should have become depleted in helium. This is exactly what measurements of the helium to hydrogen ratio in Saturn's atmosphere have shown.
Clicker questions for peer-instruction exploring this model are covered in the previous post: Astronomy clicker question: jovian coffee model (1).
Tristan Guillot, "Probing the Giant Planets," Physics Today, vol. 57 no. 4, (April 2004), p. 63.
John D. Fix, Astronomy: Journey to the Cosmic Frontier, McGraw-Hill Higher Education, New York, NY, p. 279.
20070710
Lolcat: doomed
Doomed, by menopaws
icanhascheezburger.com
July 5, 2007
That general sense of dread associated with an impending exam.
icanhascheezburger.com
July 5, 2007
That general sense of dread associated with an impending exam.
Labels:
grading,
lolcat,
study habits,
test anxiety
20070709
Godzilla, virtuoso
Woot.com contest: "Show us your dream musical supergroup performing live"
First place, cascadeking--L.A. Philharmonster
Winners' Gallery: the Best of Contest 128
June 26, 2007
Physics 8B learning goal Q1.4
Would the fundamental frequency of Godzilla's violin be infrasonic?
First place, cascadeking--L.A. Philharmonster
Winners' Gallery: the Best of Contest 128
June 26, 2007
Physics 8B learning goal Q1.4
Would the fundamental frequency of Godzilla's violin be infrasonic?
Labels:
resonance,
standing waves
20070706
Education research: overcoming initial problem-solving block
Arnold B. Arons (1997) from the University of Washington, Seattle, WA observes that an effect of firmly requiring use of a systematic problem-solving procedure is that
Physics 8AB problem rubric
Astronomy 10 (introductory astronomy, general education requirement) has a similar grading scale, where the "y" rubric provides an incentive for students to at least start writing and hopefully get past an initial "brain-lock," and the "t" rubric is the B-/C+ cutoff for the "worst of the best answers" (or "best of the worst answers").
Astronomy 10 short-answer rubric
Arnold B. Arons, Teaching Introductory Physics, John Wiley & Sons, Inc., New York, 1997, p. 39.
Wilbert J. McKeachie, Marilla Svinicki, McKeachie's Teaching Tips: Strategies, Research, and Theory for College and University Teachers (12th edition), Houghton Mifflin Company, Boston, 2006, p. 109.
Most students at this early stage in their development refuse to put pencil to paper, or to analyze the verbal-to-symbol transitions that are essential, until they "see" the solution as a whole. Requiring that they institute the procedure propels them, willy nilly, into the problem, and the momentum thus acquired frequently carries them through to the solution. The increasing satisfaction gained from such experiences gradually makes them more willing to penetrate a new problem, with pencil and paper and inquiry, without waiting until the entire solution has been perceived. This is a very large step indeed in intellectual development and capacity for abstract logical reasoning.Wilbert J. McKeachie, University of Michigan, Ann Arbor, MI, and Marilla Svinicki, University of Texas at Austin, TX (2006) have similar advice for the student:
If a question completely baffles you, start writing...anything you know that could possibly be relevant. This starts your memory functioning, and usually you'll soon find that you have some relevant ideas.For Physics 8AB (university physics, calculus-based) at Cuesta College in San Luis Obispo, CA, students are graded not so much as for obtaining the correct answer, but for how successful they are in implementing a constructive, systematic approach towards a solution. As an incentive to get students to get started by applying pencil to paper, they are given "pity points" for not leaving the page blank (e.g., writing down equations, restating knowns, or even writing "I don't know" avoids the "z" rubric for no response). However, students are warned that they can "b---s--- their way out of an 'F,' but can't b---s--- their way to an 'A.'" A throughly methodical approach that shows understanding and mastery of the essential concepts, up to the point where purely algebraic elimination and back-substitution of unknowns would begin is worth a "t" rubric, which on the grade scale is just below the B-/C+ cutoff.
Physics 8AB problem rubric
- p = 15/15: Correct.
- r = 12/15: Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
- t = 9/15: Contains right ideas, but discussion is unclear/incomplete or contains major errors.
- v = 6/15: Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
- x = 3/15: Implementation/application of ideas, but credit given for effort rather than merit.
- y = 1.5/15: Irrelevant discussion/effectively blank.
- z = 0/15: Blank.
Astronomy 10 (introductory astronomy, general education requirement) has a similar grading scale, where the "y" rubric provides an incentive for students to at least start writing and hopefully get past an initial "brain-lock," and the "t" rubric is the B-/C+ cutoff for the "worst of the best answers" (or "best of the worst answers").
Astronomy 10 short-answer rubric
- p = 20/20: Correct.
- r = 16/20: Nearly correct, but includes minor math errors.
- t = 12/20: Nearly correct, but approach has conceptual errors, and/or major/compounded math errors.
- v = 8/20: Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner.
- x = 4/20: Implementation of ideas, but credit given for effort rather than merit.
- y = 2/20: Irrelevant discussion/effectively blank.
- z = 0/20: Blank.
Arnold B. Arons, Teaching Introductory Physics, John Wiley & Sons, Inc., New York, 1997, p. 39.
Wilbert J. McKeachie, Marilla Svinicki, McKeachie's Teaching Tips: Strategies, Research, and Theory for College and University Teachers (12th edition), Houghton Mifflin Company, Boston, 2006, p. 109.
Labels:
education research,
grading,
problem-solving
20070705
Astronomy clicker question: jovian coffee model (1)
Astronomy 10, Summer Session 2007
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q6.1
Jupiter and Saturn
http://www.spirit-alembic.com/Issue5/jupiter-saturn.jpg
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the beginning of their learning cycle:
[0.3 points.] Why is there less helium in the upper atmosphere of Saturn, compared to Jupiter?
(A) Saturn does not conduct as much electricity.
(B) Saturn was made out of less helium.
(C) Helium escaped to become rings.
(D) Saturn is cooler.
Correct answer: not revealed yet (see discussion below).
The above question is asked after some preliminary discussion on mass, size, and density of Jupiter and Saturn, retention of core heat, and the generation of the belt-zone and cyclonic weather patterns. Since Saturn's rings have not yet been discussed, students are expecting it to be the cause of the lack of helium in Saturn's upper atmosphere, compared to Jupiter, which lacks prominent rings.
Student responses
Section 8027
(A) : 1 student
(B) : 0 students
(C) : 12 students
(D) : 0 students
Two coffee cups are brought out. Students are asked whether they are experienced coffee drinkers (as opposed to energy drink users), and are directed to the fact that while the same amount of sugar was stirred into either coffee cup, one cup was kept warm, while the other cup was chilled, before being asked the following clicker question.
[0.3 points.] Two cups of coffee each have three spoonfuls of sugar stirred into them. One cup is kept very hot, while the other cup is allowed to cool off. If you take a sip from just off the top of each cup, which will taste sweeter?
(A) The hotter cup of coffee will taste sweeter.
(B) The cooler cup of coffee will taste sweeter.
(C) (Both cups of coffee will taste equally sweet.)
(D) (Not enough information is given to determine which cup of coffee will taste sweeter.)
Correct answer: (A).
Student responses
Section 8027
(A) : 6 students
(B) : 7 students
(C) : 0 students
(D) : 0 students
Students offer that sip off of the top of the cold cup of coffee tastes less sweet because its sugar has settled into a sludge in the bottom of the cup. Then the first clicker question is put up again, to see if students make the connection.
[0.3 points.] Why is there less helium in the upper atmosphere of Saturn, compared to Jupiter?
(A) Saturn does not conduct as much electricity.
(B) Saturn was made out of less helium.
(C) Helium escaped to become rings.
(D) Saturn is cooler.
Correct answer: (A).
Student responses
Section 8027
(A) : 0 students
(B) : 1 student
(C) : 1 student
(D) : 11 students
In this model, Jupiter the hotter cup of coffee, while Saturn is the cooler cup of coffee (being smaller in mass, it did not have as much original formation and radioactive decay heat, and cooled off faster). Coffee represents hydrogen molecules, while sugar represents helium in the atmospheres of these jovian planets. Jupiter's warmer atmosphere allows it to keep helium suspended (sugar stays dissolved in hot coffee, due to the thermal motion of water molecules), while Saturn's atmosphere is too cool, and helium begins to fall inwards (sugar is allowed to precipitate). This explains both the lack of helium in the upper atmosphere of Saturn, and also the observation that while cooler Saturn has less active belt-zone and cyclonic weather patterns than Jupiter, Saturn is not as cool as expected for its mass, as helium condensation now provides a secondary source of energy for Saturn.
Further discussion of this model is covered in a subsequent post: Jovian coffee model (2).
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q6.1
Jupiter and Saturn
http://www.spirit-alembic.com/Issue5/jupiter-saturn.jpg
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the beginning of their learning cycle:
[0.3 points.] Why is there less helium in the upper atmosphere of Saturn, compared to Jupiter?
(A) Saturn does not conduct as much electricity.
(B) Saturn was made out of less helium.
(C) Helium escaped to become rings.
(D) Saturn is cooler.
Correct answer: not revealed yet (see discussion below).
The above question is asked after some preliminary discussion on mass, size, and density of Jupiter and Saturn, retention of core heat, and the generation of the belt-zone and cyclonic weather patterns. Since Saturn's rings have not yet been discussed, students are expecting it to be the cause of the lack of helium in Saturn's upper atmosphere, compared to Jupiter, which lacks prominent rings.
Student responses
Section 8027
(A) : 1 student
(B) : 0 students
(C) : 12 students
(D) : 0 students
Two coffee cups are brought out. Students are asked whether they are experienced coffee drinkers (as opposed to energy drink users), and are directed to the fact that while the same amount of sugar was stirred into either coffee cup, one cup was kept warm, while the other cup was chilled, before being asked the following clicker question.
[0.3 points.] Two cups of coffee each have three spoonfuls of sugar stirred into them. One cup is kept very hot, while the other cup is allowed to cool off. If you take a sip from just off the top of each cup, which will taste sweeter?
(A) The hotter cup of coffee will taste sweeter.
(B) The cooler cup of coffee will taste sweeter.
(C) (Both cups of coffee will taste equally sweet.)
(D) (Not enough information is given to determine which cup of coffee will taste sweeter.)
Correct answer: (A).
Student responses
Section 8027
(A) : 6 students
(B) : 7 students
(C) : 0 students
(D) : 0 students
Students offer that sip off of the top of the cold cup of coffee tastes less sweet because its sugar has settled into a sludge in the bottom of the cup. Then the first clicker question is put up again, to see if students make the connection.
[0.3 points.] Why is there less helium in the upper atmosphere of Saturn, compared to Jupiter?
(A) Saturn does not conduct as much electricity.
(B) Saturn was made out of less helium.
(C) Helium escaped to become rings.
(D) Saturn is cooler.
Correct answer: (A).
Student responses
Section 8027
(A) : 0 students
(B) : 1 student
(C) : 1 student
(D) : 11 students
In this model, Jupiter the hotter cup of coffee, while Saturn is the cooler cup of coffee (being smaller in mass, it did not have as much original formation and radioactive decay heat, and cooled off faster). Coffee represents hydrogen molecules, while sugar represents helium in the atmospheres of these jovian planets. Jupiter's warmer atmosphere allows it to keep helium suspended (sugar stays dissolved in hot coffee, due to the thermal motion of water molecules), while Saturn's atmosphere is too cool, and helium begins to fall inwards (sugar is allowed to precipitate). This explains both the lack of helium in the upper atmosphere of Saturn, and also the observation that while cooler Saturn has less active belt-zone and cyclonic weather patterns than Jupiter, Saturn is not as cool as expected for its mass, as helium condensation now provides a secondary source of energy for Saturn.
Further discussion of this model is covered in a subsequent post: Jovian coffee model (2).
Labels:
astronomy think-pair-share question,
coffee,
Jupiter,
Saturn
20070703
Astronomy quiz question: molecule escape from atmosphere
Astronomy 10 Quiz 4, Summer Session 2007
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q4.1
[3.0 points.] Which one of the following statements best explains why it is not possible for molecules near sea level to escape from the Earth's atmosphere into space?
(A) Speeds are too slow.
(B) Frequent collisions with other molecules.
(C) Low temperatures.
(D) Infrared radiation trapped by the greenhouse effect.
(E) Tectonic plate motion and subduction.
Correct answer: (B).
The average speed of a molecule/atom in the atmosphere depends on the mass of the molecule/atom, and on the temperature. If this speed is greater than approximately one-sixth of the escape velocity of a planet (which depends on the mass of the planet), then the atmosphere will eventually leak out over several billion years. However, molecules/atoms may only escape from the exosphere (the upper atmosphere); no matter how fast their average speed is in the lower level, frequent collisions there prevent escape into space. This is the "mosh pit effect"--you can't leave the mosh pit if you are in the middle of the moshers, but once you can make your way to the edge, escape is possible (provided you don't get pulled back in).
Student responses
Section 8027
(A) : 3 students
(B) : 7 students
(C) : 1 student
(D) : 0 students
(E) : 2 students
Sinfest, by Tatsuya Ishida
www.sinfest.net
January 25, 2001 (excerpt)
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q4.1
[3.0 points.] Which one of the following statements best explains why it is not possible for molecules near sea level to escape from the Earth's atmosphere into space?
(A) Speeds are too slow.
(B) Frequent collisions with other molecules.
(C) Low temperatures.
(D) Infrared radiation trapped by the greenhouse effect.
(E) Tectonic plate motion and subduction.
Correct answer: (B).
The average speed of a molecule/atom in the atmosphere depends on the mass of the molecule/atom, and on the temperature. If this speed is greater than approximately one-sixth of the escape velocity of a planet (which depends on the mass of the planet), then the atmosphere will eventually leak out over several billion years. However, molecules/atoms may only escape from the exosphere (the upper atmosphere); no matter how fast their average speed is in the lower level, frequent collisions there prevent escape into space. This is the "mosh pit effect"--you can't leave the mosh pit if you are in the middle of the moshers, but once you can make your way to the edge, escape is possible (provided you don't get pulled back in).
Student responses
Section 8027
(A) : 3 students
(B) : 7 students
(C) : 1 student
(D) : 0 students
(E) : 2 students
Sinfest, by Tatsuya Ishida
www.sinfest.net
January 25, 2001 (excerpt)
20070702
Astronomy midterm question: light-gathering power
Astronomy 10 Midterm 1, Summer Session 2007
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal M1.3
[3.0 points.] Which one of the following choices best explains why telescopes with larger diameter objective mirrors are better than telescopes with smaller diameter objective mirrors?
(A) More light can be collected.
(B) Wider field of view.
(C) Easier to adapt its optics for different people.
(D) It is cheaper to make mirrors as large as possible.
(E) Better atmospheric seeing.
Correct answer: (A).
Light-gathering power is the ability to see dim objects, and depends on the area of the objective mirror (and thus proportional to the square of the diameter). As for (B), if the mirror had a longer focal length (which is not necessarily related to the diameter), then it would have a higher magnification power, and thus a narrower field of view. The other choices (C)-(E) are merely garbled distractors.
Student responses
Section 8027
(A) 7 students
(B) 5 students
(C) 0 students
(D) 0 students
(E) 1 student
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal M1.3
[3.0 points.] Which one of the following choices best explains why telescopes with larger diameter objective mirrors are better than telescopes with smaller diameter objective mirrors?
(A) More light can be collected.
(B) Wider field of view.
(C) Easier to adapt its optics for different people.
(D) It is cheaper to make mirrors as large as possible.
(E) Better atmospheric seeing.
Correct answer: (A).
Light-gathering power is the ability to see dim objects, and depends on the area of the objective mirror (and thus proportional to the square of the diameter). As for (B), if the mirror had a longer focal length (which is not necessarily related to the diameter), then it would have a higher magnification power, and thus a narrower field of view. The other choices (C)-(E) are merely garbled distractors.
Student responses
Section 8027
(A) 7 students
(B) 5 students
(C) 0 students
(D) 0 students
(E) 1 student
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