Astronomy 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal F.2
Students were asked the following clicker question (Classroom Performance System, einstruction.com) in the middle of their learning cycle.
[0.3 points.] The Drake equation may predict that the Milky Way has many advanced technological civilizations. Which argument against this extremely optimistic result do you consider most valid?
(A) If there are supposed to be so many of them, then why aren't they here (yet)?
(B) Our advanced technological civilization on Earth arose due to an extraordinary number of highly unlikely coincidences and conditions.
(C) Since are no simple explanations as to why there would be zero contact from many other advanced technological civilizations (if they do exist), they do not exist.
(D) (More than one of the above arguments.)
Correct answer: (none; opinion only)
Each of these arguments was explained to the class before this question was asked. Item (A) is Fermi's Paradox; (B) is the Rare Earth Hypothesis; (C) is Occam's Razor (which is related to Fermi's Paradox). Students were instructed to choose the argument they considered the most valid, and only to select (D) if they were willing to talk about which of these arguments they considered equally valid.
Student responses
Section 4160
(A) : 5 students
(B) : 23 students
(C) : 5 students
(D) : 0 students
Section 5166
(A) : 4 students
(B) : 27 students
(C) : 12 students
(D) : 6 students
20080530
Astronomy quiz question: horizon problem
Astronomy 10 Quiz 12, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q12.4
[3.0 points.] Which one of the following statements best explains how inflation explains the "isotropy" of the universe today (equal amounts and same temperatures of cosmic background radiation coming from every direction)?
(A) Equal amounts of luminous matter and dark matter canceled each other out.
(B) The first generation of stars were pulled back apart into gaseous nebulae.
(C) Tiny quantum fluctuations in the universe expanded to become galaxy cluster formations.
(D) The curvature of the early universe expanded to become nearly flat.
(E) A very small part of an infinite universe increased to become the entire universe that we can see today.
Correct answer: (E)
This is the horizon problem of the cosmic background radiation, which is solved by the inflationary era. Since the universe is not old enough for the entire visible universe to have reached near thermal equilibrium, it must have experienced a brief, sudden expansion soon after recombination.
Student responses
Section 5166
(A) : 14 students
(B) : 5 students
(C) : 13 students
(D) : 10 students
(E) : 11 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q12.4
[3.0 points.] Which one of the following statements best explains how inflation explains the "isotropy" of the universe today (equal amounts and same temperatures of cosmic background radiation coming from every direction)?
(A) Equal amounts of luminous matter and dark matter canceled each other out.
(B) The first generation of stars were pulled back apart into gaseous nebulae.
(C) Tiny quantum fluctuations in the universe expanded to become galaxy cluster formations.
(D) The curvature of the early universe expanded to become nearly flat.
(E) A very small part of an infinite universe increased to become the entire universe that we can see today.
Correct answer: (E)
This is the horizon problem of the cosmic background radiation, which is solved by the inflationary era. Since the universe is not old enough for the entire visible universe to have reached near thermal equilibrium, it must have experienced a brief, sudden expansion soon after recombination.
Student responses
Section 5166
(A) : 14 students
(B) : 5 students
(C) : 13 students
(D) : 10 students
(E) : 11 students
20080529
Astronomy quiz question: Hubble's law
Astronomy 10 Quiz 12, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q12.2
[3.0 points.] Which one of the following statements best describes Hubble's law?
(A) Distant galaxies recede faster than nearby galaxies.
(B) Spacetime becomes more flat over time.
(C) Expansion must be followed by an equal and opposite contraction.
(D) There are equal amounts of dark matter and dark energy in the universe.
(E) The distances to other galaxies are farther away than expected.
Correct answer: (A)
Hubble's law correlates the recession speeds (v) of galaxies to their distances (d) from the Earth, where:
v = H*d.
Response (B) is related to the flatness problem of the cosmic background radiation, which is solved by the inflationary era. Response (C) is a mangled (but apparently convincing) version of Newton's third law: "For every action there must be an equal and opposite reaction." Response (E) is evidence (from type Ia supernovae) that the expansion of the universe is relatively recent.
Student responses
Section 5166
(A) : 36 students
(B) : 1 student
(C) : 6 students
(D) : 4 students
(E) : 2 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q12.2
[3.0 points.] Which one of the following statements best describes Hubble's law?
(A) Distant galaxies recede faster than nearby galaxies.
(B) Spacetime becomes more flat over time.
(C) Expansion must be followed by an equal and opposite contraction.
(D) There are equal amounts of dark matter and dark energy in the universe.
(E) The distances to other galaxies are farther away than expected.
Correct answer: (A)
Hubble's law correlates the recession speeds (v) of galaxies to their distances (d) from the Earth, where:
v = H*d.
Response (B) is related to the flatness problem of the cosmic background radiation, which is solved by the inflationary era. Response (C) is a mangled (but apparently convincing) version of Newton's third law: "For every action there must be an equal and opposite reaction." Response (E) is evidence (from type Ia supernovae) that the expansion of the universe is relatively recent.
Student responses
Section 5166
(A) : 36 students
(B) : 1 student
(C) : 6 students
(D) : 4 students
(E) : 2 students
20080528
Astronomy quiz question: universe inflation evidence
Astronomy 10 Quiz 12, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q12.4
[3.0 points.] Which one of the following choices best describes the evidence for the sudden early inflation of space in the universe?
(A) Stars in the inner and outer parts of the disk orbit the center of the Milky Way with the same speed.
(B) Each and every galaxy sees all other galaxies moving away from its location.
(C) Nearly equal amounts of cosmic background radiation are detected from all directions.
(D) Distant galaxies appear to be much younger than nearby galaxies.
(E) Type Ia supernovae in distant galaxies appear to be dimmer than expected from their redshifts.
Correct answer: (C)
Response (C) is the horizon problem of the cosmic background radiation, which may be explained by being a very small portion of the universe inflating suddenly and briefly to become the extent of the visible universe today. Response (A) is evidence for dark matter; response (B) is a reason why the expansion of the universe can be said to have no center; response (D) is a result of the finite speed of light ("lookback time"); while response (E) is evidence for the recent acceleration of the universe's expansion rate, caused by dark energy.
Student responses
Section 4160
(A) : 3 students
(B) : 4 students
(C) : 10 student
(D) : 0 students
(E) : 19 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q12.4
[3.0 points.] Which one of the following choices best describes the evidence for the sudden early inflation of space in the universe?
(A) Stars in the inner and outer parts of the disk orbit the center of the Milky Way with the same speed.
(B) Each and every galaxy sees all other galaxies moving away from its location.
(C) Nearly equal amounts of cosmic background radiation are detected from all directions.
(D) Distant galaxies appear to be much younger than nearby galaxies.
(E) Type Ia supernovae in distant galaxies appear to be dimmer than expected from their redshifts.
Correct answer: (C)
Response (C) is the horizon problem of the cosmic background radiation, which may be explained by being a very small portion of the universe inflating suddenly and briefly to become the extent of the visible universe today. Response (A) is evidence for dark matter; response (B) is a reason why the expansion of the universe can be said to have no center; response (D) is a result of the finite speed of light ("lookback time"); while response (E) is evidence for the recent acceleration of the universe's expansion rate, caused by dark energy.
Student responses
Section 4160
(A) : 3 students
(B) : 4 students
(C) : 10 student
(D) : 0 students
(E) : 19 students
20080527
Physics clicker question: blackbody versus silverbody
Physics 5A, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 14.22
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the start of their learning cycle:
[0.6 participation points.] What should the outside of your jacket be like, in order to stay as warm as possible when hiking at night (in the dark)?
(A) Shiny, reflective.
(B) Black, matte.
(C) (It depends on how cold it is outside.)
(D) (It doesn't matter what type of jacket is worn outside in the dark.)
(E) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 11 students
(B) : 5 students
(C) : 0 students
(D) : 15 students
(E) : 0 students
Correct answer: (A)
The rate of heat absorbed/emitted by an object depends on its emissivity: 1 for a perfect blackbody, and 0 for a perfect silverbody. While a blackbody will be an efficient absorber, it will also be a perfect emitter, which is desirable when cooling is an issue (e.g., car radiators, bottom tiles on the Space Shuttle), but at night, when heat loss by radiation needs to be minimized, a jacket approximating a silverbody would be better.
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 14.22
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the start of their learning cycle:
[0.6 participation points.] What should the outside of your jacket be like, in order to stay as warm as possible when hiking at night (in the dark)?
(A) Shiny, reflective.
(B) Black, matte.
(C) (It depends on how cold it is outside.)
(D) (It doesn't matter what type of jacket is worn outside in the dark.)
(E) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 11 students
(B) : 5 students
(C) : 0 students
(D) : 15 students
(E) : 0 students
Correct answer: (A)
The rate of heat absorbed/emitted by an object depends on its emissivity: 1 for a perfect blackbody, and 0 for a perfect silverbody. While a blackbody will be an efficient absorber, it will also be a perfect emitter, which is desirable when cooling is an issue (e.g., car radiators, bottom tiles on the Space Shuttle), but at night, when heat loss by radiation needs to be minimized, a jacket approximating a silverbody would be better.
20080526
Physics clicker question: lava lamps
Physics 5A, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 14.8
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the start of their learning cycle:
[0.6 participation points.] Why do wax droplets rise in a lava lamp after being heated in the base?
(A) Heat rises.
(B) Weight decreases.
(C) Density decreases.
(D) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 9 students
(B) : 0 students
(C) : 21 students
(D) : 1 student
Correct answer: (C)
A lava lamp is an excellent demonstration of the principles of thermal convection. The light bulb at the bottom of the lava lamp heats the wax, which causes it to expand, decreasing in density. The expanded wax then rises, as it is now less dense than the fluid it is suspended in. Later after cooling at the top of the fluid, the wax will decrease in volume, increasing in density, and later sink.
"How lava lamps work"
http://home.howstuffworks.com/lava-lamp.htm
http://video.google.com/videoplay?docid=1145452462666090515
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 14.8
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the start of their learning cycle:
[0.6 participation points.] Why do wax droplets rise in a lava lamp after being heated in the base?
(A) Heat rises.
(B) Weight decreases.
(C) Density decreases.
(D) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 9 students
(B) : 0 students
(C) : 21 students
(D) : 1 student
Correct answer: (C)
A lava lamp is an excellent demonstration of the principles of thermal convection. The light bulb at the bottom of the lava lamp heats the wax, which causes it to expand, decreasing in density. The expanded wax then rises, as it is now less dense than the fluid it is suspended in. Later after cooling at the top of the fluid, the wax will decrease in volume, increasing in density, and later sink.
"How lava lamps work"
http://home.howstuffworks.com/lava-lamp.htm
http://video.google.com/videoplay?docid=1145452462666090515
20080525
Physics clicker question: latent heat
Physics 5A, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Question 14.4
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the start of their learning cycle:
[0.6 participation points.] The heat capacity of a material undergoing a phase change (such as melting/freezing, or boiling/condensing) is typically:
(A) 0.
(B) ∞.
(C) some finite number.
(D) (Not enough information is given.)
(E) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 8 students
(B) : 7 students
(C) : 12 students
(D) : 1 student
(E) : 0 students
Correct answer: (B)
The heat capacity is given by:
C = Q/delta(T).
When heat is put into (or taken from) a system undergoing a phase change, the temperature remains constant, and thus the denominator is zero, making the heat capacity infinite. Thus a system undergoing a phase change cannot have a defined heat capacity, as the heat goes into breaking or making bonds, instead of increasing/decreasing the amount of thermal energy in the system. Instead, the latent heat is defined:
L = Q/m,
where m is the amount of material undergoing a phase change.
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Question 14.4
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the start of their learning cycle:
[0.6 participation points.] The heat capacity of a material undergoing a phase change (such as melting/freezing, or boiling/condensing) is typically:
(A) 0.
(B) ∞.
(C) some finite number.
(D) (Not enough information is given.)
(E) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 8 students
(B) : 7 students
(C) : 12 students
(D) : 1 student
(E) : 0 students
Correct answer: (B)
The heat capacity is given by:
C = Q/delta(T).
When heat is put into (or taken from) a system undergoing a phase change, the temperature remains constant, and thus the denominator is zero, making the heat capacity infinite. Thus a system undergoing a phase change cannot have a defined heat capacity, as the heat goes into breaking or making bonds, instead of increasing/decreasing the amount of thermal energy in the system. Instead, the latent heat is defined:
L = Q/m,
where m is the amount of material undergoing a phase change.
20080524
Astronomy free-response question: Kepler's Third Law
Astronomy 10L Lab 12, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Students were asked the following open-ended question in an online pre-lab assignment, in order to prepare for a lab on determining the mass of Jupiter from the orbital periods and semi-major axes of its Galilean satellites:
Student responses
Sections 4161, 4162, 4163, 6544
Most responses were correct or nearly correct:
Cuesta College, San Luis Obispo, CA
Students were asked the following open-ended question in an online pre-lab assignment, in order to prepare for a lab on determining the mass of Jupiter from the orbital periods and semi-major axes of its Galilean satellites:
[0.2 points.] In your own words, describe Kepler's third law. (Graded for completion.)Students were graded for any non-blank response that demonstrates a serious (or not-so serious) attempt at completing the assignment.
Student responses
Sections 4161, 4162, 4163, 6544
Most responses were correct or nearly correct:
"The squares of the orbital periods of planets are directly proportional to the cubes of the semi-major axis of the orbits."A number of responses discussed Kepler's first or second laws, Newton's laws of motion, or even laws of cooling:
--K. D.
"The farther a planet is away from the sun, the longer time it will take for that planet to orbit the sun once."
--C. R.
"Which one was that... is that the one where all planets... I give up. The first one is about ellipses,... it has something to do with how long planets take to orbit the sun, I think."
--A. R. (1)
"the close ones go faster just like michelle kwan."
--M. M.
"Equal distances are covered in equal times."Some responses went in apparently just for kicks (but still get full "completion" credit):
--A. B.
"It represents the motion of planets around the sun."
--E. D.
"An object in motion will remain in motion."
--J. A.
"Area to volume ratio! yea."
--S. M.
"its a law that kelper made."
--A. R. (2)
"My words dont do this law Justice."
--R. G.
"Drawin' a blank."
--B. M.
"Keplers third law is never talk about keplers first two laws."
--D. F.
20080523
Planetary rings: tire debris
Photoshop Phriday: World Without Wheels, by Cuddly Dansomethingawful.com
May 16, 2008
Oddly enough, the plane of the ring does not coincide with the equatorial plane of the Earth.
Astronomy quiz question: globular cluster star formation
Astronomy 10 Quiz 11, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.5
[3.0 points.] Which one of the following statements best explains why there are no young stars in the halo clusters of the Milky Way?
(A) There are no longer enough heavy elements remaining in the halo.
(B) There are not enough gravitational forces in the halo for new stars to form.
(C) The dark matter in the halo prevents new stars from forming there.
(D) All of the metal-rich stars in the halo clusters have already become metal-poor stars.
(E) All of the interstellar hydrogen in the halo moved into the disk.
Correct answer: (E)
According to the monolithic formation theory of the Milky Way, the gas and dust in our galaxy was originally spherically shaped. Then as the first-generation stars of the globular clusters formed, hydrogen in the Milky Way gradually collapsed into the thin disk structure that it has today. Thus no new stars can form in the relatively empty spherical halo that surrounds the flattened Milky Way.
Student responses
Section 5166
(A) : 3 students
(B) : 16 students
(C) : 6 students
(D) : 7 students
(E) : 18 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.5
[3.0 points.] Which one of the following statements best explains why there are no young stars in the halo clusters of the Milky Way?
(A) There are no longer enough heavy elements remaining in the halo.
(B) There are not enough gravitational forces in the halo for new stars to form.
(C) The dark matter in the halo prevents new stars from forming there.
(D) All of the metal-rich stars in the halo clusters have already become metal-poor stars.
(E) All of the interstellar hydrogen in the halo moved into the disk.
Correct answer: (E)
According to the monolithic formation theory of the Milky Way, the gas and dust in our galaxy was originally spherically shaped. Then as the first-generation stars of the globular clusters formed, hydrogen in the Milky Way gradually collapsed into the thin disk structure that it has today. Thus no new stars can form in the relatively empty spherical halo that surrounds the flattened Milky Way.
Student responses
Section 5166
(A) : 3 students
(B) : 16 students
(C) : 6 students
(D) : 7 students
(E) : 18 students
20080522
Astronomy quiz question: spiral arm tracers
Astronomy 10 Quiz 11, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.2
[3.0 points.] Which one of the following statements best explains why massive main sequence stars are used to map the spiral arm structure of the Milky Way?
(A) They emit 21 cm radio waves.
(B) They are extremely metal-poor.
(C) They will explode as type II supernovae.
(D) They will become neutron stars or black holes.
(E) They are very luminous and short-lived.
Correct answer: (E)
Due to the dust and gas in the disk of the Milky Way, only a very small fraction of our galaxy is visible from the Earth. Thus massive main sequence stars, which are very luminous, would be most visible through the interstellar medium. These massive stars also have the property of forming as they enter the spiral arms, and due to their short lives die as they leave the spiral arms, making them de facto "spiral arm tracers."
Student responses
Section 5166
(A) : 10 students
(B) : 6 students
(C) : 1 student
(D) : 2 students
(E) : 31 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.2
[3.0 points.] Which one of the following statements best explains why massive main sequence stars are used to map the spiral arm structure of the Milky Way?
(A) They emit 21 cm radio waves.
(B) They are extremely metal-poor.
(C) They will explode as type II supernovae.
(D) They will become neutron stars or black holes.
(E) They are very luminous and short-lived.
Correct answer: (E)
Due to the dust and gas in the disk of the Milky Way, only a very small fraction of our galaxy is visible from the Earth. Thus massive main sequence stars, which are very luminous, would be most visible through the interstellar medium. These massive stars also have the property of forming as they enter the spiral arms, and due to their short lives die as they leave the spiral arms, making them de facto "spiral arm tracers."
Student responses
Section 5166
(A) : 10 students
(B) : 6 students
(C) : 1 student
(D) : 2 students
(E) : 31 students
20080521
Astronomy quiz question: globular cluster age
Astronomy 10 Quiz 11, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.5
[3.0 points.] Which one of the following statements best supports the theory that globular cluster stars in the halo of the Milky Way must be older than stars in the disk?
(A) Globular cluster stars are more abundant.
(B) Globular cluster stars are more luminous.
(C) Globular cluster stars are metal-rich.
(D) Globular clusters do not contain massive main sequence stars.
(E) Globular clusters do not contain dark matter.
Correct answer: (D)
Massive stars have much shorter main sequence lives than medium mass and low mass stars, thus globular clusters, being older, do not have massive main sequence stars. Note that responses (B) and (C) would correspond to very young stars.
Student responses
Section 4160
(A) : 1 student
(B) : 3 students
(C) : 12 students
(D) : 16 students
(E) : 1 student
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.5
[3.0 points.] Which one of the following statements best supports the theory that globular cluster stars in the halo of the Milky Way must be older than stars in the disk?
(A) Globular cluster stars are more abundant.
(B) Globular cluster stars are more luminous.
(C) Globular cluster stars are metal-rich.
(D) Globular clusters do not contain massive main sequence stars.
(E) Globular clusters do not contain dark matter.
Correct answer: (D)
Massive stars have much shorter main sequence lives than medium mass and low mass stars, thus globular clusters, being older, do not have massive main sequence stars. Note that responses (B) and (C) would correspond to very young stars.
Student responses
Section 4160
(A) : 1 student
(B) : 3 students
(C) : 12 students
(D) : 16 students
(E) : 1 student
20080520
Astronomy quiz question: globular cluster locations
Astronomy 10 Quiz 11, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.2
[3.0 points.] Which one of the following statements best describes what can be deduced from observing the positions of globular clusters in the halo of the Milky Way?
(A) The location of the center of the Milky Way.
(B) Globular clusters are composed of dark matter.
(C) The universe is steadily expanding.
(D) The spiral arm structure of the Milky Way.
(E) The location of metal-rich stars in the Milky Way.
Correct answer: (A)
Due to the dust and gas in the disk of the Milky Way, only a very small fraction of our galaxy is visible from the Earth. Thus to determine the overall size of the Milky Way, and the distance from the Earth to its center, it is necessary to observe the positions of globular clusters in the halo, which are visible well above and below the plane of the disk. Students who chose response (D) are probably confusing globular clusters with spiral arm tracers.
Student responses
Section 4160
(A) : 11 students
(B) : 1 student
(C) : 0 students
(D) : 22 students
(E) : 0 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.2
[3.0 points.] Which one of the following statements best describes what can be deduced from observing the positions of globular clusters in the halo of the Milky Way?
(A) The location of the center of the Milky Way.
(B) Globular clusters are composed of dark matter.
(C) The universe is steadily expanding.
(D) The spiral arm structure of the Milky Way.
(E) The location of metal-rich stars in the Milky Way.
Correct answer: (A)
Due to the dust and gas in the disk of the Milky Way, only a very small fraction of our galaxy is visible from the Earth. Thus to determine the overall size of the Milky Way, and the distance from the Earth to its center, it is necessary to observe the positions of globular clusters in the halo, which are visible well above and below the plane of the disk. Students who chose response (D) are probably confusing globular clusters with spiral arm tracers.
Student responses
Section 4160
(A) : 11 students
(B) : 1 student
(C) : 0 students
(D) : 22 students
(E) : 0 students
20080518
Physics quiz question: moles of H_2
Physics 5A Quiz 7, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 13.14
[3.0 points.] How many moles of H2 molecules are there in 0.40 g of H_2? The atomic mass of a single hydrogen atom is 1.0 u.
(A) 0.20 moles.
(B) 0.40 moles.
(C) 4.0 moles.
(D) 5.0 moles.
Correct answer: (A)
A mole of H_2 molecules will have a mass of 2.0 g. Thus 0.40 g of H_2 molecules must be less than a mole, and thus corresponds to (0.40 g)*((1 mole)/(2.0 g)) = 0.2 moles. Response (B) is using a molecular mass of 1.0 g per mole; response (B) is 1,000*(0.4), while response (D) is the molecular mass divided by moles.
Student responses
Sections 4987, 4988
(A) : 12 students
(B) : 12 students
(C) : 7 students
(D) : 0 students
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 13.14
[3.0 points.] How many moles of H2 molecules are there in 0.40 g of H_2? The atomic mass of a single hydrogen atom is 1.0 u.
(A) 0.20 moles.
(B) 0.40 moles.
(C) 4.0 moles.
(D) 5.0 moles.
Correct answer: (A)
A mole of H_2 molecules will have a mass of 2.0 g. Thus 0.40 g of H_2 molecules must be less than a mole, and thus corresponds to (0.40 g)*((1 mole)/(2.0 g)) = 0.2 moles. Response (B) is using a molecular mass of 1.0 g per mole; response (B) is 1,000*(0.4), while response (D) is the molecular mass divided by moles.
Student responses
Sections 4987, 4988
(A) : 12 students
(B) : 12 students
(C) : 7 students
(D) : 0 students
20080517
Physics quiz question: cooling a brim-filled container
Physics 5A Quiz 7, spring semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 13.14
[3.0 points.] A cylindrical aluminum container is filled to the brim with water when the system is at 30.0° C. The temperature of the container and the water is lowered to 5.0° C. The volume expansion coefficient of aluminum is 69 ×10-6 K–1. The volume expansion coefficient of water is 210×10-6 K–1. What happens to the water when the temperature of the container and the water is lowered to 5.0° C?
(A) Some water will overflow the container.
(B) No water will overflow, and the water level will still be at the brim of the container.
(C) No water will overflow, and the water level will be below the brim of the container.
(D) (Not enough information is given to determine this.)
Correct answer: (C)
The volume expansion coefficient for water is greater than the volume expansion coefficient for aluminum. If they experience the same decrease in temperature, the water will decrease in volume more than the interior of the aluminum container, and thus will no longer be up at the brim of the container. Response (A) would occur if the water and the aluminum container experience the same increase in temperature, starting with the water at the brim of the container. Response (B) would only occur if water and aluminum had the same volume expansion coefficient.
Student responses
Sections 4987, 4988
(A) : 12 students
(B) : 8 students
(C) : 12 students
(D) : 0 students
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 13.14
[3.0 points.] A cylindrical aluminum container is filled to the brim with water when the system is at 30.0° C. The temperature of the container and the water is lowered to 5.0° C. The volume expansion coefficient of aluminum is 69 ×10-6 K–1. The volume expansion coefficient of water is 210×10-6 K–1. What happens to the water when the temperature of the container and the water is lowered to 5.0° C?
(A) Some water will overflow the container.
(B) No water will overflow, and the water level will still be at the brim of the container.
(C) No water will overflow, and the water level will be below the brim of the container.
(D) (Not enough information is given to determine this.)
Correct answer: (C)
The volume expansion coefficient for water is greater than the volume expansion coefficient for aluminum. If they experience the same decrease in temperature, the water will decrease in volume more than the interior of the aluminum container, and thus will no longer be up at the brim of the container. Response (A) would occur if the water and the aluminum container experience the same increase in temperature, starting with the water at the brim of the container. Response (B) would only occur if water and aluminum had the same volume expansion coefficient.
Student responses
Sections 4987, 4988
(A) : 12 students
(B) : 8 students
(C) : 12 students
(D) : 0 students
20080516
Physics clicker question: temperature and temperature changes
Physics 5A, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problems 13.2, 13.4, Conceptual Question 13.3
Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the start of their learning cycle:
[0.6 participation points.] Which one of the following temperatures is the warmest?
(A) 10° C.
(B) 10 K.
(C) 10° F.
(D) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 5 students
(B) : 14 students
(C) : 0 students
(D) : 0 students
Correct answer: (A)
The coldest temperature is (B), which is most nearest to absolute zero (perhaps demonstrating the students' unfamiliarity with the Kelvin scale). Between (A) and (C), 10° F is below freezing (32° F), while 10° C is above freezing (0° C), so response (A) is the warmest temperature.
[Follow-up question]
[0.6 participation points.] Which one of the following temperature changes represents the largest amount of cooling?
(A) –10° C.
(B) –10 K.
(C) (Both changes in temperature corresponds to the same amount of cooling.)
(D) (Both the initial and final temperatures need to be given.)
(E) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 10 students
(B) : 10 students
(C) : 2 students
(D) : 1 student
(E) : 0 students
Correct answer: (C)
Apparently these students are not at all familiar with temperature scales and changes in temperatures (preoccupied on an impending semi-comprehensive midterm on chapters previous to this introduction to thermodynamics?). It is pointed out to them after these clicker questions that T (°C) ≠ T (K), but they are related by an "offset":
T (°C) = T (K) - 273.15.
However, even though these temperatures are not equivalent, changes in these temperatures are, such that:
delta_T (°C) = delta_T (K).
Students are warned that this property of temperature changes will come in handy throughout thermodynamics (expansion, heat, etc. calculations).
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problems 13.2, 13.4, Conceptual Question 13.3
Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the start of their learning cycle:
[0.6 participation points.] Which one of the following temperatures is the warmest?
(A) 10° C.
(B) 10 K.
(C) 10° F.
(D) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 5 students
(B) : 14 students
(C) : 0 students
(D) : 0 students
Correct answer: (A)
The coldest temperature is (B), which is most nearest to absolute zero (perhaps demonstrating the students' unfamiliarity with the Kelvin scale). Between (A) and (C), 10° F is below freezing (32° F), while 10° C is above freezing (0° C), so response (A) is the warmest temperature.
[Follow-up question]
[0.6 participation points.] Which one of the following temperature changes represents the largest amount of cooling?
(A) –10° C.
(B) –10 K.
(C) (Both changes in temperature corresponds to the same amount of cooling.)
(D) (Both the initial and final temperatures need to be given.)
(E) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 10 students
(B) : 10 students
(C) : 2 students
(D) : 1 student
(E) : 0 students
Correct answer: (C)
Apparently these students are not at all familiar with temperature scales and changes in temperatures (preoccupied on an impending semi-comprehensive midterm on chapters previous to this introduction to thermodynamics?). It is pointed out to them after these clicker questions that T (°C) ≠ T (K), but they are related by an "offset":
T (°C) = T (K) - 273.15.
However, even though these temperatures are not equivalent, changes in these temperatures are, such that:
delta_T (°C) = delta_T (K).
Students are warned that this property of temperature changes will come in handy throughout thermodynamics (expansion, heat, etc. calculations).
20080515
Astronomy clicker question: dark matter evidence
Astronomy 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.3
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the start of their learning cycle.
[0.3 points.] How do we know that there is dark matter in the halo of the Milky Way (above/below its disk)?
(A) The temperature of empty space is not really absolute zero.
(B) Stars just outside of the Milky Way appear dimmer than they should be.
(C) Stars in the inner and outer parts of the disk orbit the center of the Milky Way at the same speed.
(D) The Milky Way has more black holes than expected.
Correct answer: (withheld until after whole-class discussion)
Student responses
Section 4160
(A) : 5 students
(B) : 19 students
(C) : 4 students
(D) : 3 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q11.3
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the start of their learning cycle.
[0.3 points.] How do we know that there is dark matter in the halo of the Milky Way (above/below its disk)?
(A) The temperature of empty space is not really absolute zero.
(B) Stars just outside of the Milky Way appear dimmer than they should be.
(C) Stars in the inner and outer parts of the disk orbit the center of the Milky Way at the same speed.
(D) The Milky Way has more black holes than expected.
Correct answer: (withheld until after whole-class discussion)
Student responses
Section 4160
(A) : 5 students
(B) : 19 students
(C) : 4 students
(D) : 3 students
Whole-class discussionCorrect answer: (C)
Instructor: "So why did you guys choose (B)?"
Student 1: "Because dark matter is dark?"
Instructor: "Fine. And what's the deal with (C)--does this look like Kepler's third law?"
Student 2: (Explains why it is not.)
Instructor: "Yes. That would certainly be strange, to see inner and outer orbits having the same speed. But it is exactly what is seen for the stars that orbit around the center of the Milky Way."
Class: "What?!?"
Instructor: "...And that is the evidence for dark matter."
Class: "What?!?"
Well, that's the way it might have been said.
20080514
FCI post-test comparison: Cuesta College versus UC-Davis
Students at both Cuesta College (San Luis Obispo, CA) and the University of California at Davis were administered the Force Concept Inventory (Doug Hestenes, et al.) during the last week of instruction, in order to follow up on the pre-test results from the first week of instruction (which showed no statistical difference between pre-test scores).
The pre- to post-test gain for this semester at Cuesta College is:
Notable about this Physics 5A class at Cuesta College is the implementation of electronic response system "clickers" (Classroom Performance System, einstruction.com), compared to the traditional lecture of Physics 8A at Cuesta College, and the reformed peer-instruction centered approach at UC-Davis. More analysis on the impact of using clickers on this introductory physics class will be forthcoming.
Cuesta College UC-DavisA "Student" t-test of the null hypothesis results in p = 0.34, thus there is no significant difference between Cuesta College and UC-Davis FCI post-test scores.
Physics 5A Physics 7B
Spring Semester Summer Session II
2008 2002
N 30 students 76 students
low 6 3
mean 14.0 +/- 5.0 12.9 +/- 5.5
high 27 26
The pre- to post-test gain for this semester at Cuesta College is:
Physics 5A Spring Semester 2008 sections 4987, 4988This is comparable to the gains for algebra-based introductory physics at Cuesta College (0.23), UC-Davis (0.16), and for calculus-based introductory physics at Cuesta College (0.14-0.16), as discussed in previous postings on this blog.
<initial%> = 33% +/- 16% (N = 46)
<final%> = 47% +/- 17% (N = 30)
<g> = 0.21 +/- 0.15 (matched-pairs); 0.20 (class-wise)
Notable about this Physics 5A class at Cuesta College is the implementation of electronic response system "clickers" (Classroom Performance System, einstruction.com), compared to the traditional lecture of Physics 8A at Cuesta College, and the reformed peer-instruction centered approach at UC-Davis. More analysis on the impact of using clickers on this introductory physics class will be forthcoming.
Astronomy quiz question: type Ia supernovae
Astronomy 10 Quiz 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.3
[3.0 points.] Which one of the following choices best explains why an isolated white dwarf cannot explode as a type Ia supernova?
(A) There is no companion star to heat up.
(B) There is no external source of hydrogen.
(C) There is no companion star to distort spacetime.
(D) It expended all of its extra energy during the planetary nebula phase.
(E) There is not enough degeneracy pressure.
Correct answer: (B)
If there is no companion star to transfer hydrogen to a white dwarf, then the white dwarf star cannot collect and compact this material onto its surface, making it undergo fusion to undergo a type Ia supernova (or a mere nova) explosion.
Student responses
Section 5166
(A) : 8 students
(B) : 17 students
(C) : 4 students
(D) : 6 students
(E) : 7 students
Previous post:
Astronomy in-class activity: compact objects with companion stars.
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.3
[3.0 points.] Which one of the following choices best explains why an isolated white dwarf cannot explode as a type Ia supernova?
(A) There is no companion star to heat up.
(B) There is no external source of hydrogen.
(C) There is no companion star to distort spacetime.
(D) It expended all of its extra energy during the planetary nebula phase.
(E) There is not enough degeneracy pressure.
Correct answer: (B)
If there is no companion star to transfer hydrogen to a white dwarf, then the white dwarf star cannot collect and compact this material onto its surface, making it undergo fusion to undergo a type Ia supernova (or a mere nova) explosion.
Student responses
Section 5166
(A) : 8 students
(B) : 17 students
(C) : 4 students
(D) : 6 students
(E) : 7 students
Previous post:
Astronomy in-class activity: compact objects with companion stars.
20080513
Astronomy quiz question: x-ray bursts
Astronomy 10 Quiz 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.3
[3.0 points.] Which one of the following choices best explains why an isolated neutron star cannot have repeated x-ray bursts?
(A) There is no companion star to heat up.
(B) There is no companion star to distort spacetime.
(C) It expended all of its extra energy during its type II supernova.
(D) There is no external source of hydrogen.
(E) There is not enough degeneracy pressure.
Correct answer: (D)
If there is no companion star to transfer hydrogen to a neutron star, then the neutron star cannot collect and compact this material onto its surface, making it undergo fusion to produce an x-ray burst.
Student responses
Section 4160
(A) : 7 students
(B) : 1 student
(C) : 10 students
(D) : 16 students
(E) : 0 students
Previous post:
Astronomy in-class activity: compact objects with companion stars.
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.3
[3.0 points.] Which one of the following choices best explains why an isolated neutron star cannot have repeated x-ray bursts?
(A) There is no companion star to heat up.
(B) There is no companion star to distort spacetime.
(C) It expended all of its extra energy during its type II supernova.
(D) There is no external source of hydrogen.
(E) There is not enough degeneracy pressure.
Correct answer: (D)
If there is no companion star to transfer hydrogen to a neutron star, then the neutron star cannot collect and compact this material onto its surface, making it undergo fusion to produce an x-ray burst.
Student responses
Section 4160
(A) : 7 students
(B) : 1 student
(C) : 10 students
(D) : 16 students
(E) : 0 students
Previous post:
Astronomy in-class activity: compact objects with companion stars.
20080512
Astronomy quiz question: accretion disk blackbody radiation
Astronomy 10 Quiz 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.3
[3.0 points.] Which one of the following choices best explains how accretion disks emit intense ultraviolet, and/or x-ray blackbody radiation?
(A) Heat generated from friction.
(B) Electrons jump down to lower orbits.
(C) Electrons flip their spins.
(D) Curvature of spacetime.
(E) Hydrogen fuses into helium.
Correct answer: (A)
A companion star of a compact object (white dwarf, neutron star, or black hole) may be close enough such that it will overflow its Roche lobe when it becomes a giant or superigant, thus transferring hydrogen to the compact object. This infalling hydrogen collects into an accretion disk surrounding the compact object, and due to Kepler's third law, the outer part of the disk orbits slower than the inner part of the disk. Due to the density of the disk, the difference in speeds between adjacent parts causes friction, producing enough heat for the disk to become incandescent, thus emitting blackbody radiation.
Student responses
Section 5166
(A) : 9 students
(B) : 17 students
(C) : 2 students
(D) : 3 students
(E) : 5 students
Previous post:
Astronomy in-class activity: compact objects with companion stars.
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.3
[3.0 points.] Which one of the following choices best explains how accretion disks emit intense ultraviolet, and/or x-ray blackbody radiation?
(A) Heat generated from friction.
(B) Electrons jump down to lower orbits.
(C) Electrons flip their spins.
(D) Curvature of spacetime.
(E) Hydrogen fuses into helium.
Correct answer: (A)
A companion star of a compact object (white dwarf, neutron star, or black hole) may be close enough such that it will overflow its Roche lobe when it becomes a giant or superigant, thus transferring hydrogen to the compact object. This infalling hydrogen collects into an accretion disk surrounding the compact object, and due to Kepler's third law, the outer part of the disk orbits slower than the inner part of the disk. Due to the density of the disk, the difference in speeds between adjacent parts causes friction, producing enough heat for the disk to become incandescent, thus emitting blackbody radiation.
Student responses
Section 5166
(A) : 9 students
(B) : 17 students
(C) : 2 students
(D) : 3 students
(E) : 5 students
Previous post:
Astronomy in-class activity: compact objects with companion stars.
20080511
Astronomy quiz question: black holes with companion stars
Astronomy 10 Quiz 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.3
[3.0 points.] Which one of the following statements best explains why x-rays are observed from close binary systems comprised of black holes with companion stars?
(A) Only x-ray photons are fast enough to escape from black holes.
(B) Electrons falling into lower energy orbitals release x-ray photons.
(C) Black holes can only emit x-ray photons.
(D) Accretion disks surrounding black holes emit very hot blackbody radiation.
(E) Companion stars orbiting black holes emit very hot blackbody radiation.
Correct answer: (D)
A companion star of a black hole may be close enough such that it will overflow its Roche lobe when it becomes a giant or superigant, thus transferring hydrogen to the black hole. This infalling hydrogen collects into an accretion disk surrounding the black hole, heating up in the process to become incandescent, emitting blackbody radiation.
Student responses
Section 4160
(A) : 3 students
(B) : 5 students
(C) : 6 students
(D) : 16 students
(E) : 4 students
Previous post:
Astronomy in-class activity: compact objects with companion stars.
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.3
[3.0 points.] Which one of the following statements best explains why x-rays are observed from close binary systems comprised of black holes with companion stars?
(A) Only x-ray photons are fast enough to escape from black holes.
(B) Electrons falling into lower energy orbitals release x-ray photons.
(C) Black holes can only emit x-ray photons.
(D) Accretion disks surrounding black holes emit very hot blackbody radiation.
(E) Companion stars orbiting black holes emit very hot blackbody radiation.
Correct answer: (D)
A companion star of a black hole may be close enough such that it will overflow its Roche lobe when it becomes a giant or superigant, thus transferring hydrogen to the black hole. This infalling hydrogen collects into an accretion disk surrounding the black hole, heating up in the process to become incandescent, emitting blackbody radiation.
Student responses
Section 4160
(A) : 3 students
(B) : 5 students
(C) : 6 students
(D) : 16 students
(E) : 4 students
Previous post:
Astronomy in-class activity: compact objects with companion stars.
20080510
Astronomy quiz question: mass transfer in close-pair binaries
Astronomy 10 Quiz 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
[3.0 points.] Which one of the following choices best explains why the Roche lobes of two stars in a close-pair (mass-exchanging) binary system get smaller as they begin to orbit closer to each other?
(A) Gravitational forces get weaker.
(B) Centrifugal forces get stronger.
(C) Degeneracy pressure increases.
(D) Hydrogen is transferred slowly.
(E) Repeated nova explosions disrupts hydrogen transfer.
Correct answer: (B)
Due to conservation of angular momentum, the orbital speeds of the stars will increase as their separation distance decreases (as they would during hydrogen transfer from a more massive star to a less massive star). This will increase the centrifugal forces exerted on them, decreasing the size of their Roche lobes.
Student responses
Section 5166
(A) : 5 students
(B) : 29 students
(C) : 4 students
(D) : 7 students
(E) : 2 students
Related post:
Astronomy in-class activity: mass transfer in close-pair binaries.
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
[3.0 points.] Which one of the following choices best explains why the Roche lobes of two stars in a close-pair (mass-exchanging) binary system get smaller as they begin to orbit closer to each other?
(A) Gravitational forces get weaker.
(B) Centrifugal forces get stronger.
(C) Degeneracy pressure increases.
(D) Hydrogen is transferred slowly.
(E) Repeated nova explosions disrupts hydrogen transfer.
Correct answer: (B)
Due to conservation of angular momentum, the orbital speeds of the stars will increase as their separation distance decreases (as they would during hydrogen transfer from a more massive star to a less massive star). This will increase the centrifugal forces exerted on them, decreasing the size of their Roche lobes.
Student responses
Section 5166
(A) : 5 students
(B) : 29 students
(C) : 4 students
(D) : 7 students
(E) : 2 students
Related post:
Astronomy in-class activity: mass transfer in close-pair binaries.
20080509
Astronomy quiz question: mass transfer in close-pair binaries
Astronomy 10 Quiz 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
[3.0 points.] Which one of the following statements best explains why the transfer of hydrogen in a close-pair binary system from a less massive star to a more massive star happens very slowly?
(A) There is not much hydrogen left in the less massive star.
(B) They move farther apart from each other.
(C) The less massive star has not yet ended its main sequence lifetime.
(D) The more massive star produces strong winds.
(E) (None of above choices (A)-(D), as it is not possible for hydrogen to be transferred from a less massive star to a more massive star.)
Correct answer: (B)
When a less massive star is transferring hydrogen to a more massive star, their masses become more unequal, such that their separation distance increases. This decreases their orbital speeds, decreasing centrifugal forces, resulting in enlarging their Roche lobes, making it more difficult for the less-massive star to overflow its Roche lobe. This will slow down, and eventually stop the transfer of hydrogen from the less massive to the more massive star.
Student responses
Section 4160
(A) : 11 students
(B) : 11 students
(C) : 6 students
(D) : 1 student
(E) : 5 students
Related post:
Astronomy in-class activity: mass transfer in close-pair binaries.
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
[3.0 points.] Which one of the following statements best explains why the transfer of hydrogen in a close-pair binary system from a less massive star to a more massive star happens very slowly?
(A) There is not much hydrogen left in the less massive star.
(B) They move farther apart from each other.
(C) The less massive star has not yet ended its main sequence lifetime.
(D) The more massive star produces strong winds.
(E) (None of above choices (A)-(D), as it is not possible for hydrogen to be transferred from a less massive star to a more massive star.)
Correct answer: (B)
When a less massive star is transferring hydrogen to a more massive star, their masses become more unequal, such that their separation distance increases. This decreases their orbital speeds, decreasing centrifugal forces, resulting in enlarging their Roche lobes, making it more difficult for the less-massive star to overflow its Roche lobe. This will slow down, and eventually stop the transfer of hydrogen from the less massive to the more massive star.
Student responses
Section 4160
(A) : 11 students
(B) : 11 students
(C) : 6 students
(D) : 1 student
(E) : 5 students
Related post:
Astronomy in-class activity: mass transfer in close-pair binaries.
20080508
Education research: "clicker zombies"
*POOF* Mah sleeping spell wrkd!, by Firestar779
icanhascheezburger.com
May 8, 2008
With 16 clicker questions in a weekly, three hour evening lecture in introductory astronomy, students stay awake only enough to click in. With so many clicker questions interspersed throughout lecture (approximately 10 minutes per clicker question) with not enough student-student interaction, results in the above dramatization--many students with heads down, too exhausted to take notes, only rousing themselves to make the obligatory effort to click in (for participation credit).
Astronomy quiz question: mass, densities, and event horizons
Astronomy 10 Quiz 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.1
[3.0 points.] Which one of the following choices best explains why a black hole is the only object that can have an event horizon?
(A) It has zero density.
(B) It has infinite mass.
(C) It has zero size.
(D) It has infinite energy.
(E) It can distort spacetime.
Correct answer: (C)
Section 4160
(A) : 5 students
(B) : 7 students
(C) : 15 students
(D) : 1 students
(E) : 2 students
Section 5166
(A) : 6 students
(B) : 16 students
(C) : 11 students
(D) : 3 students
(E) : 5 students
Previous posts:
Astronomy clicker question: masses, densities, and event horizons.
Astronomy in-class activity: masses, densities, and escape velocities.
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.1
[3.0 points.] Which one of the following choices best explains why a black hole is the only object that can have an event horizon?
(A) It has zero density.
(B) It has infinite mass.
(C) It has zero size.
(D) It has infinite energy.
(E) It can distort spacetime.
Correct answer: (C)
Section 4160
(A) : 5 students
(B) : 7 students
(C) : 15 students
(D) : 1 students
(E) : 2 students
Section 5166
(A) : 6 students
(B) : 16 students
(C) : 11 students
(D) : 3 students
(E) : 5 students
Previous posts:
Astronomy clicker question: masses, densities, and event horizons.
Astronomy in-class activity: masses, densities, and escape velocities.
20080507
Astronomy in-class activity: compact objects with companion stars
Astronomy 210 In-class activity 23 v.07.04.28, spring semester 2008
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 compare and contrast the different features of compact objects with companion stars. Students are instructed to look for connections and similarities, then to concentrate on specific differences.
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 compare and contrast the different features of compact objects with companion stars. Students are instructed to look for connections and similarities, then to concentrate on specific differences.
20080506
Astronomy clicker question: close-pair binary star system observations
Astronomy 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle (specifically, following the astronomy in-class activity: mass transfer in close-pair binaries).
[0.3 points.] If you were to survey every visible close pair (mass-exchanging) binary star system, which one of the choices (A)-(D) would best describe your observations?
(B) You would see (I) just as often as (II).
(C) You would see (II) more often than (I).
(D) You would only be able to see (I), as it is impossible for (II) to occur.
Correct answer: (C)
Student responses
Section 4160
(A) : 7 students
(B) : 13 students
(C) : 10 students
(D) : 0 students
Section 5166
(A) : 4 students
(B) : 10 students
(C) : 8 students
(D) : 10 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle (specifically, following the astronomy in-class activity: mass transfer in close-pair binaries).
[0.3 points.] If you were to survey every visible close pair (mass-exchanging) binary star system, which one of the choices (A)-(D) would best describe your observations?
I. Mass being transferred from a more-massive star to a less-massive star.(A) You would see (I) more often than (II).
II. Mass being transferred from a less-massive star to a more-massive star.
(B) You would see (I) just as often as (II).
(C) You would see (II) more often than (I).
(D) You would only be able to see (I), as it is impossible for (II) to occur.
Correct answer: (C)
Student responses
Section 4160
(A) : 7 students
(B) : 13 students
(C) : 10 students
(D) : 0 students
Section 5166
(A) : 4 students
(B) : 10 students
(C) : 8 students
(D) : 10 students
Physics midterm problem: disk rolling uphill
Physics 5A Midterm 2, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 8.62
[20 points.] A solid disk of mass 1.90 kg and radius 0.0500 m rolls without slipping along a horizontal surface with a translational speed of 0.240 m/s. (The rotational inertia of a disk is given by (1/2)*M*R^2.) It comes to an incline that makes an angle of 35.0 degrees with the horizontal surface. Neglecting energy losses due to friction, to what height above the horizontal surface does the disk rise on the incline? Show your work and explain your reasoning.
Solution and grading rubric:
Grading distribution:
p: 4 students
r: 9 students
t: 12 students
v: 9 students
x: 1 student
y: 1 student
z: 0 students
A sample of a "p" response (from student 1125) is shown below:
A sample of a "t" response (from student 1239) that calculates both K_rot and K_tr, but only applies K_tr to U_grav in energy conservation:
Another "t" response sample (from student 7137) that applies only K_rot to U_grav in energy conservation:
One more "t" response (from student 1337) that also applies only K_rot to U_grav in energy conservation, with additional editorial comments:
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 8.62
[20 points.] A solid disk of mass 1.90 kg and radius 0.0500 m rolls without slipping along a horizontal surface with a translational speed of 0.240 m/s. (The rotational inertia of a disk is given by (1/2)*M*R^2.) It comes to an incline that makes an angle of 35.0 degrees with the horizontal surface. Neglecting energy losses due to friction, to what height above the horizontal surface does the disk rise on the incline? Show your work and explain your reasoning. Solution and grading rubric:
- p = 20/20:
Correct. Recognizes that W_nc = 0, such that when the disk has reached its highest height, all of its K_tr and K_rot has gone into U_grav. Writes out an energy balance equation 0 = delta(K_rot) + delta(K_tr) + delta(U_grav) and solves for y_f = 0.00441 m; or solves for K_tr,i and K_rot,i separately, then sets K_tot,i = K_tr,i + K_rot,i = U_grav,f to solve for y_f. Note that since the disk rolls without slipping, w_i = v_i/R. - r = 16/20:
Nearly correct, but includes minor math errors. Applies rolling without slipping condition, and recognizes both K_rot and K_tr must be accounted for separately in energy conservation. - t = 12/20:
Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Attempts to apply energy conservation, but one missing energy term out of K_rot, K_tr, or U_grav. - v = 8/20:
Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at finding I = (1/2)*m*R^2, energy terms, Newton's laws, or applying angular momentum conservation. - 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.
Grading distribution:
p: 4 students
r: 9 students
t: 12 students
v: 9 students
x: 1 student
y: 1 student
z: 0 students
A sample of a "p" response (from student 1125) is shown below:
A sample of a "t" response (from student 1239) that calculates both K_rot and K_tr, but only applies K_tr to U_grav in energy conservation:
Another "t" response sample (from student 7137) that applies only K_rot to U_grav in energy conservation:
One more "t" response (from student 1337) that also applies only K_rot to U_grav in energy conservation, with additional editorial comments:
Physics midterm problem: boom-suspended package
Physics 5A Midterm 2, spring semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 8.36
[20 points.] A box is suspended by a uniform boom, of length 2.20 m and weight 80.0 N, as shown at right. A cable is attached perpendicular to the boom at its midpoint, and has a breaking strength of 200 N. What is the maximum weight of the box that can be suspended by this system before the cable breaks? Show your work and explain your reasoning.
Solution and grading rubric:
Grading distribution:
p: 5 students
r: 15 students
t: 8 students
v: 6 students
x: 2 students
y: 0 students
z: 0 students
A sample of a "p" response (from student 1468) is shown below:
Another "p" response (from student 2102):
An "r" response (from student 5913), recognizing the clockwise and counterclockwise torques that must be set equal to each other (but confusing the force of the 200 N tension cable with its torque):
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 8.36
[20 points.] A box is suspended by a uniform boom, of length 2.20 m and weight 80.0 N, as shown at right. A cable is attached perpendicular to the boom at its midpoint, and has a breaking strength of 200 N. What is the maximum weight of the box that can be suspended by this system before the cable breaks? Show your work and explain your reasoning.Solution and grading rubric:
- p = 20/20:
Correct. Sums all torques to zero (or sets counterclockwise torque of cable equal to the clockwise torques of the boom's weight, and the box) with lever arms measured from the pivot point to find that the maximum weight of the box is 116 N. - r = 16/20:
Nearly correct, but includes minor math errors. Applies rotational equilibrium condition to the boom, and at least recognizes and attempts to systematically calculate these three torque terms. - t = 12/20:
Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Attempts to apply rotational equilibrium, but one missing torque out of those exerted by the cable, weight of the boom, and weight of the box. - v = 8/20:
Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at applying equilibrium condition to forces or components of forces. - 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.
Grading distribution:
p: 5 students
r: 15 students
t: 8 students
v: 6 students
x: 2 students
y: 0 students
z: 0 students
A sample of a "p" response (from student 1468) is shown below:
Another "p" response (from student 2102):
An "r" response (from student 5913), recognizing the clockwise and counterclockwise torques that must be set equal to each other (but confusing the force of the 200 N tension cable with its torque):
Physics midterm question: wind-blown Doppler shift
Physics 5A Midterm 2, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 12.14
[10 points.] The source and observer of a sound wave are both at rest with respect to the ground. The wind blows in the direction from the source to observer. Discuss whether the observed frequency is higher, the same as, or lower than the source frequency. Explain your reasoning using the properties of the Doppler effect.
Solution and grading rubric:
Grading distribution:
p: 11 students
r: 3 students
t: 21 students
v: 1 student
x: 0 students
y: 0 students
z: 0 students
A somewhat fanciful, overwrought sample of a "p" response (from student 1337) is shown below:
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 12.14
[10 points.] The source and observer of a sound wave are both at rest with respect to the ground. The wind blows in the direction from the source to observer. Discuss whether the observed frequency is higher, the same as, or lower than the source frequency. Explain your reasoning using the properties of the Doppler effect.
Solution and grading rubric:
- p = 10/10:
Correct. Either of two answer suffices: (a) Doppler shifts only occur if the source and/or frequency move relative to each other, thus it can be shown from the frequency shift equation that the numerator and denominator are both equal to one, and thus no shift occurs; or (b) the wind increases the wavelength of the waves traveling from the source to the observer, but these spaced-out wavefronts reach the observer quicker, resulting in no net change in frequency. Explanation includes the frequency shift equation (with v_obs = 0, v_source = 0, but with v_sound increased) and/or illustration of wavefronts. - r = 8/10:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. - t = 6/10:
Nearly correct, but argument has conceptual errors, or is incomplete. Typically motivates the observed frequency as being higher or lower than the source frequency. - v = 4/10:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. - x = 2/10:
Implementation/application of ideas, but credit given for effort rather than merit. - y = 1/10:
Irrelevant discussion/effectively blank. - z = 0/10:
Blank.
Grading distribution:
p: 11 students
r: 3 students
t: 21 students
v: 1 student
x: 0 students
y: 0 students
z: 0 students
A somewhat fanciful, overwrought sample of a "p" response (from student 1337) is shown below:
Physics midterm question: roller coaster hills
Physics 5A (currently Physics 205A) Midterm 2, spring semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 6.6
In the design of a roller coaster, is it possible for any hill of the ride to be higher than the first one? If so, discuss how. If not, then discuss why this would be impossible. Explain your reasoning using the properties of energy conservation.
Solution and grading rubric:
Grading distribution:
p: 27 students
r: 0 students
t: 7 students
v: 2 students
x: 0 students
y: 0 students
z: 0 students
A sample of a "p" response (from student 2607) is shown below:
Another "p" response (from student 5711):
A sample of a "v" response (from student 7937), where it is argued that each subsequent hill can be higher than the first:
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 6.6
In the design of a roller coaster, is it possible for any hill of the ride to be higher than the first one? If so, discuss how. If not, then discuss why this would be impossible. Explain your reasoning using the properties of energy conservation.
Solution and grading rubric:
- p:
Correct. Either of two answers suffices using properties of energy conservation: (a) if the roller coaster car starts at rest from the first hill, and no additional energy is put into the system after it is released, then the second hill can at most be as high as the first (under ideal conditions only), or lower; or (b) if the car has been given sufficient kinetic energy at the top of the first hill, then it may make it over a second hill that is higher. - 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. Argument somehow based on momentum conservation; but at least recognizes that the second hill must be at most as high as the first hill unless something is done. - v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Typically states that the second hill can be higher, but does not clearly explain how this would be possible. - x:
Implementation/application of ideas, but credit given for effort rather than merit. - y:
Irrelevant discussion/effectively blank. - z:
Blank.
Grading distribution:
p: 27 students
r: 0 students
t: 7 students
v: 2 students
x: 0 students
y: 0 students
z: 0 students
A sample of a "p" response (from student 2607) is shown below:
Another "p" response (from student 5711):
A sample of a "v" response (from student 7937), where it is argued that each subsequent hill can be higher than the first:
Physics midterm question: sound wave intensity
Physics 5A Midterm 2, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 11.3
[3.0 points.] The intensity of a sound wave from a jet airplane as it is taking off is 12 Watts/m^2 at a distance of 10 m. What happens to the intensity of the sound waves as they travel out to a distance of 20 m from the jet airplane?
(A) It decreases by a factor of 1.4.
(B) It decreases by a factor of 2.0.
(C) It decreases by a factor of 4.0.
(D) (The intensity of the sound waves remains the same.)
Correct answer: (C)
The intensity of a spherical sound wave, with no reflections or absorption, is inversely proportional to the square of the radius, as measured from the (isotropic) source. Thus doubling the distance from the source results in reducing the intensity by a factor of four.
Student responses
Sections 4987, 4988
(A) : 7 students
(B) : 10 students
(C) : 15 students
(D) : 3 students
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 11.3
[3.0 points.] The intensity of a sound wave from a jet airplane as it is taking off is 12 Watts/m^2 at a distance of 10 m. What happens to the intensity of the sound waves as they travel out to a distance of 20 m from the jet airplane?
(A) It decreases by a factor of 1.4.
(B) It decreases by a factor of 2.0.
(C) It decreases by a factor of 4.0.
(D) (The intensity of the sound waves remains the same.)
Correct answer: (C)
The intensity of a spherical sound wave, with no reflections or absorption, is inversely proportional to the square of the radius, as measured from the (isotropic) source. Thus doubling the distance from the source results in reducing the intensity by a factor of four.
Student responses
Sections 4987, 4988
(A) : 7 students
(B) : 10 students
(C) : 15 students
(D) : 3 students
20080505
Astronomy in-class activity: mass transfer in close-pair binaries
Astronomy 10 In-class activity 23 v.07.04.30, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
Students find their assigned groups of three to four students, and work cooperatively on an in-class activity worksheet to summarize the three stages of initial mass transfer in a close-pair (mass transferring) binary star system.
Star B is more massive than Star A, due to its being closer to the center of mass, and also because of its larger Roche lobes. As a result, Star B will end its main sequence lifetime earlier than Star A, and become a giant or supergiant.
As a result of Star B expanding and filling its Roche lobe up to the "neck" or "pinch point," hydrogen will then be transferred to Star A. This makes their masses become more equal, making their separation distance decrease, while increasing their orbital speeds. As a result, centrifugal forces increase, such that the size of Star B's lobe shrinks (while it is expanding in its giant/supergiant phase), making the "spillage" from Star B to Star A rapid.
Eventually the masses of Star B and Star A equalize; this is when their separation distance is the smallest, and their lobes are equal in size, but the physical size of Star B (still) fills its (smaller) Roche lobe.
As Star B still transfers hydrogen to Star A, this makes their masses become unequal, making their separation distance increase, while decreasing their orbital speeds. As a result, centrifugal forces decrease, such that the size of Star B's lobe expands. This makes it harder for Star B to "spill over" hydrogen out of its Roche lobe, and the transfer of material slows, and eventually stops when Star B is unable to exceed its much larger Roche lobe.
(Transfer from Star A, when it eventually ends its main sequence lifetime, to Star B will eventually occur, but this discussion is outside the scope of this course.)
Thus with all observations of close-pair binary star systems where mass transfer is taking place, it is more likely to see a less-massive star feeding a more-massive star than vice versa.
Follow-up post:
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.2
Students find their assigned groups of three to four students, and work cooperatively on an in-class activity worksheet to summarize the three stages of initial mass transfer in a close-pair (mass transferring) binary star system.
Star B is more massive than Star A, due to its being closer to the center of mass, and also because of its larger Roche lobes. As a result, Star B will end its main sequence lifetime earlier than Star A, and become a giant or supergiant.As a result of Star B expanding and filling its Roche lobe up to the "neck" or "pinch point," hydrogen will then be transferred to Star A. This makes their masses become more equal, making their separation distance decrease, while increasing their orbital speeds. As a result, centrifugal forces increase, such that the size of Star B's lobe shrinks (while it is expanding in its giant/supergiant phase), making the "spillage" from Star B to Star A rapid.
Eventually the masses of Star B and Star A equalize; this is when their separation distance is the smallest, and their lobes are equal in size, but the physical size of Star B (still) fills its (smaller) Roche lobe. As Star B still transfers hydrogen to Star A, this makes their masses become unequal, making their separation distance increase, while decreasing their orbital speeds. As a result, centrifugal forces decrease, such that the size of Star B's lobe expands. This makes it harder for Star B to "spill over" hydrogen out of its Roche lobe, and the transfer of material slows, and eventually stops when Star B is unable to exceed its much larger Roche lobe.
(Transfer from Star A, when it eventually ends its main sequence lifetime, to Star B will eventually occur, but this discussion is outside the scope of this course.)
Thus with all observations of close-pair binary star systems where mass transfer is taking place, it is more likely to see a less-massive star feeding a more-massive star than vice versa.
Follow-up post:
20080504
Astronomy clicker question: masses, densities, and event horizons
Astronomy 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.1
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle (specifically, following the astronomy in-class activity: masses, densities, and escape velocities).
[0.3 points.] Why is a black hole the only object that has an event horizon?
(A) Because of its mass.
(B) Because of its density.
(C) Because of both its mass and density.
(D) (Neither its mass nor density matters.)
Correct answer: (B)
Student responses
Section 5166
(A) : 1 student
(B) : 21 students
(C) : 14 students
(D) : 3 students
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.1
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle (specifically, following the astronomy in-class activity: masses, densities, and escape velocities).
[0.3 points.] Why is a black hole the only object that has an event horizon?
(A) Because of its mass.
(B) Because of its density.
(C) Because of both its mass and density.
(D) (Neither its mass nor density matters.)
Correct answer: (B)
Student responses
Section 5166
(A) : 1 student
(B) : 21 students
(C) : 14 students
(D) : 3 students
20080503
Astronomy in-class activity: masses, densities, and escape velocities
Astronomy 10 In-class activity 23 v.07.04.25, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.1
Students find their assigned groups of three to four students, and work cooperatively on an in-class activity worksheet to determine the relationship between the escape velocity with mass and/or density.
Start off with filling some of the entries for the students, before they start working in their groups. The main sequence star on the left is the smallest and least massive of the main sequence stars: "G2," "1.0 M_Sun." Point out the progression in increasing size and mass of the main sequence stars, from right-to-left. Also point out the progression in decreasing size from left-to-right for the compact objects, and also the increasing escape velocities. Undoubtably the compact object with zero size and an infinite escape velocity is the black hole.
There will be several ties for the list of increasing masses:
G2 = white dwarf; A3 = neutron star; B3 = black hole.
However, there is no trend in escape velocities for this list, such that there is no relationship between the mass and escape velocity.
There are also several (approximate) ties for the list of increasing densities:
G2 = A3 = B3; white dwarf, neutron star; black hole.
Since the escape velocities also increase from left-to-right along this list, there is a direct relationship between the density of an object and its escape velocity.
Follow-up post: Astronomy clicker question: masses, densities, and event horizons.
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.1
Students find their assigned groups of three to four students, and work cooperatively on an in-class activity worksheet to determine the relationship between the escape velocity with mass and/or density.
Start off with filling some of the entries for the students, before they start working in their groups. The main sequence star on the left is the smallest and least massive of the main sequence stars: "G2," "1.0 M_Sun." Point out the progression in increasing size and mass of the main sequence stars, from right-to-left. Also point out the progression in decreasing size from left-to-right for the compact objects, and also the increasing escape velocities. Undoubtably the compact object with zero size and an infinite escape velocity is the black hole.
There will be several ties for the list of increasing masses:
G2 = white dwarf; A3 = neutron star; B3 = black hole.
However, there is no trend in escape velocities for this list, such that there is no relationship between the mass and escape velocity.
There are also several (approximate) ties for the list of increasing densities:
G2 = A3 = B3; white dwarf, neutron star; black hole.
Since the escape velocities also increase from left-to-right along this list, there is a direct relationship between the density of an object and its escape velocity.
Follow-up post: Astronomy clicker question: masses, densities, and event horizons.
Physics clicker question: Doppler effect
Physics 5A, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Question 12.8
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle:
[0.6 participation points.] A source emits a 440.0 Hz sound. In which situation would an observer hear the highest frequency? (Take the speed of sound to be 340.0 m/s.)
(A) Source is stationary, observer moves at 5.0 m/s towards source.
(B) Source moves at 5.0 m/s towards observer, observer is stationary.
(C) Source and observer both move at 5.0 m/s towards each other.
(D) (More than one of the above choices.)
(E) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 3 students
(B) : 5 students
(C) : 17 students
(D) : 4 students
(E) : 1 student
Correct answer: (C)
The frequency detected by the observer is given by:
f_observer = ((1 - (v_observer/v_sound))/(1 - (v_source/v_sound))*f_source.
In each of cases (A)-(C), the numerator in the parenthesis above is less than the numerator, so f_observer is always higher than f_source (i.e., in each case the motions of the source and observer relative to each other are towards each other). However, the Doppler shift is the greatest when both the source and observer are moving towards each other with respect to the lab frame. Thus for (C), f_observer = (345.0/335.0)*f_source = 453.1 Hz, although most students had determined this without having done any calculations.
[Follow-up question]
[0.6 participation points.] In which situation would the observer hear the second highest frequency? (Use the same (A)-(E) choices from question (10).)
Sections 4987, 4988
(A) : 4 students
(B) : 6 students
(C) : 1 student
(D) : 18 students
(E) : 0 students
Correct answer: (B)
Inexplicably, one student chose (C) as the second highest frequency heard by the observer.
The modal response (D) means that most students said that (A) and (B) would both give the same higher f_observer. However, for (A) v_observer = -5 m/s, v_source = 0, then f_observer = (345.0/340.0)*f_source = 446.5 Hz; and for (B) v_observer = 0, v_source = +5 m/s, then f_observer = (340.0/335.0)*f_source = 446.6 Hz.
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Question 12.8
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle:
[0.6 participation points.] A source emits a 440.0 Hz sound. In which situation would an observer hear the highest frequency? (Take the speed of sound to be 340.0 m/s.)
(A) Source is stationary, observer moves at 5.0 m/s towards source.
(B) Source moves at 5.0 m/s towards observer, observer is stationary.
(C) Source and observer both move at 5.0 m/s towards each other.
(D) (More than one of the above choices.)
(E) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 3 students
(B) : 5 students
(C) : 17 students
(D) : 4 students
(E) : 1 student
Correct answer: (C)
The frequency detected by the observer is given by:
f_observer = ((1 - (v_observer/v_sound))/(1 - (v_source/v_sound))*f_source.
In each of cases (A)-(C), the numerator in the parenthesis above is less than the numerator, so f_observer is always higher than f_source (i.e., in each case the motions of the source and observer relative to each other are towards each other). However, the Doppler shift is the greatest when both the source and observer are moving towards each other with respect to the lab frame. Thus for (C), f_observer = (345.0/335.0)*f_source = 453.1 Hz, although most students had determined this without having done any calculations.
[Follow-up question]
[0.6 participation points.] In which situation would the observer hear the second highest frequency? (Use the same (A)-(E) choices from question (10).)
Sections 4987, 4988
(A) : 4 students
(B) : 6 students
(C) : 1 student
(D) : 18 students
(E) : 0 students
Correct answer: (B)
Inexplicably, one student chose (C) as the second highest frequency heard by the observer.
The modal response (D) means that most students said that (A) and (B) would both give the same higher f_observer. However, for (A) v_observer = -5 m/s, v_source = 0, then f_observer = (345.0/340.0)*f_source = 446.5 Hz; and for (B) v_observer = 0, v_source = +5 m/s, then f_observer = (340.0/335.0)*f_source = 446.6 Hz.
20080502
Physics quiz question: independent versus dependent wave parameters
Physics 5A Quiz 6, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Questions 11.5, 11.6, and 11.8
[3.0 points.] A wave has a frequency and wavelength of 200 Hz and 0.750 m, respectively. If the frequency of the wave source is changed, which wave parameters will also change as a result?
(A) The wavelength.
(B) The speed of the wave.
(C) (Both the wavelength and the speed of the wave will change.)
(D) (Neither the wavelength nor the speed of the wave will change.)
Correct answer: (B)
The speed of a wave depends on the properties of the medium, and not on frequency, which is a source parameter. However, wavelength is the parameter that is dependent on both speed and frequency, and thus changes as a result of changing the frequency of the source.
Student responses
Sections 4987, 4988
(A) : 9 students
(B) : 4 students
(C) : 20 students
(D) : 1 student
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Questions 11.5, 11.6, and 11.8
[3.0 points.] A wave has a frequency and wavelength of 200 Hz and 0.750 m, respectively. If the frequency of the wave source is changed, which wave parameters will also change as a result?
(A) The wavelength.
(B) The speed of the wave.
(C) (Both the wavelength and the speed of the wave will change.)
(D) (Neither the wavelength nor the speed of the wave will change.)
Correct answer: (B)
The speed of a wave depends on the properties of the medium, and not on frequency, which is a source parameter. However, wavelength is the parameter that is dependent on both speed and frequency, and thus changes as a result of changing the frequency of the source.
Student responses
Sections 4987, 4988
(A) : 9 students
(B) : 4 students
(C) : 20 students
(D) : 1 student
Physics quiz question: stress and strain of different cross-section bars
Physics 5A Quiz 6, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 10.4(b)
[3.0 points.] A cylindrical steel bar of radius 2.50 cm is compressed by the application of forces of magnitude 1.50 x 10^3 N at each end. What magnitude forces (at each end) would be required to compress by the same amount a steel bar of the same length, but a radius of 1.25 cm?
(A) 375 N.
(B) 750 N.
(C) 1.06 x 10^3 N.
(D) 1.50 x 10^3 N.
Correct answer: (A)
Hooke's law relating stress (F/A) and strain (delta(L)/L)) is:
(F/A) = Y*(delta(L)/L),
where Y is the Young's modulus for the material. Since the radius is reduced by a factor of two, then the area is reduced by a factor of four, such that the force required to cause the same amount of strain in this smaller cross section bar would be one-fourth the original force. Response (B) is one-half of the original force, while response (C) is sqrt(2) = 0.707 times the original force.
Student responses
Sections 4987, 4988
(A) : 8 students
(B) : 19 students
(C) : 3 students
(D) : 3 students
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Conceptual Question 10.4(b)
[3.0 points.] A cylindrical steel bar of radius 2.50 cm is compressed by the application of forces of magnitude 1.50 x 10^3 N at each end. What magnitude forces (at each end) would be required to compress by the same amount a steel bar of the same length, but a radius of 1.25 cm?
(A) 375 N.
(B) 750 N.
(C) 1.06 x 10^3 N.
(D) 1.50 x 10^3 N.
Correct answer: (A)
Hooke's law relating stress (F/A) and strain (delta(L)/L)) is:
(F/A) = Y*(delta(L)/L),
where Y is the Young's modulus for the material. Since the radius is reduced by a factor of two, then the area is reduced by a factor of four, such that the force required to cause the same amount of strain in this smaller cross section bar would be one-fourth the original force. Response (B) is one-half of the original force, while response (C) is sqrt(2) = 0.707 times the original force.
Student responses
Sections 4987, 4988
(A) : 8 students
(B) : 19 students
(C) : 3 students
(D) : 3 students
Astronomy clicker question: entering the event horizon
Get into vortex you say, by ?rico Lopez
icanhascheezburger.com
January 3, 2008
Astronomy 10, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Astronomy 10 learning goal Q10.1
Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the start of their learning cycle:
[0.3 points.] Why is entering the event horizon (the "Schwarzchild radius") surrounding a black hole considered a "point of no return?"
(A) All matter is crushed into nothingness.
(B) All matter is instantly converted into energy.
(C) The escape velocity is faster than the speed of light.
(D) Time runs backwards.
Correct answer: (C)
If students ask about what's inside the event horizon of a black hole, point out that they could certainly try to enter it and find out for themselves. However, in the very unlikely event that would be able to survive the journey, they would not be able to tell anyone in the outside universe about their discoveries, as nothing, not even light would be able to escape from within the event horizon. So they should bring a friend to share in the experience. Otherwise, it'd be lonely in there...
And yes, the expression on the middle kitten is priceless.
Student responses
Section 4160
(A) : 9 students
(B) : 4 students
(C) : 17 students
(D) : 0 students
Section 5166
(A) : 7 students
(B) : 7 students
(C) : 23 students
(D) : 0 students
20080501
Physics clicker question: sound wave frequency
Physics 5A, Spring Semester 2008
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 11.52(c)
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle:
[0.6 participation points.] Consider a cello string, which vibrates at its fundamental frequency of 65.4 Hz. The velocity of sound waves in air is 340 m/s. The frequency of sound waves created by the cello string is:
(A) lower than 65.4 Hz.
(B) equal to 65.4 Hz.
(C) higher than 65.4 Hz.
(D) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 2 students
(B) : 12 students
(C) : 15 students
(D) : 0 students
Correct answer: (B)
While the wavelengths of the string standing wave and sound wave will be different, their frequencies must be the same. It is instructive to think about the fact that the string is vibrating transversely 65.4 times per second, and it is "slapping" the air (and thus generating a sound wave) with this same frequency.
Cuesta College, San Luis Obispo, CA
Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Problem 11.52(c)
Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle:
[0.6 participation points.] Consider a cello string, which vibrates at its fundamental frequency of 65.4 Hz. The velocity of sound waves in air is 340 m/s. The frequency of sound waves created by the cello string is:
(A) lower than 65.4 Hz.
(B) equal to 65.4 Hz.
(C) higher than 65.4 Hz.
(D) (I'm lost, and don't know how to answer this.)
Sections 4987, 4988
(A) : 2 students
(B) : 12 students
(C) : 15 students
(D) : 0 students
Correct answer: (B)
While the wavelengths of the string standing wave and sound wave will be different, their frequencies must be the same. It is instructive to think about the fact that the string is vibrating transversely 65.4 times per second, and it is "slapping" the air (and thus generating a sound wave) with this same frequency.
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