## 20151130

### Online reading assignment: internal energy conservation

Physics 205A, fall semester 2015
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters and previewing a presentation on internal energy conservation.

Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.
"Internal thermal energy is not concerned with the macroscopic movement or rotation of an object, rather, the microscopic motion of the individual atoms and molecules within an object. Looks like we will be able to understand the physics of how objects heat up internally due to separate and internal forces of energy."

"Energy cannot be created or destroyed, thus the energy lost by a system (-Q) is equal to the energy gained by its surroundings (+Q), and vice-versa. When that energy is being transferred between objects of different temperatures it is called 'heat.' Heat comes from the internal energy of a substance, which is the sum of molecular energies, including kinetic and potential molecular energies. When bonds are broken within a substance its internal energy increases, and when bonds are formed its internal energy decreases. So for a particular substance, the gas form has more energy than the liquid form, which has more energy than the solid form."

"If there is no energy transferred into or out of the thermal internal energy of a system, then it is effectively thermally isolated from the environment, and the heat exchanged between the system and the external environment is zero. Also, if the thermal internal energy of a system increases, its temperature increases, and thus external heat from the environment is positive, being added into the system. In contrast, if the thermal internal energy of a system decreases, its temperature decreases, and thus external heat from the environment is negative, being removed from the system."

"Thermal internal energy is the measurement of movement of a systems individual molecules. This measurement relies on the objects mass, specific heat capacity and changes in temperature. The lower thermal internal energy the lower the temperature and vice versa. We are more concerned with changes in thermal internal energy rather than measuring amounts. Bond internal energy is much like gravitational potential energy in that it can store energy. Heat is the transfer of internal energy on a microscopic scale. Objects cool when hear is lost. Thermal equilibrium is reached when two objects become the same temperature and flow stops."

"The internal energy of a system is the total energy of all of the molecules in the system except for the macroscopic kinetic energy."

"I am familiar with using Etherm equation from chemistry."

"The idea that "heat: only refers to the energy actually in transit form hot to cold. Something cannot contain heat just internal energy the heat is the energy moving form hot to cold."

"There does not exist 'cold,' but only 'lack of heat' helps with understanding this topic."

"During vaporization, bonds are broken and the bond internal energy decreases. Similarly, during the reverse processes bonds are formed and bond internal energy decreases."

"We should be careful to not confuse this topic with the thermodynamics we learn in chemistry. Also, the total thermal internal energy of an object's atomic and molecular motion depends on its mass (m) and its temperature T in Kelvins, and has units of joules (J), not calories or kilocalories like in chemistry class. We are going to look at the initial-to-final changes in the internal energies of systems."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.
"We covered thermodynamics extensively in my chemistry class earlier this semester, so most of the material in these sections was familiar. I'd like to see some applications of the transfer/balance equation, though."

"I got confused while I was reading about bond internal energy when you mentioned about atoms and molecules need to get closer in order to connect."

"The only thing that I found a little confusing was the the transfer balance equation and exactly what it meant. Everything else made sense to me, though."

"Almost everything."

"I could use more practice with the transfer balance equation."

"I did not have anything confusing with the reading assignment. But I did have trouble applying those concepts to the examples."

Two objects that are brought into contact with each other will reach thermal equilibrium when they have the same:
 internal energy. ************* [13] temperature. *********************** [23] (Both of the above choices.) ************ [12] (Neither of the above choices.) [0] (Unsure/lost/guessing/help!) ** [2]

Raw seafood is placed on a block of salt that has already been heated up. The energy contained in the high-temperature block of salt is then transferred to the seafood, cooking it. While it is being cooked, the internal thermal energy of the seafood __________, while the thermal internal energy of the salt block __________.
 increases; decreases. ********************************************* [45] decreases; increases. [0] does not change; does not change. * [1] (Unsure/lost/guessing/help!) **** [4]

For the seafood cooking on the salt block (ignoring heat transfers with the environment), the object that experienced the greatest amount of change (increase or decrease) in thermal internal energy was the:
 seafood. ******** [13] salt block. *** [3] (There is a tie.) ****************************** [30] (Unsure/lost/guessing/help!) **** [4]

Frozen meat is placed in a water bath, in order to defrost it. At the very start of this defrosting process (where the frozen meat just begins to warm up from its below-freezing temperature, and the ice crystals inside have not yet reached the melting point), the internal thermal energy of the meat __________, while the thermal internal energy of the water __________.
 increases; decreases. ******************************** [32] decreases; increases. ******* [7] does not change; does not change. ****** [5] (Unsure/lost/guessing/help!) ****** [6]

For the frozen meat in the water bath (ignoring heat transfers with the environment), the object that experienced the greatest amount of change (increase or decrease) in thermal internal energy was the:
 frozen meat. ***** [5] water bath. ******** [8] (There is a tie.) ******************************* [31] (Unsure/lost/guessing/help!) ****** [6]

A shot of whiskey is mixed with a pint of beer to make a boilermaker. Assuming that the whiskey and beer have approximately the same temperature before they are mixed together, the internal thermal energy of the whiskey __________, while the thermal internal energy of the beer __________.
 increases; decreases. *** [3] decreases; increases. ** [2] does not change; does not change. **************************************** [40] (Unsure/lost/guessing/help!) ***** [5]

For the shot of whiskey being mixed with the pint of beer (ignoring heat transfers with the environment), the object that experienced the greatest amount of change (increase or decrease) in thermal internal energy was the:
 shot of whiskey. ** [2] pint of beer. ** [2] (There is a tie.) *************************************** [39] (Unsure/lost/guessing/help!) ******* [7]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.
"Can we conduct an experiment with boilermakers in class?" (Already did that, after finishing grading your midterms.)

"How is internal thermal energy related to temperature?" (The same way translational kinetic energy is related to speed. And gravitational potential energy is related to height. These are all types of energies, each of which changes when a certain observable parameter (temperature, speed, height) changes.)

"How do we know what the specific heat is for stuff? Is there an equation for 'c' that we are supposed to know? Or would that value be given/have to solve for in most equations regarding internal thermal energy." (Generally specific heat values for different materials are given to you, unless you need to solve for it. Treat it like any other property of a material that you can look up, or have to solve for.)

"The only part I find confusing, is when you asked which object experienced the greatest amount of change (increase or decrease) in thermal energy, I said that they were all tied because I believe that whatever one object loses, the other one should pick it up--but I believe I am getting confused about this part." (Assuming that the objects that are interacting are isolated from the environment, then you are correct.)

"Do you enjoy teaching?" (Yes. Enough to teach sixty more semesters of physics.)

"What were we supposed to read?" (#smh.)

## 20151128

### Astronomy midterm question: Star Wars "That's no moon..." analysis

Astronomy 210 Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

In the movie Star Wars (20th Century Fox, 1977), the Death Star was a giant spherical traveling space station that was initially mistaken as a "small moon." In the webcomic panel at right[*] it is further argued that the Death Star is "too small to be a moon." Discuss why the Death Star cannot be considered a moon, despite its small size. Explain using the International Astronomical Union classification scheme.

[*] Randall Munroe, "Small Moon," xkcd.com/1458/.

• p:
Correct. Of the three IAU requirements (orbits the sun directly, has a rounded shape, cleared/dominates its orbit), if an object does not orbit the sun (or presumably any star) directly, but orbits a planet, then it is classified as a moon/satellite. Since the Death Star can travel through space, then it would not be a moon/satellite while it is not in orbit around a planet.
• r:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. May argue that the Death Star is not a moon because it does not orbit the sun (when the IAU classification scheme asks whether the object directly orbits the sun, or is in orbit around another object such as a planet), but understands the importance of the first classification question.
• t:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. May also involve the shape of the Death Star and/or its ability to gravitationally dominate other objects as important, other than its orbiting another object in orbit around the sun (or any star).
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Discussion only tangentially related to the IAU classification scheme.
• x:
Implementation/application of ideas, but credit given for effort rather than merit. Discussion unrelated to the IAU classification scheme.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Section 70160
Exam code: midterm02n1N0
p: 16 students
r: 9 students
t: 5 students
v: 1 student
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 7063), also sharing tangentially personal opinions on Star Wars:

### Astronomy midterm question: new two-rule planet classification rules

Astronomy 210 Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

An astronomer at the Southwest Research Institute in Boulder, CO proposed an alternate scheme[*] for defining a planet, paraphrased here as:
Rule 1: Must be small enough that it is not a star.
Rule 2: Must also be large enough that it formed itself into a spherical shape.
Discuss one example of something in our solar system that is currently not a planet according to the International Astronomical Union that would now be considered a planet under this new two-rule scheme. Explain using both the International Astronomical Union classification scheme, and this new two-rule scheme.

[*] Mark W. Buie, "Definition of a Planet," boulder.swri.edu/~buie/pluto/planetdefn.html.

• p:
Correct. Selects a round object such as a dwarf planet (such a Ceres, or Pluto) or the moon that is not a planet (due to not dominating its orbit around the sun; or because it does not orbit the sun directly), and shows that these objects would be considered a planet under the new two-rule scheme, as they would be small enough to not be a star, and also be rounded in the shape.
• r:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
• t:
Contains right ideas, but discussion is unclear/incomplete or contains major errors.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Discussion only tangentially related to the IAU classification scheme.
• x:
Implementation/application of ideas, but credit given for effort rather than merit. Discussion unrelated to the IAU classification scheme.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Section 70158
Exam code: midterm02sP3c
p: 18 students
r: 4 students
t: 11 students
v: 0 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 3695) discussing how a dwarf planet such as Pluto or Ceres would then be considered a planet:

A sample "p" response (from student 1022) discussing how the moon would then be considered a planet:

### Astronomy midterm question: ranking distances given apparent magnitudes, absolute magnitudes

Astronomy 210 Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

The apparent magnitudes and absolute magnitudes of three stars are listed below.
 mapparentmagnitude Mabsolutemagnitude Avior +1.9 –4.5 Benetnash +1.9 –3.0 Tau Ceti +3.5 +5.7

Determine which star is farthest away (or indicate a tie, if any). Explain using the relationships between apparent magnitude, absolute magnitude, and distance.

• p:
Correct. Understands difference between apparent magnitude m (brightness as seen from Earth, when placed at their actual distance from Earth) and absolute magnitude M (brightness as seen from Earth, when placed 10 parsecs away), and compares:
1. how star A (Avior) and star B (Benetash) appear equally bright (same apparent magnitude m), but star A is brighter than star B when both are placed at 10 parsecs away from Earth (brighter absolute magnitude), thus star A had to be moved a greater distance closer to Earth than star B; and
2. how star C (Tau Ceti) appears dimmer than star A and star B (dimmer apparent magnitude m), and when star C is moved further back to 10 parsecs away from Earth, it will be even dimmer than star A and star B at 10 parsecs (and thus star C is the closest of these three stars).
• r:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Discusses (1), but does not discuss (2).
• t:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least discussion demonstrates understanding of relationships between apparent magnitudes, absolute magnitudes, and distances. May have order of brightnesses reversed, picking star C as having the brightest absolute magnitude.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least attempts to use relationships between apparent magnitudes, absolute magnitudes, and distances.
• x:
Implementation/application of ideas, but credit given for effort rather than merit. Discussion based on garbled definitions of, or not based on proper relationships between apparent magnitudes, absolute magnitudes,
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Section 70160
Exam code: midterm02sP3c
p: 27 students
r: 3 students
t: 9 students
v: 2 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 1022):

### Astronomy midterm question: ranking absolute magnitudes given apparent magnitudes, distances

Astronomy 210 Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

The apparent magnitudes and distances of three stars are listed below.
 mapparentmagnitude ddistancefrom Earth Avior +1.9 190 pcs Benetnash +1.9 32 pcs Tau Ceti +3.5 4 pcs

Determine which star has the brightest absolute magnitude (or indicate a tie, if any). Explain using the relationships between apparent magnitude, absolute magnitude, and distance.

• p:
Correct. Understands difference between apparent magnitude m (brightness as seen from Earth, when placed at their actual distance from Earth) and absolute magnitude M (brightness as seen from Earth, when placed 10 parsecs away), and compares:
1. how star A (Avior) and star B (Benetash) appear equally bright (same apparent magnitude m), but star A is brighter than star B when both are placed at 10 parsecs away from Earth (and thus has a brighter absolute magnitude), as star A will be moved a greater distance closer to Earth than star B; and
2. how star C (Tau Ceti) appears dimmer than star A and star B (dimmer apparent magnitude m), and when star C is moved further back to 10 parsecs away from Earth, it will be even dimmer than star A and star B at 10 parsecs (and thus star C has the dimmest absolute magnitude of these three stars).
• r:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors. Discusses (1), but does not discuss (2).
• t:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least discussion demonstrates understanding of relationships between apparent magnitudes, absolute magnitudes, and distances. May have order of brightnesses reversed, picking star C as having the brightest absolute magnitude.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least attempts to use relationships between apparent magnitudes, absolute magnitudes, and distances.
• x:
Implementation/application of ideas, but credit given for effort rather than merit. Discussion based on garbled definitions of, or not based on proper relationships between apparent magnitudes, absolute magnitudes,
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Section 70160
Exam code: midterm02n1N0
p: 18 students
r: 3 students
t: 5 students
v: 1 student
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 3030):

A sample "p" response (from student 5555):

### Astronomy midterm question: same luminosity, same size, different temperature stars?

Astronomy 210 Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

An astronomy question on an online discussion board[*] was asked and answered:
??: Can two stars with the same luminosity have the same size, but different temperatures?
Lodar: If the temperatures are different, the sizes would also have to be different for their luminosities to be the same.
Discuss why this answer is correct, and how you know this. Explain using Wien's law, the Stefan-Boltzmann law and/or an H-R diagram.

• p:
Correct. Uses Wien's law, the Stefan-Boltzmann law and/or interprets H-R diagram to demonstrate how same luminosity stars with different temperatures would have to have different sizes. May also demonstrate that same luminosity stars with different temperatures cannot have the same temperatures.
• r:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
• t:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least discussion demonstrates understanding of Wien's law, H-R diagram and/or the Stefan-Boltzmann law.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least attempts to use Wien's law, H-R diagram and/or the Stefan-Boltzmann law.
• x:
Implementation/application of ideas, but credit given for effort rather than merit. Discussion not clearly based on Wien's law, H-R diagram and/or the Stefan-Boltzmann law.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Section 70158
Exam code: midterm02sP3c
p: 34 students
r: 3 students
t: 1 student
v: 2 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 5635) discussing how same luminosity stars with different temperatures would have to have different sizes:

A sample "p" response (from student 1022), discussing how same luminosity stars with different temperatures cannot have the same temperatures:

### Astronomy midterm question: hotter giant same size as cooler supergiant?

Astronomy 210 Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

An astronomy question on an online discussion board[*] was asked and answered:
??: Can a giant be the same size as a supergiant if the giant were hotter than the supergiant?
MirJ: No.
Discuss why this answer is correct, and how you know this. Explain using Wien's law, the Stefan-Boltzmann law and/or an H-R diagram.

• p:
Correct. Uses Wien's law, the Stefan-Boltzmann law and/or interprets H-R diagram to demonstrate how a supergiant and a giant with the same size is not possible if the giant is hotter by discussing either that:
1. for a supergiant and giant with the same size, the supergiant must be hotter than the giant (thus the giant cannot be hotter than the supergiant);
2. for a giant hotter than a supergiant of the same size, the giant would be more luminous than the supergiant (which is not possible, as all supergiants are more luminous than giants);
3. for a giant hotter than a supergiant, the giant must be smaller than the supergiant (thus the giant cannot be the same size as the supergiant).
• r:
Nearly correct (explanation weak, unclear or only nearly complete); includes extraneous/tangential information; or has minor errors.
• t:
Contains right ideas, but discussion is unclear/incomplete or contains major errors. At least discussion demonstrates understanding of Wien's law, H-R diagram and/or the Stefan-Boltzmann law.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least attempts to use Wien's law, H-R diagram and/or the Stefan-Boltzmann law.
• x:
Implementation/application of ideas, but credit given for effort rather than merit. Discussion not clearly based on Wien's law, H-R diagram and/or the Stefan-Boltzmann law.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Section 70160
Exam code: midterm02n1N0
p: 18 students
r: 2 students
t: 1 student
v: 3 students
x: 6 students
y: 1 student
z: 0 students

A sample "p" response (from student 4321) discussing how a supergiant must be hotter than the giant of the same size (thus the giant cannot be hotter than the supergiant):

A sample "p" response (from student 6392) how giant that is hotter than a supergiant of the same size would be more luminous than the supergiant (which is not possible, as all supergiants are more luminous than giants):

A sample "p" response (from student 1996) how a giant hotter than a supergiant must be smaller than the supergiant (thus the giant cannot be the same size as the supergiant):

## 20151127

### Astronomy current events question: oldest known Milky Way stars

Astronomy 210L, fall semester 2015
Cuesta College, San Luis Obispo, CA

Students are assigned to read online articles on current astronomy events, and take a short current events quiz during the first 10 minutes of lab. (This motivates students to show up promptly to lab, as the time cut-off for the quiz is strictly enforced!)
Charles Q. Choi, "These Ancient Stars May Be the Oldest Ever Seen in the Milky Way" (November 11, 2015)
space.com/31083-oldest-stars-milky-way-galaxy.html
The Australia National University's SkyMapper telescope discovered the oldest known stars in the Milky Way, based on their:
(A) lack of metals.
(B) erratic orbits.
(C) type II supernova explosions.
(D) gravitational redshifts.
(E) dark matter content.

Student responses
Sections 70178, 70186
(A) : 13 students
(B) : 1 student
(C) : 4 students
(D) : 3 students
(E) : 1 student

### Astronomy current events question: discovery of dwarf planet V774104

Astronomy 210L, fall semester 2015
Cuesta College, San Luis Obispo, CA

Students are assigned to read online articles on current astronomy events, and take a short current events quiz during the first 10 minutes of lab. (This motivates students to show up promptly to lab, as the time cut-off for the quiz is strictly enforced!)
Geoff Brumfiel, "Astronomers Spot Most Distant Object So Far In The Solar System" (November 16, 2015)
npr.org/sections/thetwo-way/2015/11/11/455643251/astronomers-spot-most-distant-object-in-the-solar-system
Dwarf planet V774104, the most distant solar system object discovered so far was detected by:
(A) gravitational lensing.
(B) gamma ray bursts.
(C) motion in photographs.
(E) gravitational forces on inner planets.

Student responses
Sections 70178, 70186
(A) : 1 students
(B) : 2 students
(C) : 14 students
(D) : 2 students
(E) : 1 students

### Astronomy current events question: Phobos' surface egrooves

Astronomy 210L, fall semester 2015
Cuesta College, San Luis Obispo, CA

Students are assigned to read online articles on current astronomy events, and take a short current events quiz during the first 10 minutes of lab. (This motivates students to show up promptly to lab, as the time cut-off for the quiz is strictly enforced!)
Elizabeth Zubritsky, "Mars’ Moon Phobos is Slowly Falling Apart" (November 10, 2015)
nasa.gov/feature/goddard/phobos-is-falling-apart
Computer models indicate that the grooves on Mars' moon Phobos may be produced by:
(A) solar heating.
(B) convection currents.
(C) cryovolcanoes.
(D) Mars' gravity.
(E) its rapid rotation.

Student responses
Sections 70178, 70186
(A) : 1 student
(B) : 3 students
(C) : 1 student
(D) : 15 students
(E) : 0 students

### Physics midterm question: comparing tensions in suspended beams

Physics 205A Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

Two horizontal uniform beams are each mounted on pivots and suspended by 45° angle cables. The short beam is one-half the length of the long beam, but the short and long beams have the same mass. Discuss why the tension in the short beam cable is equal to the tension in the long beam cable. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

• p:
Correct. Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
1. Newton's first law applies to each case, such that the cw torque of the cable equals the ccw torque of the weight;
2. for either case, the perpendicular lever arm for the weight force is (L/2) and the perpendicular lever arm for the cable is L·sin(45°), where L is the length of either beam;
3. from equating torques, for either beam, the tension force T = w/(2·sin(45°)) independent of the length of the beam, and thus is the same for either beam.
May also instead argue (2)-(3) by recognizing similar triangles, such that tension force T must be the same for either beam given that their weights are the same.
• r:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May have incorrect or missing cable lever arm.
• t:
Nearly correct, but argument has conceptual errors, or is incomplete. Typically equates similiar lever arms (short ℓw = long ℓw, short ℓT = long ℓT), when it is actually the ratios of these lever arms that are equal (short ℓw/short ℓT) = (long ℓw/long ℓT).
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's first law to torques, forces, and perpendicular lever arms.
• x:
Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Sections 70854, 70855, 73320
Exam code: midterm02h4W6
p: 13 students
r: 15 students
t: 21 students
v: 20 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student 2934), comparing relative ratios of lever arms:

A sample "p" response (from student 1793), putting numbers in to find absolute equivalence of cable tension forces:

### Physics midterm question: rising water level in dry dock

Physics 205A Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

“Cruise Ship timelapse - Extension of Braemar at Blohm+Voss”
MK timelapse GmbH
youtu.be/QirVr-pEVU4

A cruise ship rests on the floor of a dry dock with a small amount of water. As the water level rises (but not enough water to "float" the ship yet, so it is still resting on the bottom of the dry dock), discuss why the normal force of the floor on the ship decreases. Explain your reasoning using the properties of densities, volumes, forces, Newton's laws, Archimedes' principle (buoyant forces), and free-body diagrams.

• p:
Correct. Recognizes that:
1. from Newton's first law the downwards weight force (which has a constant magnitude, as its mass remains constant) on the ship is equal to the two upwards buoyant and normal forces on the ship;
2. as the water level rises, the amount of the ship's submerged volume increases, increasing the amount of buoyant force on the ship (Archimedes' principle);
such that the normal force must decrease in magnitude in order for the two upwards forces to still balance the downwards weight force.
• r:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some constructive attempt at relating the buoyant force to the density of the fluid and volume displaced (Archimedes' principle) and/or Newton's first law.
• x:
Implementation of ideas, but credit given for effort rather than merit. Appeal to some other properties of fluids and densities other than Archimedes' principle and Newton's laws.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Sections 70854, 70855, 73320
Exam code: midterm02h4W6
p: 47 students
r: 10 students
t: 4 students
v: 9 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 3158):

### Physics midterm question: different mass boxes attached to comparable springs

Physics 205A Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

Two horizontal springs are attached to boxes of different masses. The springs have the same strength, and the oscillation amplitudes are the same. Discuss why the boxes will have the same amount of translational kinetic energy at x = 0. Ignore friction/drag. Explain your reasoning using the properties of translational kinetic energy, elastic potential energy, energy conservation and mass-spring systems.

• p:
Correct. Discusses/demonstrates the application of energy conservation:
1. at x = +A, both boxes have the same amount of elastic potential energy (same-strength springs, same amplitudes);
2. since there is zero translational kinetic energy at x = +A, then both boxes have the same amount of total energy (equal to the elastic potential energy at x = +A);
3. at x = 0, both boxes have zero elastic potential energy, such that all of their (same) total energy is in the form of translational kinetic energy at that location, which must then be the same.
May have also discussed how box 1 (with presumably less mass) will have a shorter period, higher frequency, or faster maximum speed than box 2.
• r:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May confuse changes in energy forms with amounts of energy (e.g., 0 = KEtr + PEelas, E = ∆KEtr + ∆PEelas, etc.), and/or not explicitly noting that maximum elastic potential energy and maximum translational kinetic energy values occur at different locations, etc.
• t:
Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x:
Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying energy conservation.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Sections 70854, 70855, 73320
Exam code: midterm02h4W6
p: 15 students
r: 31 students
t: 12 students
v: 10 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 1337):

### Physics midterm question: comparable transverse wave speeds

Physics 205A Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

“Transverse wave travel along a bungee cord”
The Orchard - ćžść¨ąĺś’
youtu.be/g49mahYeNgc

A transverse wave pulse is created by plucking one end such that it travels down along a cable or a wire. The cable and the wire are made of the same material, but the cable is thicker than the wire. The cable is shorter than the wire, and has more mass hanging off the end. Discuss how it is possible that waves could travel with the same speed along the cable and the wire. Explain your reasoning using the properties of waves.

• p:
Correct. The thicker cable has more tension (F) and a greater linear mass density (m/L) than the thinner wire, which has less tension and a lesser linear mass density. Transverse wave speeds depend on the square root of tension/linear mass density, such that it is plausible that wave can travel at the same speed if the ratio of the numerator and denominator in the square root is the same for cable and wire. (The length of the cable and wire does not directly influence wave speeds, given that the ratio of mass per unit length of the cable and of the wire cannot be changed by using different lengths.)
• r:
As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May involve length-dependence in linear mass density term as affecting wave speed.
• t:
Nearly correct, but argument has conceptual errors, or is incomplete. Claims that for the cable, the greater mass (whether due to the thickness of the cable, or the mass hanging off of it) is compensated by its shorter length; while for the wire less mass is compensated by its longer length, and since the tension is the same (whether assumed, implied, or ascribed to the same amount of "pluck" in creating a transverse wave pulse), then the wave speeds are the same for the cable and wire.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner.
• x:
Implementation of ideas, but credit given for effort rather than merit. Approach other than that of relating wave speeds with tensions and linear mass densities.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Sections 70854, 70855, 73320
Exam code: midterm02h4W6
p: 38 students
r: 3 students
t: 18 students
v: 4 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 2820):

### Physics midterm problem: A-10 Thunderbolt II "Hawg-wash"

Physics 205A Midterm 2, fall semester 2015
Cuesta College, San Luis Obispo, CA

"30mm GAU 8 Avenger Impacts - Slowmo (IMAX Fighter Pilot Movie)"
XNeo27564
youtu.be/nk1HU5WShpU

The possible effect of recoil on an airplane is described[*] below:
Affectionately known as the "Hawg," the A-10 Thunderbolt II is essentially a flying gun designed to provide close air support to troops on the ground... An old rumor that a reader recently asked about is the A-10's gun is so powerful that when fired the recoil slows the plane down almost to a complete stop....
The myth that its cannon drastically decelerates the aircraft is pure "Hawg-wash," said retired Air Force Col. Steve Ruehl. "I have fired as many as 500 rounds in one trigger burst..."
The mass of the aircraft[**] is 2.30×104 kg, and the combined mass of 500 bullets[***] is approximately 200 kg. Assume that the aircraft (and bullets inside) are initially flying horizontally in the forward direction at a speed of 190 m/s, and then 500 bullets are fired all at the same time with a speed of 1,200 m/s out from the front of the aircraft (as measured by an outside observer). Ignore the effects of friction, drag, and the thrust of the aircraft engines (assuming that aircraft is just gliding) during the brief amount of time that the bullets were fired. Find (a) the final speed of the aircraft after firing 500 bullets (all at the same time); then (b) discuss whether the claim that "the recoil slows the plane down almost to a complete stop" is plausible. Show your work and explain your reasoning using properties of collisions, energy (non-)conservation, and momentum conservation.

[*] stripes.com/blogs/the-rumor-doctor/the-rumor-doctor-1.104348/does-the-a-10-s-gun-slow-the-plane-when-fired-1.152557.
[**] globalsecurity.org/military/systems/aircraft/a-10-specs.htm.
[***] orbitalatk.com/defense-systems/small-caliber-systems/30mm/docs/GAU-8A_Fact_Sheet.pdf.

• p:
Correct. Momentum is conserved as external impulses from drag and the aircraft engines are to be ignored, and:
1. sets up momentum conservation for the aircraft and the bullets to solve for the final velocity of the aircraft;
2. discusses implausibility of claim that the "recoil slows the plane down almost to a complete stop" from the very slight decrease in the speed of the aircraft.
• r:
Nearly correct, but includes minor math errors in (1), or tangential/minor conceptual errors in (2).
• t:
Nearly correct, but argument has conceptual errors, or is incomplete. Certain parameters are misplaced or misidentified (typically sets initial velocity of bullets is zero), but at least successfully applied momentum conservation to find final velocity of aircraft, and discussion is consistent with result.
• v:
Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least some systematic attempt at using momentum conservation.
• x:
Implementation of ideas, but credit given for effort rather than merit. No clear attempt at applying momentum conservation.
• y:
Irrelevant discussion/effectively blank.
• z:
Blank.
Sections 70854, 70855, 73320
Exam code: midterm02h4W6
p: 33 students
r: 6 students
t: 3 students
v: 8 students
x: 5 students
y: 1 student
z: 0 students

A sample "p" response (from student 2060):

## 20151122

### Physics quiz question: femur compression

Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

An average femur bone (Young's modulus 9.4×109 N/m2) has a length of 0.48 m and an approximate cross-sectional area of 1.72×10–3 m2, and reportedly can support a maximum 2.4×104 N of force.[*][**] When this force is applied, the femur would compress by:
(A) 2.1×10–9 m.
(B) 9.1×10–9 m.
(C) 7.2×10–8 m.
(D) 7.1×10–4 m.

[*] "30 times the weight of an adult," orthopaedicsone.com/display/Review/Femur.
[**] "Weight of average adult: 178 lbs," wolfr.am/8dc2Ohi7.

Correct answer (highlight to unhide): (D)

Hooke's law is given by:

(F/A) = Y·(∆L/L),

such that the amount that the femur would be compressed is:

L = (F·L)/(A·Y),

L = ((2.4×104 N)·(0.48 m))/((1.72×10–3 m2)·(9.4×109 N/m2)),

L = 0.0007125185552 m,

or to two significant figures, the femur would compress by 7.1×10–4 m.

(Response (A) is (F·L·A)/Y; response (B) is (F·A)/(L·Y); response (C) is A/F.)

Sections 70854, 70855, 73320
Exam code: quiz06m45S
(A) : 3 students
(B) : 0 students
(C) : 3 students
(D) : 67 students

Success level: 92%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.17

### Physics quiz question: mass-spring translational kinetic energy

Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

A 0.40 kg mass attached to a horizontal spring oscillates with an amplitude of 0.20 m. Ignore friction/drag. This mass-spring system has zero translational kinetic energy at:
(A) x = 0.
(B) x = +0.20 m.
(C) (Both of the above choices.)
(D) (None of the above choices.)

Correct answer (highlight to unhide): (B)

Since there is no work done on or by this mass-spring system, its total mechanical energy is a constant:

E = KEtr + PEelas,

E = (1/2)·m·v2 + (1/2)·k·x2.

If there is zero translational kinetic energy, then KEtr = 0, and the total mechanical energy is solely comprised of elastic potential energy, which would be a maximum at x = +0.20 m (as well as at x = –0.20 m).

(If there is zero elastic potential energy, then PEelas = 0, and the total mechanical energy is solely comprised of translational kinetic energy, which would be a maximum at x = 0.)

Sections 70854, 70855, 73320
Exam code: quiz06m45S
(A) : 10 students
(B) : 63 students
(C) : 0 students
(D) : 0 students

Success level: 85%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.26

### Physics quiz question: Home Alone paint can pendulum

Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

Home Alone
Twentieth Century Fox (1990)

A full paint can (mass[*] 5.6 kg) tied to a string, starting at one side from rest, swings for 3.0 s to reach its fastest speed when the string is vertically downwards. Approximate this as a simple pendulum, and ignore friction/drag. The oscillation period of the paint can is:
(A) 1.5 s.
(B) 1.9 s.
(C) 4.7 s.
(D) 12.0 s.

[*] cockeyed.com/science/weight/paint-gallon.html.

Correct answer (highlight to unhide): (D)

The period of a pendulum is the time for it to complete one cycle of motion, returning to its starting point with the same velocity that it started with. Since it takes 3.0 s for the paint can to start from rest and swing to hang straight down, it will take an additional 3.0 s for the paint can to swing to the other side, and an additional 6.0 s for it to swing back to its original position, for a total of 12.0 s seconds.

(The amount of time that the paint can takes to swing downwards in this movie is exaggerated for dramatic effect.)

(Response (A) is one-half of the time to swing downwards; response (B) is the mass m divided by the time to swing downwards; response (C) is 2·Ď€·√(m/g).)

Sections 70854, 70855, 73320
Exam code: quiz06m45S
(A) : 9 students
(B) : 6 students
(C) : 8 students
(D) : 50 students

Success level: 68%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.61

### Physics quiz question: body mass measurement device

Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

"InsideISS #ISS #weight (Guess Reid Wiseman's Weight)"
space_station
vine.co/v/OKUDDvHgaJD

An astronaut (mass of 80.17 kg) aboard the International Space Station oscillates with a period of 0.93 s on a body mass measurement device. Approximate this as a mass-spring system (where the astronaut is the only mass attached to the spring), and ignore friction/drag. Take the gravitational constant g to be effectively zero aboard the International Space Station. The spring constant of the device is:
(A) 18 N/m.
(B) 58 N/m.
(C) 5.4×102 N/m.
(D) 3.7×103 N/m.

Correct answer (highlight to unhide): (D)

The period T of a mass m attached to a spring with spring constant k is given by:

T = 2·π·√(m/k),

such that the spring constant k will be:

k = m·(2·π/T)2 = (80.17 kg)·(2·π/(0.93 s))2 = 3,659.3649431 kg/s2,

or as expressed in more conventional units to two significant figures, the spring constant is 3.7×103 (kg·m/s2)·(1/m) = 170 N/m.

(Response (A) is 2·Ď€·√(m/g); response (B) is 2·Ď€·√(m/T); and response (C) is m·(2·π/T).)

Sections 70854, 70855, 73320
Exam code: quiz06m45S
(A) : 3 students
(B) : 6 students
(C) : 6 students
(D) : 56 students
(No response: 2 students)

Success level: 77%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.52

## 20151121

### Physics quiz question: moment of inertia of LP record

Physics 205A Quiz 5, fall semester 2015
Cuesta College, San Luis Obispo, CA

"I have a vinyl fetich"
Christmas Junkie
flic.kr/p/fFwATx

An LP record[*] has a rotational inertia of 1.46×10–3 kg·m2, and rotates at a rate of 3.5 rad/s. The record can be considered a solid disk (Idisk = (1/2)·M·R2). If the LP record then rotates at a slower rate, it would have __________ rotational inertia.
(A) more.
(B) the same.
(C) less.
(D) (Not enough information is given.)

[*] Peter Atkins, Julio de Paula, Physical Chemistry (9th ed.), W.H. Freeman and Co. (2009), p. 318.

Correct answer (highlight to unhide): (B)

Since rotational inertia of the LP record depends on its mass M and radius R, which do not physically change whether rotating at a faster or slower rate, then the rotational inertia remains constant.

Sections 70854, 70855, 73320
Exam code: quiz05bL8D
(A) : 24 students
(B) : 32 students
(C) : 9 students
(D) : 0 students

Success level: 48%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.74

### Physics quiz question: tension force of string on suspended beam

Physics 205A Quiz 5, fall semester 2015
Cuesta College, San Luis Obispo, CA

A uniform beam (0.60 m length) is suspended by a string that pulls vertically upwards, exerting a torque of magnitude 1.1 N·m. (Calculate all torques with respect to the pivot, located at the other end.) The tension in the string is:
(A) 3.7 N.
(B) 0.92 N.
(C) 0.66 N.
(D) 1.8 N.

Correct answer (highlight to unhide): (A)

The tension force T of the string acts at the top of the beam, along the string. The perpendicular lever arm ℓ for the string tension force must extend from the pivot to perpendicularly intercept the tension force line of action (which lies along the string itself), such that this will be a horizontal line of length:

ℓ = L·cos(60.0°).

The (clockwise) torque exerted by the string on the beam is then:

τ = T·ℓ,

such that the magnitude of the tension force T can be solved for:

T = τ/ℓ/ = τ/(L·cos(60.0°)),

T = (1.1 N·m)/((0.60 m)·cos(60.0°)) = 3.66666666... N,

or to two significant figures, the tension in the string is 3.7 N.

(Response (B) is τ/(2·L); response (C) is τ/2; and response (D) is τ/L.)

Sections 70854, 70855, 73320
Exam code: quiz05bL8D
(A) : 14 students
(B) : 2 students
(C) : 19 students
(D) : 30 students

Success level: 45%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.47

### Physics quiz question: Mt. Whitney vs. Badwater Basin pressure difference

Physics 205A Quiz 5, fall semester 2015
Cuesta College, San Luis Obispo, CA

"Ain't no mountain high enough!"
Christine
flic.kr/p/h9k5r4

The highest and lowest points[*] in the 48 contiguous states of the U.S are Mt. Whitney (4,421 m above sea level) and Badwater Basin (85 m below sea level). Assume that the variation of gravitational constant g and the density of air (ρair = 1.3 kg/m3) with elevation is negligible. The air pressure on top of Mt. Whitney is __________ less than the air pressure in Badwater Basin.
(A) 6.6×102 Pa.
(B) 4.5×104 Pa.
(C) 5.3×104 Pa.
(D) 1.024×105 Pa.

[*] wki.pe/Extreme_points_of_the_United_States.

Correct answer (highlight to unhide): (C)

For static fluids, the energy density relation between pressure difference ∆P and elevation change ∆y is given by:

0 = ∆P + ρ·g·∆y.

The difference in pressure ∆P between Mt. Whitney and Badwater Basin is then:

P = –ρ·g·∆y,

P = –(1.3 kg/m3)·(9.80 m/s2)·((+4,421 m) – (–85 m)),

P = –(1.3 kg/m3)·(9.80 m/s2)·(+4,506 m),

P = –57,406.44 Pa,

such that the difference in pressure between Mt. Whitney and Badwater Basin (to two significant figures) is 5.7×104 Pa (the negative sign in the above calculation denotes that the air pressure at Mt. Whitney is less than the air pressure in Badwater Basin).

(Response (A) is ρ·g·(yWhitney/yBadwater); response (B) is the absolute air pressure atop Mt. Whitney; and response (D) is the absolute air pressure in Badwater Basin.)

Sections 70854, 70855, 73320
Exam code: quiz05bL8D
(A) : 17 students
(B) : 0 students
(C) : 43 students
(D) : 5 students

Success level: 66%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.63

### Physics quiz question: hanging vs. anchored submerged objects

Physics 205A Quiz 5, fall semester 2015
Cuesta College, San Luis Obispo, CA

Two objects of different volumes and different densities are in a tank of water. One object hangs down from a string, while the other object is anchored to the bottom with a string. The amount of tension in each string is the same. The __________ object has a greater magnitude buoyant force exerted on it.
(A) hanging.
(B) anchored.
(C) (Not enough information is given.)
(D) (There is a tie.)

Correct answer (highlight to unhide): (A)

The magnitude of the buoyant force is given by:

FB = ρfluid·g·(Volume submerged),

where the density ρ is of the surrounding fluid (water) and the gravitational constant g = 9.80 m/s2. Both objects are fully submerged, and since the volume of water displaced by the hanging object is greater than the volume of water displaced by the anchored object, then the hanging object has a greater magnitude buoyant force exerted on it.

(The list of the forces acting on the hanging object are:
Buoyant force of water FB = ρfluid·g·(Volume submerged) (upwards),
Tension force of string T (upwards),
Weight force of Earth w = m·g (downwards).
The list of the forces acting on the hanging object are:
Buoyant force of water FB = ρfluid·g·(Volume submerged) (upwards),
Tension force of string T (downwards),
Weight force of Earth w = m·g (downwards).
Newton's first law applies to both objects, as they are stationary, and thus the forces acting on either object must sum to zero.

For the hanging object, the magnitude of the downward weight force w must equal the sum of the magnitudes of the upwards tension force T and (larger) upwards buoyant force FB, and thus the hanging object will have a larger magnitude weight force than the anchored object (which has a downward weight force w equal to the (smaller) upwards buoyant force FB minus the tension force T.)

Sections 70854, 70855, 73320
Exam code: quiz05bL8D
(A) : 18 students
(B) : 29 students
(C) : 17 students
(D) : 1 student

Success level: 27%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.80

### Astronomy current events question: black hole x-ray flares

Astronomy 210L, fall semester 2015
Cuesta College, San Luis Obispo, CA

Students are assigned to read online articles on current astronomy events, and take a short current events quiz during the first 10 minutes of lab. (This motivates students to show up promptly to lab, as the time cut-off for the quiz is strictly enforced!)
Whitney Clavin, "Black Hole Has Major Flare" (October 27, 2015)
jpl.nasa.gov/news/news.php?feature=4753
Two NASA space telescopes detected an x-ray flare from a supermassive black hole, which may have been triggered by __________ in its surrounding corona.
(B) disruptions.
(C) antimatter.
(D) a supernova.
(E) gravity waves.

Student responses
Sections 70178, 70186
(A) : 4 students
(B) : 13 students
(C) : 3 students
(D) : 6 students
(E) : 5 students

### Astronomy current events question: martian atmosphere stripped by solar wind

Astronomy 210L, fall semester 2015
Cuesta College, San Luis Obispo, CA

Students are assigned to read online articles on current astronomy events, and take a short current events quiz during the first 10 minutes of lab. (This motivates students to show up promptly to lab, as the time cut-off for the quiz is strictly enforced!)
Dwayne Brown, Laurie Cantillo, Nancy Neal-Jones, Bill Steigerwald, and Jim Scott, "NASA Mission Reveals Speed of Solar Wind Stripping Martian Atmosphere" (November 5, 2015)
nasa.gov/press-release/nasa-mission-reveals-speed-of-solar-wind-stripping-martian-atmosphere
Based on data from NASA’s Mars Atmosphere and Volatile Evolution (MAVEN) spacecraft, much of Mars' original atmosphere may have been removed by:
(A) the greenhouse effect.
(B) glaciers.
(C) volcanoes.
(D) solar winds.
(E) its ancient oceans and rivers.

Student responses
Sections 70178, 70186
(A) : 4 students
(B) : 0 students
(C) : 0 students
(D) : 27 students
(E) : 0 students

### Astronomy current events quiz: spiral dust patterns around newborn stars

Astronomy 210L, fall semester 2015
Cuesta College, San Luis Obispo, CA

Students are assigned to read online articles on current astronomy events, and take a short current events quiz during the first 10 minutes of lab. (This motivates students to show up promptly to lab, as the time cut-off for the quiz is strictly enforced!)
Ray Villard, "Spirals in Dust Around Young Stars May Betray Presence Of Massive Planets" (October 29, 2015)
nasa.gov/feature/goddard/spirals-dust-around-young-stars-may-betray-presence-of-massive-planets
Huge spiral patterns in the dust and gas disks surrounding newborn stars may signal the presence of giant planet formation, based on:
(A) computer simulations.
(B) Hubble Space Telescope observations.
(D) x-ray emissions.
(E) data from neutrino detectors.

Student responses
Sections 70178, 70186
(A) : 15 students
(B) : 7 students
(C) : 3 students
(D) : 6 students
(E) : 0 students

### Physics quiz archive: simple harmonic motion

Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz06m45S

Sections 70854, 70855, 73320 results
 0- 6 : 7-12 : ** [low = 12] 13-18 : ********** 19-24 : ***************************** [mean = 24.2 +/- 4.7] 25-30 : ******************************** [high = 30]

### Astronomy quiz question: old star cluster medium-mass stars

Astronomy 210 Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

Medium-mass stars in an old star cluster will be in their __________ stage.
(A) protostar.
(B) red dwarf.
(C) white dwarf.
(D) supergiant.

Correct answer (highlight to unhide): (C)

All stars in a star cluster are born at the same time, but undergo stellar evolution at different rates depending on their masses. In an old star cluster the medium-mass stars would have gone through their protostar, main-sequence and giant stages, such that they will have ultimately become white dwarfs.

Section 70158
Exam code: quiz06s7ll
(A) : 3 students
(B) : 12 students
(C) : 19 students
(D) : 6 students

Success level: 52% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.31

### Astronomy quiz question: brighter luminosity stars

Astronomy 210 Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

A main-sequence star that is more massive than the sun would have a brighter luminosity due to __________ in its core.
(A) faster fusion rate.
(B) more convection.
(C) heavy elements.

Correct answer (highlight to unhide): (A)

The more massive a main-sequence star is, the higher pressures and temperatures there will be in core, such that the rate of hydrogen fusion will be faster, releasing more energy to make the star brighter.

Section 70160
Exam code: quiz06nUm6
(A) : 23 students
(B) : 3 students
(C) : 2 students
(D) : 0 students

Success level: 83% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.15

### Astronomy quiz question: old star cluster low-mass stars

Astronomy 210 Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

Low-mass stars in an old star cluster will be in their __________ stage.
(A) protostar.
(B) red dwarf.
(C) white dwarf.
(D) supergiant.

Correct answer (highlight to unhide): (B)

All stars in a star cluster are born at the same time, but undergo stellar evolution at different rates depending on their masses. Low-mass stars evolve much more slowly than medium-mass and massive stars, so in an old star cluster they would be red dwarfs on the main-sequence.

Section 70160
Exam code: quiz06nUm6
(A) : 0 students
(B) : 17 students
(C) : 11 students
(D) : 0 students

Success level: 64% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.43

### Astronomy quiz archive: stellar evolution

Astronomy 210 Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

Section 70158, version 1
Exam code: quiz06s7ll

Section 70158
 0- 8.0 : 8.5-16.0 : **** [low = 12.0] 16.5-24.0 : ************ 24.5-32.0 : ************** [mean = 26.7 +/- 8.1] 32.5-40.0 : ********** [high = 40.0]

Section 70160, version 1
Exam code: quiz06nUm6

Section 70160
 0- 8.0 : * [low = 8.0] 8.5-16.0 : * 16.5-24.0 : ******** 24.5-32.0 : ************* [mean = 25.8 +/- 7.6] 32.5-40.0 : ***** [high = 40.0]