20090430

Physics quiz question: guitar string wave speed

Physics 205A Quiz 6, spring semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 11.9

A metal guitar string has a linear mass density of 3.6 g/m, and is stretched with a tension of 130 N. What is the speed of transverse waves on this string?
(A) 5.3×10–3 m/s.
(B) 0.68 m/s.
(C) 190 m/s.
(D) 3.6×104 m/s.

Correct answer (highlight to unhide): (C)

For transverse waves on a string, the wave speed is given by:

v = √(F/(m/L)),

where F is the tension in the string, and (m/L) is the linear mass density (mass per unit length) of the string. Plugging in the given values, where (m/L) must be properly given in units of kg/m:

v = √((130 N)/(0.0036 kg/m)) = 190.0292375 m/s,

or to two significant figures, v = 1.9×102 m/s.

(Response (A) is √((m/L)/F); response (B) is sqrt(F·(m/L)); and response (D) is F/(m/L).)

Student responses
Sections 30880, 30881
(A) : 5 students
(B) : 1 student
(C) : 28 students
(D) : 6 students

Success level: 70%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.62

20090429

Physics quiz question: extending length of pendulum string

Physics 205A Quiz 6, spring semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 10.58

A 0.50 kg bob is suspended from a 1.6 m string, forming a pendulum. The period of this pendulum is measured. If the mass of the bob is decreased to 0.25 kg while the length of the string is increased to 3.2 m, the period will change by a factor of:
(A) 1.0.
(B) 1.4.
(C) 2.0.
(D) 4.0.

Correct answer: (B)

For a simple pendulum, the period T is given by:

T = 2·π·sqrt(L/g),

where g is the gravitational acceleration constant, and L is the length of the string attached to the pendulum bob, assumed to be an ideal point mass. Increasing the length by a factor of two will increase the period by a factor of sqrt(2.0) = 1.4.

Student responses
Sections 30880, 30881
(A) : 4 students
(B) : 29 students
(C) : 5 students
(D) : 2 students

Success level: 72%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.50

20090428

Physics quiz question: kinetic energy of SHM object at equilibrium

Physics 205A Quiz 6, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Multiple-Choice Questions 10.11-10.20

[Version 1]
[3.0 points.] A graph of kinetic energy versus time for an object in SHM is shown at right. What is the earliest time that the object will be located at equilibrium?
(A) 0 s.
(B) 1 s.
(C) 2 s.
(D) 3 s.

Correct answer: (A)

An object in simple harmonic motion will be momentarily stationary (and thus have zero kinetic energy) at its x = +/- A amplitude endpoints, and have its fastest speeds as it passes through equilibrium (x = 0). Since the kinetic energy graph already has it maximum value at t = 0 s, that is the earliest time that the object is at equilibrium.

Student responses
Sections 30880, 30881
(A) : 14 students
(B) : 0 students
(C) : 7 students
(D) : 0 students

[Version 2]
[3.0 points.] A graph of kinetic energy versus time for an object in SHM is shown at right. What is the earliest time that the object will be located at equilibrium?
(A) 0 s.
(B) 1 s.
(C) 2 s.
(D) 3 s.

Correct answer: (C)

An object in simple harmonic motion will be momentarily stationary (and thus have zero kinetic energy) at its x = +/- A amplitude endpoints, and have its fastest speeds as it passes through equilibrium (x = 0). Since the kinetic energy graph does not have its maximum value until t = 2 s, that is the earliest time that the object is at equilibrium.

Student responses
Sections 30880, 30881
(A) : 4 students
(B) : 1 student
(C) : 14 students
(D) : 0 students

"Difficulty level": 70%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.69

20090427

Physics clicker question: standing wave on tightened string

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 11.46

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the end of their learning cycle:

A guitar's A-string of a length 0.80 m and is stretched to a tension of 40 N. It vibrates at a fundamental frequency of 220 Hz. If the tension in the string is increased by a factor of 2.0, the fundamental frequency of this string increases by a factor of:
(A) 1.4.
(B) 2.0.
(C) 4.0.
(D) (The fundamental frequency remains the same.)
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 19 students
(B) : 7 students
(C) : 4 students
(D) : 0 students
(E) : 1 student

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 21 students
(B) : 1 student
(C) : 3 students
(D) : 0 students
(E) : 0 students

Correct answer: (A)

The speed v of transverse waves along a string is:

v = sqrt(F/mu),

where F is the tension in the string, while mu is the linear mass density (mass per unit length) of the string.

The fundamental frequency f_1 for a standing wave on a string is given by:

f_1 = v/(2*L),

where L is the length of the rope. Thus doubling the tension F would increase the wave speed v by a factor of sqrt(2) = 1.4, and thus the fundamental frequency f_1 would also increase by the same 1.4 factor.

Pre- to post- peer-interaction gains:
pre-interaction correct = 61%
post-interaction correct = 84%
Hake, or normalized gain = 59%

20090426

Astronomy quiz question: globular cluster star age

Astronomy 210 Quiz 6, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

The ages of globular cluster stars in the Milky Way halo are determined from:
(A) parallax.
(B) the turn-off point on an H-R diagram.
(C) width of their absorption spectra lines.
(D) their colors.

Section 30676
(A) : 8 students
(B) : 13 students
(C) : 19 students
(D) : 14 students

Correct answer: (B)

The most massive stars left on the main sequence line on an H-R diagram gives the age of the star cluster, because all the stars started their formation at the same time, and the more massive stars have shorter lifetimes (and leave the main sequence line faster) than the less massive stars. Response (A) would determine the distance to stars; response (C) would give an indication of their densities/luminosity class; response (D) would be correlated with the surface temperatures of these stars.

"Difficulty level": 31% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.06

20090425

Physics clicker question: changing tension of a rope wave

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

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

Students were asked the following clicker question (Classroom Performance System, einstruction.com) in the middle of their learning cycle:

Cord tension F = 75 N
Wave speed v = 140 m/s
Frequency f = 20 Hz (from a vibrating source)

Which wave parameter(s) increase if the cord were made more taut?
(A) Wave speed v.
(B) Wavelength lambda.
(C) Period T.
(D) (More than one of above choices.)
(E) (None of above choices.)
(F) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 7 students
(B) : 5 students
(C) : 2 students
(D) : 19 students
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 4 students
(B) : 0 students
(C) : 0 students
(D) : 27 students
(E) : 0 students

Correct answer: (D)

The frequency f of a wave is determined solely by the properties of the source, while the wave speed v is determined solely by the properties of the medium (thus frequency and wave speed are said to be the independent wave parameters). For rope waves:

v = sqrt(F/mu),

where F is the tension in the rope, while mu is the linear mass density (mass per unit length) of the rope.

The wavelength is the dependent wave parameter that is determined by both the source and medium, given by:

lambda = v/f or v*T,

where T is the period (which is the inverse of frequency). So if tension F increases, the wave speed v will increase (while the frequency remains constant), and also the wavelength will increase as result of the increase in v.

Pre- to post- peer-interaction gains:
pre-interaction correct = 58%
post-interaction correct = 87%
Hake, or normalized gain = 70%

20090424

Physics clicker question: changing frequency of a wave source

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

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

Students were asked the following clicker question (Classroom Performance System, einstruction.com) in the middle of their learning cycle:

A wave has a frequency and wavelength of 200 Hz and 0.75 m, respectively. If the frequency of the wave source is changed, which wave parameters(s) will change as a result?
(A The wavelength.
(B) The speed of the wave.
(C) (Both of the above choices.)
(D) (None of the above choices.)
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 20 students
(B) : 4 students
(C) : 9 students
(D) : 1 students
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 25 students
(B) : 2 students
(C) : 6 students
(D) : 0 students
(E) : 0 students

Correct answer: (A)

The frequency f of a wave is determined solely by the properties of the source, while the wave speed v is determined solely by the properties of the medium (thus frequency and wave speed are said to be the independent wave parameters). The wavelength is the dependent wave parameter that is determined by both the source and medium, given by:

lambda = v/f or v*T,

where T is the period (which is the inverse of frequency). So if frequency changes, the wave speed will still remain constant, but the wavelength will as a result change.

Pre- to post- peer-interaction gains:
pre-interaction correct = 59%
post-interaction correct = 76%
Hake, or normalized gain = 41%

20090423

Astronomy quiz question: distant galaxies and lookback time

Astronomy 210 Quiz 6, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

As seen from the Milky Way, a distant galaxy will:
(A) not appear to contain dark matter.
(B) not have absorption lines that are redshifted.
(C) have type Ia supernovae that explode slowly.
(D) appear as it did in its past.

Section 30674
(A) : 2 students
(B) : 3 students
(C) : 2 students
(D) : 25 students

Correct answer: (D)

Due to the finite speed of light, a galaxy that is n light years away will appear to us as it did n years in the past, when that light first left that galaxy.

"Difficulty level": 79% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.31

20090422

Astronomy quiz question: expansion, not explosion

Astronomy 210 Quiz 6, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

__________ is evidence that the universe is not expanding like an explosion.
(A) The finite speed of light.
(B) The night sky is dark, and not blindingly bright.
(C) Galaxy redshifts are proportional to galaxy distances.
(D) Matter and antimatter annihilate to convert into energy.

Section 30674
(A) : 5 students
(B) : 3 students
(C) : 19 students
(D) : 3 students

Correct answer: (C)

In an explosion, debris travels slower as it gets farther from the explosion, which is the opposite of what is actually observed, which is Hubble's law (response (C). Response (A) is the cause of lookback time, and response (B) is Olber's explanation of why the universe cannot be infinite in extent.

A similar version of this question was asked on a Final Exam in Spring 2008.

"Difficulty level": 63% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.53

20090421

Overheard: radioactive element counting

Astronomy 210, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

(Overheard in lecture during an in-class activity on comparing ratios of radioactive and daughter elements in samples to determine relative radioactive ages.)

Instructor: (Seeing a student quickly counting up all elements on the worksheet) "Look at you--getting all Rain Man on this."

Student 1: "I am an excellent driver."

(Beat.)

Student 2: "You so did not just quote that movie, did you?"

20090420

Physics clicker question: maximum SHM speed

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 10.30

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the end of their learning cycle:

A 0.40 kg object on a 99 N/m spring oscillates left to right on a frictionless surface, with an amplitude of 0.10 m. What is the speed of the object at the equilibrium point?
(A) 0 m/s.
(B) 0.40 m/s.
(C) 1.57 m/s.
(D) 24.8 m/s.
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 2 students
(B) : 14 students
(C) : 9 students
(D) : 3 students
(E) : 4 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 1 student
(B) : 19 students
(C) : 0 students
(D) : 0 students
(E) : 5 students

Correct answer: (C)

From energy conservation, the potential energy of the mass when the spring is fully compressed or stretched at its maximum displacement from equilibrium x = +/- A can be set equal to the kinetic energy of the mass as it passes through equilibrium at x = 0:

U_max = K_max,
(1/2)*k*A^2 = (1/2)*m*v_max^2,

such that v_max = sqrt(k/m). Response (B) is the period T = 2*pi*sqrt(m/k) of this mass-spring system, and (D) is the maximum acceleration a_max = (k/m*)*A.

Pre- to post- peer-interaction gains:
pre-interaction correct = 28%
post-interaction correct = 0%
Hake, or normalized gain = -39% (!)

20090413

Overheard: I see dead stars

Astronomy 210, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

(Overheard in lecture during an in-class activity on main sequence lifetimes and lookback times.)

Student: "...so can we still see these stars, even though they're already dead?"

Instructor: "What? You want to watch the stars when they die?"

(Beat.)

Student: "Well...yeah."

Instructor: "You are sick."

20090412

Physics quiz question: horizontal, narrowing pipe

Physics 205A Quiz 5, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.50

[3.0 points.] Water at point [1] flows with a speed of 0.75 m/s through a pipe of 0.20 m inner radius. The pipe at point [2] tapers down to an inner radius of 0.050 m. Assume ideal fluid flow. The pressure of the water __________ as it flows from point [1] to point [2].
(A) decreases.
(B) increases.
(C) remains constant.
(D) (Not enough information is given.)

Correct answer: (A)

From applying the continuity equation:

A_1*v_1 = A_2*v_2,

the water speed at point [2] is greater than at point [1], because the cross-sectional area at point [2] is smaller than at point [1].

Then from Bernoulli's equation:

0 = delta(P) + (1/2)*rho*delta(v^2) + rho*g*delta(y),

the kinetic head (the second term on the right-hand side) increases due to the increase in speed from point [1] to point [2], while the gravitational head (the third term on the right-hand side) remains constant due to no change in elevation along the horizontal pipe, thus the pressure must decrease as water flows from point [1] to point [2].

Student responses
Sections 30880, 30881
(A) : 22 students
(B) : 16 students
(C) : 1 student
(D) : 0 students

"Difficulty level": 56%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.65

20090411

Physics quiz question: tipping a square plate

Physics 205A Quiz 5, spring semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Multiple-Choice Question 8.1

A Physics 205A student would like to turn a heavy square plate clockwise around a pivot by pushing or pulling on it using different forces separately. The force on the square plate that would produce the largest magnitude torque is:
(A) FA (applied horizontally at corner diagonal to the pivot).
(B) FB (applied horizontally at corner above the pivot).
(C) FC (applied upwards at corner left of pivot).
(D) (There is a tie.)
(E) (Not enough information is given to determine this.)

Correct answer (highlight to unhide): (D)

All three applied forces have the same magnitude. Also, they all have the same perpendicular lever arm ℓ, drawn from the pivot to intercept their respective lines of forces perpendicularly. Thus for each of the three forces, they have the same magnitude torque τ = ℓ·F.

Student responses
Sections 30880, 30881
(A) : 1 student
(B) : 13 students
(C) : 8 students
(D) : 17 students
(E) : 0 students

Success level: 43%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46

20090410

Physics quiz question: two- versus four-bladed fan

Physics 205A Quiz 5, spring semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 8.5

(Version 1)
[3.0 points.] Four uniform rods with identical masses, and equal lengths are arranged as a long two-bladed fan, or as a shorter four-bladed fan, as shown at right. Which fan blade arrangement has the largest rotational inertia?
(A) The two-bladed fan.
(B) The four-bladed fan.
(C) (There is a tie.)
(D) (Not enough information is given to determine this.)

Correct answer: (A)

For a thin, uniform rod of mass M and length L, students were given that the rotational inertia about an axis through its center, perpendicular to the length is given by Icenter = (1/12)·M·L2; while the rotational inertia about an axis at one end, perpendicular to the length is given by Iend = (1/3)·M·L2.

If each individual rod segment has mass m and length l, then the rotational inertia of the two-bladed fan is:

Itwo-blade = 2·Iend (where M = 2·m, L = 2·l) = 2·(1/3)·(2·m)·(2·l)2 = (16/3)·m·l2.

This is larger than the rotational inertia of the four-bladed fan:

Ifour-blade = 4·Iend (where M = m, L = l) = 4·(1/3)·m·l2 = (4/3)·m·l2.

(The four-bladed fan rotational inertia could also be derived from:

Ifour-blade = 2·Icenter (where M = 2·m, L = 2·l) = 2·(1/12)·(2·m)·(2·l)2 = (4/3)·m·l2.)

Student responses
Sections 30880, 30881
(A) : 13 students
(B) : 1 student
(C) : 6 students
(D) : 0 students

(Version 2)
[3.0 points.] Four uniform rods with identical masses, and equal lengths are arranged as a long two-bladed fan, or as a shorter four-bladed fan, as shown at right. Which fan blade arrangement has the smallest rotational inertia?
(A) The two-bladed fan.
(B) The four-bladed fan.
(C) (There is a tie.)
(D) (Not enough information is given to determine this.)

Correct answer: (B)

Student responses
Sections 30880, 30881
(A) : 5 students
(B) : 12 students
(C) : 3 students
(D) : 0 students

"Difficulty level": 64%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.29

20090409

Astronomy clicker question: star cluster lookback time and evolution rates

Astronomy 210, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the beginning of their learning cycle:


Which star's light takes the shortest time to travel to us?
(A) O5 star.
(B) B0 star.
(C) A5 star.
(D) (There is a tie.)
(E) (I'm lost, and don't know how to answer this.)

Section 30674 (pre-)
(A) : 12 students
(B) : 2 students
(C) : 1 student
(D) : 6 students
(E) : 3 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Section 30674 (post-)
(A) : 10 students
(B) : 0 students
(C) : 0 students
(D) : 13 students
(E) : 1 student

Correct answer: (B)

Since these stars are in the same cluster, and are all located at the same distance away, light from all three stars will take the same amount of time to travel to us.

Pre- to post- peer-interaction gains:
pre-interaction correct = 25%
post-interaction correct = 54%
Hake (normalized) gain <g> = 39%

[Follow-up question.]
Which star's supernova explosion (type II or type Ia) was/will be observed last by us?
(A) O5 star.
(B) B0 star.
(C) A5 star.
(D) (There is a tie.)
(E) (I'm lost, and don't know how to answer this.)

Section 30674 (pre-)
(A) : 1 students
(B) : 2 students
(C) : 18 students
(D) : 1 students
(E) : 3 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Section 30674 (post-)
(A) : 2 students
(B) : 0 students
(C) : 21 students
(D) : 0 students
(E) : 2 students

Correct answer: (C)

Massive stars will have shorter main sequence lifetimes, while low-mass stars will have longer main sequence lifetimes. Since these stars are located in the same cluster, and thus were born at the same time, and are located the same distance away from us, the lowest-mass star will be seen to end its main sequence lifetime last.

Pre- to post- peer-interaction gains:
pre-interaction correct = 72%
post-interaction correct = 84%
Hake (normalized) gain <g> = 43%

20090408

Physics clicker question: partially- and fully-submerged objects

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

(Adapted from Fig. 2.1, p. 11, Mazur, Peer Instruction: A User's Manual, Prentice-Hall, Inc., 1997)

Students were asked the following clicker question (Classroom Performance System, einstruction.com) at the end of their learning cycle:

Two objects (same volumes, same densities) are held either half- or fully-submerged in a tank of water. It is not known whether these two objects are both lighter or denser than water.

Which object has the greatest bouyant force exerted on it?
(A) Object X.
(B) Object Y.
(C) (There is a tie.)
(D) (Depends on whether both these two objects are lighter or denser than water.)
(E) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 5 students
(B) : 25 students
(C) : 2 students
(D) : 1 student
(E) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 1 student
(B) : 30 students
(C) : 1 student
(D) : 1 student
(E) : 0 students

Correct answer: (A)

The bouyant force on an object is given by:

F_B = rho*g*V,

where rho is the density of the fluid, and V is the volume displaced by the object--the volume of the object that is submerged under the surface of the liquid. Thus regardless of the density of the objects, and whether the hands are lifting or pushing down on these objects, the submerged object experiences a greater bouyant force than the partially submerged object.

Pre- to post- peer-interaction gains:
pre-interaction correct = 76%
post-interaction correct = 91%
Hake, or normalized gain = 63%

20090407

Found physics: Speculative Enceladus visualizations

20090403068
http://www.flickr.com/photos/waiferx/3421648711/
Originally uploaded by Waifer X

"Surface of Enceladus (Speculative)" (hoarfrost on a car trunk, Atascadero, CA). Photo by Cuesta College Physical Sciences Division instruction Dr. Patrick M. Len.


20090404075
http://www.flickr.com/photos/waiferx/3421648887/
Originally uploaded by Waifer X

"Enceladan Life (Speculative)" (live crab at Pier 46 seafood market, Templeton, CA). Photo by Cuesta College Physical Sciences Division instruction Dr. Patrick M. Len.

20090406

Central Coast Astronomical Society monthly meeting: CubeSat presentation

20090326050
http://www.flickr.com/photos/waiferx/3422446706/
Originally uploaded by Waifer X

CubeSat on display, at the March 2009 meeting of the Central Coast Astronomical Society, Cuesta College, San Luis Obispo, CA. Photo by Cuesta College Physical Sciences Division instruction Dr. Patrick M. Len.


20090326052
http://www.flickr.com/photos/waiferx/3422446964/
Originally uploaded by Waifer X

Central Coast Astronomical Society members and bystanders inspect a CubeSat P-Pod payload deployer, with a single CubeSat on display on the tabletop, brought in by Cal Poly student Marissa Brummitt (visible second from left). Photo by Cuesta College Physical Sciences Division instruction Dr. Patrick M. Len.


20090326053
http://www.flickr.com/photos/waiferx/3421640967/
Originally uploaded by Waifer X

Central Coast Astronomical Society members and bystanders inspect a CubeSat P-Pod payload deployer, with a single CubeSat on display on the tabletop, brought in by Cal Poly student Marissa Brummitt (fourth from left). Photo by Cuesta College Physical Sciences Division instruction Dr. Patrick M. Len.


20090326054
http://www.flickr.com/photos/waiferx/3422447406/
Originally uploaded by Waifer X

Central Coast Astronomical Society members and bystanders inspect a CubeSat, under the watchful eye of Cal Poly student Marissa Brummitt (left). Photo by Cuesta College Physical Sciences Division instruction Dr. Patrick M. Len.

20090326055
http://www.flickr.com/photos/waiferx/3421641339/
Originally uploaded by Waifer X

Cal Poly student Marissa Brummitt displays a CubeSat at a Central Coast Astronomical Meeting. Photo by Cuesta College Physical Sciences Division instruction Dr. Patrick M. Len.

20090405

Found physics: sphygmomanometer

20090324048
http://www.flickr.com/photos/waiferx/3422440296/
Originally uploaded by Waifer X

Baumanometer® blood pressure gauge (sphygmomanometer), La Posada Medical Plaza, Templeton, CA. Photo by Cuesta College Physical Sciences Division instruction Dr. Patrick M. Len.

20090404

Physics clicker question: bag of chips, from sea level to the mountains

Physics 205A, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Conceptual Question 9.9

Gas Bag
Posted by witch532
http://www.youtube.com/watch?v=wRFZXy-999I

Students were asked the following clicker questions (Classroom Performance System, einstruction.com) at the start of their learning cycle:

An unopened bag of potato chips is purchased at sea level, and then is brought up on a car trip into the mountains. As a result, the bag "inflates" and becomes rigid. This is because of a(n):
(A) increase in pressure inside the bag.
(B) decrease in pressure outside the bag.
(C) (Both of the above choices.)
(D) (I'm lost, and don't know how to answer this.)

Sections 30880, 30881
(A) : 5 students
(B) : 22 students
(C) : 5 students
(D) : 0 students

This question was asked again after displaying the tallied results with the lack of consensus, with the following results. No comments were made by the instructor, in order to see if students were going to be able to discuss and determine the correct answer among themselves.

Sections 30880, 30881
(A) : 0 students
(B) : 31 students
(C) : 3 students
(D) : 0 students

Correct answer: (B)

The principal cause of the bag "inflating" is the decrease in pressure in the air surrounding the bag as it is brought to a higher elevation. The pressure inside the bag then is greater than the ambient pressure outside. The absolute pressure inside the bag could increase due to an increase in temperature (using the ideal gas law), which is what some students discussed in choosing (C).

Pre- to post- peer-interaction gains:
pre-interaction correct = 69%
post-interaction correct = 91%
Hake, or normalized gain <g> = 72%

20090403

Astronomy quiz question: possible/impossible star clusters

Astronomy 210 Quiz 5, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

[Version 1]
The H-R diagrams of three possible or impossible star clusters are shown below.

How many of these star cluster H-R diagram(s) is/are possible? (Do not circle the H-R diagrams; clearly circle a response below.)
(A) Only one.
(B) Two.
(C) Three.
(D) (None of the star clusters.)

Section 30676
(A) : 7 students
(B) : 6 students
(C) : 5 students
(D) : 0 students

Correct answer: (B) (Star clusters X and Y)

Massive stars will evolve the quickest, reaching the upper part of the main sequence line first, staying there the shortest amount of time, and then leaving first, compared to the low-mass stars, which evolve the slowest, reaching the lower part of the main sequence line last, and then staying there...and remain there, as the universe has not been around long enough for low-mass stars to have exhausted their main-sequence lifetimes. Thus star cluster Z is impossible, as it still has massive stars remaining on the main-sequence at the same time that the low-mass stars are on the main sequence.

[Version 2]
The H-R diagrams of three possible or impossible star clusters are shown below.

How many of these star cluster H-R diagram(s) is/are possible? (Do not circle the H-R diagrams; clearly circle a response below.)
(A) Only one.
(B) Two.
(C) Three.
(D) (None of the star clusters.)

Section 30676
(A) : 11 students
(B) : 17 students
(C) : 1 student
(D) : 0 students

Correct answer: (C) (All star clusters are possible)

"Difficulty level": 42% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.33

20090402

Astronomy quiz question: star cluster age

Astronomy 210 Quiz 5, Spring Semester 2009
Cuesta College, San Luis Obispo, CA

The H-R diagrams of three star clusters are shown below.

[Version 1]
Which star cluster is the youngest? (Do not circle the H-R diagrams; clearly circle a response below.)
(A) Star cluster X.
(B) Star cluster Y.
(C) Star cluster Z.
(D) (There is a tie.)

Section 30674
(A) : 7 students
(B) : 6 students
(C) : 5 students
(D) : 0 students

Correct answer: (A) (Star cluster X)

Massive stars will evolve the quickest, reaching the upper part of the main sequence line first, staying there the shortest amount of time, and then leaving first, compared to the low-mass stars, which evolve the slowest, reaching the lower part of the main sequence line last, and then staying there...and remain there, as the universe has not been around long enough for low-mass stars to have exhausted their main-sequence lifetimes. Thus star cluster X is younger than both star clusters Y and Z, which have massive stars that already left the main-sequence, and low-mass stars that already reached the main sequence.

[Version 2]
Which star cluster is the oldest? (Do not circle the H-R diagrams; clearly circle a response below.)
(A) Star cluster X.
(B) Star cluster Y.
(C) Star cluster Z.
(D) (There is a tie.)

Section 30674
(A) : 0 students
(B) : 12 students
(C) : 5 students
(D) : 0 students

Correct answer: (B) (Star cluster Y)

"Difficulty level": 56% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.69

20090401

Found physics: choose an appropriate coordinate system

090302-1080356 ("Choose An Appropriate Coordinate System")
http://www.flickr.com/photos/waiferx/3347461833/
Originally uploaded by Waifer X

Costco fuel station pavement arrow, San Luis Obispo, CA. To eventually accompany a planned photo, "Resolve Diagonal Vectors Into Orthogonal Components"). Photo by Cuesta College Physical Sciences Division instructor Dr. Patrick M. Len.