Monday, April 6, 2009
All Juniors Scheduling
I have received word from a guidance counselor that Level V Honors courses are not available to seniors. You may take only CP or AP Level V. Tomorrow, someone can let me know what Level V means
Sunday, April 5, 2009
HonorsThermo quiz Quiz
Here is the link for the Thermo quiz. Assignment is as posted yesterday.
http://h1.ripway.com/DrCherdack/HnrsthermoquizB09r1.doc
http://h1.ripway.com/DrCherdack/HnrsthermoquizB09r1.doc
Saturday, April 4, 2009
AP Geometrical Optics Unit 14 Schedule
Here are new schedule and worksheets. Note the quiz next Thursday. The notes and summary for geometrical optics were posted March 30. I've posted the modern physics stuff you'll need later here.I also threw a couple of separate derivations you can look over.
http://h1.ripway.com/DrCherdack/APUnit14BLightsched-09.doc
http://h1.ripway.com/DrCherdack/Unit14problemsheetr1.doc
http://h1.ripway.com/DrCherdack/ModernPhysicsReview.doc
http://h1.ripway.com/DrCherdack/ModernPhysAPSummary08.doc
http://h1.ripway.com/DrCherdack/Derivlenslawsphmirr.doc
http://h1.ripway.com/DrCherdack/Derivoffforsphmirr.doc
http://h1.ripway.com/DrCherdack/APUnit14BLightsched-09.doc
http://h1.ripway.com/DrCherdack/Unit14problemsheetr1.doc
http://h1.ripway.com/DrCherdack/ModernPhysicsReview.doc
http://h1.ripway.com/DrCherdack/ModernPhysAPSummary08.doc
http://h1.ripway.com/DrCherdack/Derivlenslawsphmirr.doc
http://h1.ripway.com/DrCherdack/Derivoffforsphmirr.doc
Honors Thermo Quiz
Here is last Wednesday's quiz. Correct the version you took and then take the other version and turn in that version showing all work done neatly on Monday
Friday, April 3, 2009
All Students including Topics in Physics
You first assignment for the new quarter is the following. Print out the following message and return it signed.
All assignments must be turned in no later than the end of a unit, or in the case of labs, no more than three school days after the completion of the lab. No credit for any late work (other than submission delayed by absence ) will be granted.
_______________
Student
_______________
Parent or Guardian
All assignments must be turned in no later than the end of a unit, or in the case of labs, no more than three school days after the completion of the lab. No credit for any late work (other than submission delayed by absence ) will be granted.
_______________
Student
_______________
Parent or Guardian
Thursday, April 2, 2009
AP Problem sheet
Problem number 10 seems to have a problem. The answer given assumes no phase shift from the reflection of the ray on the soap water interface. I think this is wrong. If Nwater>Nsoap>air both reflections result in a phase shift and so for a destructive interference 2 Nsoap x thickness = lambda/2. This gives lambda longer than the maximum for visible light. If one looks for the next destructive interfernce, it would require 2 Nsoap x thickness = 3 x lambda/2. This puts lambda at the edge or beow the visble light limit.
There may also be a problem for the oil drop problem,#13. I'll try to check it tonight.
There may also be a problem for the oil drop problem,#13. I'll try to check it tonight.
Group Reports
Well, I got one reminder to publish the list of group reports from about 80 students.
Here it is. These reports are due tomorrow. They are REQUIRED.
http://h1.ripway.com/DrCherdack/Grpreports4-2-09qrtr3.xls
Here it is. These reports are due tomorrow. They are REQUIRED.
http://h1.ripway.com/DrCherdack/Grpreports4-2-09qrtr3.xls
Wednesday, April 1, 2009
AP Three important items
1) I post a lot items so don't think seeing one is seeing 'em all. Scroll down to older posts.
2) Make sure you go to the gridding fest in the auditorium lecture halls on Tuesday.
3) Remind me to discuss polarization tomorrow.
2) Make sure you go to the gridding fest in the auditorium lecture halls on Tuesday.
3) Remind me to discuss polarization tomorrow.
Thermal Energy conservation lab
I expect, or at least, would like, a brief but formal writeup showing that you know you were trying to demonstrate the conservation of thermal energy. You did this by collecting data and using it in a conservation based equation to determine the specific heat of your sample. If the specific heat came out close to the standard value, it confirms that your equation which was based on conservation of thermal energy was correct.
You should definitely include both the original equation relating loss of energy of the meatl to gain of energy by water/container AND the rearranged equation for cmetal = etc.
Standard values: brass about 390 J/ (K kg); iron 440 J/ (K kg); Aluminum 920 J/ (K kg)
You should definitely include both the original equation relating loss of energy of the meatl to gain of energy by water/container AND the rearranged equation for cmetal = etc.
Standard values: brass about 390 J/ (K kg); iron 440 J/ (K kg); Aluminum 920 J/ (K kg)
Honors period 6 thermo quiz
I haven't graded it yet, but I'm guessing the results will not be great. Since there were some serious typos on the test, I will allow a mass retest on Thursday if I receive enough emails to warrant it.
AP Quiz
After some soul searching (always a difficult exercise for those of us without one), I have decided to postpone the quiz on wave behavior of light until Friday. It will include gratings and you will have to know something about the electromagnetic spectrum and dispersion. Dispersion is the fact that light of different frequencies have different indices of refraction ( travel at different speeds in matter) so they bend different amounts, thus rainbows and some of the reults from your Snel's law lab.
AP currents labs
A large number of you are laboring under serious misconceptions. Currents flow in circuits because fields are created in conductors which cause the electrons present in the conductors to gain small average velocities. The amount of current that flows depends on the field and on how many electrons are available to move and how easily they can move. The field , in turn, is determined by how much voltage difference is placed across the conductor. Thus the amount of current that flows depends on the nature and geometry of the conductor and the voltage across it. The total current in a circuit is DEFINTELY NOT some fixed quantity, it IS dependent on the number of conductors available to carry it.
Batteries impose a voltage difference across a circuit. The current arises in response to this voltage difference and, once again, depends on the number and quality of the conductors experiencing the voltage. By removing a parallel resistor you have removed a conductor (resistors are just imperfect conductors) and thus reduced the total current. If the voltage remains the same, each remaining parallel elements will carry its own original current. If the voltage available to these elements rises, as it sometimes does because the lower total current reduces voltage losses elsewhere in the circuit, there will be an increase in the current in the remaining elements, but this is NOT because the current from the removed element has to find somewhere to go. The current that was flowing in the removed element simply ceases to exist because the electrons that were flowing in that element are no longer in the circuit.
Batteries impose a voltage difference across a circuit. The current arises in response to this voltage difference and, once again, depends on the number and quality of the conductors experiencing the voltage. By removing a parallel resistor you have removed a conductor (resistors are just imperfect conductors) and thus reduced the total current. If the voltage remains the same, each remaining parallel elements will carry its own original current. If the voltage available to these elements rises, as it sometimes does because the lower total current reduces voltage losses elsewhere in the circuit, there will be an increase in the current in the remaining elements, but this is NOT because the current from the removed element has to find somewhere to go. The current that was flowing in the removed element simply ceases to exist because the electrons that were flowing in that element are no longer in the circuit.
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