Sunday, January 27, 2008
January 25, 08 -Last scribe for physics!
Physics exam:
Monday, January 28 /08
Gym, 1-4
For Grade 12 Physics.
Thursday, January 17, 2008
Today...
Physics exam is coming soon, so study lots!
Tuesday, January 15, 2008
Today in Physics....
Reminder that the exam is on the 28th in the afternoon. STUDY STUDY STUDY!
The scribe for tomorrow is...sandy!
The, super late friday scribe
We went over some work sheets, the Chapter 25 study guide, and we were assigned the end of the chapter questions.
Exam is in the Afternoon, Jan 28. 1pm.
Test for this unit is on Wednesday, Jan 16.
Next scribe... Steven.
Thursday, January 10, 2008
Wednesday, January 9, 2008
Wednesday, January 8, 2008
Florjeval can scribe for tomorrow.
Monday, January 7, 2008
Scribe #4
The next scribe is going to be Renan.
Friday, January 4, 2008
Hey guys! (and Mrs. Kozoriz!) Did you have a great winter break? I finally got this post up. I have been procrastinating on everything. I had so much time that I didn't know when to actually do this post. I didn't realize that we go back to school on Monday until today. Pretty sad... I hope everyone's all pumped up for the new year even though it means GOING BACK TO SCHOOL. No more sleeping in - BOOO!! I hope you guys are happy with all the presents you got from your loved ones. However, we should turn our switches back on to let the current flow in our circuits because guess what? MORE CIRCUITS when we get back! HAHAHA! Anyway...
Here are the answers for those who didn't take them down last year.
1.a)

b) Resistancetotal = 10Ω + 10Ω + 10Ω = 30Ω (The Greek letter, Ω, represents OHMS, which is the unit for resistance.)
c)
(The unit for current, I, is A or amperes.)
d) Potential Drop = 0.4A x 10Ω = 4V (The unit for potential drop is volts, or V.)
e) Current = 12V / 30Ω = 0.4A
(Note that the total current circulating a SERIES CIRCUIT is the same as the current that goes through each of the resistor.)
f) Power = 4V x 0.4A = 1.6W (Watt is the unit for power.)
c) Currenttotal = 12V / 3.3Ω = 3.6A
d) Potential Drop = 12 V (In a parallel circuit, each resistor has the same potential drop.)
e) Current in each resistor = 3.6A / 3 = 1.2A
f) Power = 1.2A x 12V = 14.4W
3) Series circuit = 2.4 watts; Parallel circuit = 11 watts
4.a) 1440Ω
b) 0.16A
REVIEW:
- Electric Potential - is the electric potential energy per unit of charge
- E = qV
- Electric Potential Difference (voltage) - ∆V
*** It is necessary to determine the total or equivalent resistance of the circuit before other elctrical quantities can be calculated. It is also important to identify the resistors that are in series with each other and those that are parallel with each other.
*** Add the resistors to reduce the number of resistors in the circuit. Begin by looking at the resistors furthest from the source (battery).
- Kirchoff's Current Law - the sum of the currents entering a junction must equal the sum of the currents leaving the same junction
- Kirchoff's Voltage Law - the sum of the potential drops around a circuit must euqal the sum of potential rises around the circuit
- Factors that affect resistance :
- Length - the longer the conductor, the greater the resistance
- Cross Sectional Area- the larger the cross sectional area, the less resistant it is to charge flow
- Temperature - the greater the molecular motion, at higher temperatures, the greater the resistance (not for all substances)
- Type of Material - some materials are better conductors than others.
- RESISTIVITY - general measure of the resistance of a substance, has units Ωm
- where R is the resistance in ohms, L is the length in meters, A is the cross-sectional area, in meters2 and ρ is the resistivity in Ωm2/ m
- Five steps to follow to successfully analyze a compound circuit:
- Each set of parallel conductors must be replaced by a single equivalent resistance
- Each series resistance must be combined to find the total.
- Total current in the circuit can be found using the equation: I = V / R
- The voltage across each series group is found using: V = IR; to check, the total voltage must equal the sum of the individual voltage
- The current in each branch of the parallel conductors can be found using: I = V / R; to check, the total current must equal the sum of individual currents
Oh and I remember now. We got more worksheets for this unit. Make sure you do them! Be careful because they are tricky. You can easily get confused and mix up the rules for series and parallel circuits. See you guys on January 7, 2008. Lastly, the next scribe is DINO!
HAPPY NEW YEAR TO ALL!
HAVE A GREAT 2008!!!
Friday, December 21, 2007
Physics 40S 3rd Period
Next Scribe for Friday: Vincent!!!
Wednesday, December 19, 2007
Scribe Post
We also were given a lab to do which didn't go all so well because most of us had a hard time understanding exactly what it was asking. I'm guessing that tomorrow we will probably be going over the lab.
Also the test for Electric Circuits has been moved to the second week that we come back from holidays; it will be joined to our next unit as one big test. This happened after lots of complaining of the lack of knowledge in this unit.
That was basically all of today's class, and the tomorrow's scribe will be .........Hall of Fame!
Hope everyone enjoyed the festive colours=)
Tuesday, December 18, 2007
Electric Circuits

1 - December 17 - Scribe
That basically sums up the class. The next scribe is kim.
Saturday, December 15, 2007
Thursday, Dec 13 and Friday, Dec 14
Next is 1
Wednesday, December 12, 2007
Scribe Post
The next scribe is 111111111 !!!!!!! what an awesome name.
magnetic fields and current
1) an electron is forced in the opposite direction of a proton.
2) the charge must be moving. A magnetic field will not influence the motion of a charged particle at rest.
3) the velocity of the moving charge must be perpendicular to that of the direction of the magnetic field.
here are easy steps to determine force direction.*use this for proton
1) open your hand. Now point your fingers in the direction of the magnetic field.
2)rotate your hand so that your thumb points in the direction of the current.
3) look at your palm. the direction your palm is facing is the direction of the force acting upon the charge.
*use for electron
do steps 1-3 above but use your left hand! amazing it works! Because the electron acts opposite to the proton the left hand will work. Founded By Professor Cadonic.
next scribe is oliver
Monday, December 10, 2007
Voltage
For electric potential:
1. "Just as PE (potential energy) transforms to KE (kinetic energy) for a mass lifted against the gravitational field, the electric PE of an electric charge transforms to other forms of energy when it changes location in an electric field. When released, how does the KE acquired by each compare to the decrease in PE?"
The answer: equal (since both require work to move an object while still under the influence of the field whether it's gravitational or electric. Once work has been applied some potential energy has been transformed into kinetic energy in both cases.)
2. "Similarly, a force pushes the charge (called a test charge) closer to a charged sphere. The work done in moving the test charge is the product of the average force and the distance moved. W = FD. This work is equal to the PE of the test charge. If the test charge is released, it will be repelled and fly past the starting point. Its gain in KE at this point is equal to its decrease in PE." (note: terms in bold font are the answers since it was fill in the blanks.)
3. "complete the statements.
-Electric PE/charge has the special name Electric potential.
-Since it is measured in volts it is commonly called voltage."
4. "When a charge of 1 C has an electric PE of 1 J, it has an electric potential of 1 V. When a charge of 2 C has an electric PE of 2 J, its potential is = 1 V." (This is due to the fact that electric potential is the electric potential energy divided by the charge, as stated above in question 3)
5. "If a conductor connected to the terminal of a battery has a potential of 12 volts, then each coulomb of charge on the conductor has a PE of 12."
6. "If a charge of 1 C has a PE of 5000 J, its voltage is 5000 V."
7. "If a charge of 0.001 C has a PE of 5 J, its voltage is 5000 V."
8. "If a charge of 0.0001 C has a PE of 0.5 J, its voltage is 5000 V."
9. "If a rubber balloon is charged to 5000 V, and the quantity of charge on the balloon is 1 millionth coulomb (0.000001 C) then the PE of this charge is only 0.005 J.
10. "Some people get mixed up between force and pressure. Recall that pressure is force per area. Similarly, some people get mixed up between electric PE and voltage. According to this chapter, voltage is electric PE per unit charge."
Next, we moved on to a lab which consumed the majority of today's class time. The lab was essentially the assembly of a figure that embodied the main idea of increasing potential energy within an electric field (raising a charge AKA applying work to raise the charge against the electric field). The procedure and materials required for the lab are approximately similar to the following (since I don't have a green book with me right now):
Materials:
A Ruler
Clay
Steel balls (Any object can substitute for this, such as pennies in this case)
Masking tape
Procedure:
1. Insert a ruler within a molded ball of clay so that the ruler can stand upright without much independent mobility.
2. Take a 2 cm x 8 cm piece of tape and label it "3 V - 3 J/C" and repeat this process for three more equal sized pieces of tape with the labels "6 V - 6 J/C", "9 V - 9 J/C" and "12 V - 12 J/C."
3. Attach these pieces of tape to the ruler approximately at each 3 cm interval indicated on the ruler.
4. Once each piece of tape is attached to the ruler, attach the steel balls (or pennies) to each piece of tape, starting with 4 at the bottom on the 3V piece of tape, then with 3 on the 6V, and so on.
Results
Basically, we were given questions accompanying the lab as usual, and this was our overall analysis for the lab. Now essentially for the lab there was a structure which indicated that the amount of charges decreased as the voltage increased while you move farther up. If you think about the surface the structure is resting on in terms of the base of the electric field, you move the charges upwards and the voltage increases while the charges decrease. This also implicates an energy change, which can be calculated by multiplying the voltage with the amount of charges present at that level. So for this lab, the following data table was completed as the lab progressed:
"1. How much energy is required to lift each coulomb of charge from the tabletop to the 9-V level?"
The energy required to raise one coulomb of charge from the tabletop to the 9-V level is 9 J of energy. This is due to the fact that there are 2 charges at a voltage of 9 J/C accumulatively giving an energy value of 18 J, but since the question only inquires about the energy of one charge this 18 J is divided by 2 to give 9.
"2. What is the total potential energy stored in the 9-V level?"
(Note that this is basically explaining the calculations made in the table for each energy value)The total potential energy at a given location when you raise the object further up against the electric field can be found by using the electric potential of the object / charge. Since V = PE/q, and both q and V are known, solving for PE gives PE = V*q, PE = (2 C)*(9 J/C) = 18 J.
"3. The total energy of the charges in the 6-V level is not 6 J. Explain this"
The total energy of the charges in this level is not equivalent to 6 J since there are 3 charges present at a voltage of 6 J/C, rather than just one charge at a voltage of 6 J/C. Since there are 3 charges at a voltage of 6 J/C giving an overall energy for the level as 18 J.
"4. How much energy would be given off if the charges in the 9-V level fell to the 6-V level? Explain"
Now since the charges are dropping to a lower voltage at 6 J/C from 9 J/C, an energy equivalent to 3 joules would be emitted from the system since the charges are falling from 9 to 6. * not sure about this one since I can't remember what I actually put and am not in the mentality to reason out the answers momentarily.
Application question
"1. A 9-V battery is very small. A 12-V car battery is very big. Use your model to help explain why two 9-V batteries would not start your car."
Now if you think about it, the more charges that are present then the higher the energy value for that level. Now since these two cases involve voltages that are quite similar, there are obviously much more charges found in the large car battery than in the small 9-V battery, therefore allowing for quite a bit more energy available for use by applying the battery correctly.
Once all the labs were in, Ms. K went over the answers to the other sheet we were given on friday. The answers to the questions are as follows:
1. V = 31 V
2. a] PE = 3.6 x 10-14 J
2. b] V = 180 V
3. 1.9 x 10 7 m/s
4. 3.2 x 10 -9 J
5. a] w = 1.92 x 10 -18 J = 12 eV
5. b] E = 1.92 x 10 -18 J = 12 eV
5. c] V = 2.0 x 10 6 m/s
Sunday, December 9, 2007
Scribe
We had a dose of the Mars Mission from Oliver...
and a presentation on telescopes from Sergio.
After we moved on to a couple of the sheets that Kizoriz handed out.
Well that pretty much sums it up.
Have a good weekend.
The next scribe will be...
Mr SiwWy
PS. Afternoon class worked on the Charges, Voltage, Energy lab from the green book.
Make sure you read the notes on potential, potential energy, potential difference.
Note the difference among them (pardon the pun :))
Thursday, December 6, 2007
More Presentations...
Remember, all groups should hand in a self evaluation which includes a mark out of 10 for Presentation, a mark out of 40 for the total Project, and explanations as to why they deserve such a mark.
That's all for today... I have to work now, so I'll see you all tomorrow where we will finish the remaining presentations and GREY-M will be scribing about them =D
Wednesday, December 5, 2007
Scribing time!
Tuesday, December 4, 2007
Scriiiibe
In today's class, we were assigned page 423 questions 7-12 in the green book! YAY! We basically had the whole class to go over that, so I assume we all should have finished! We also viewed Kim's lovely powerpoint presentation on Mars today because she will not be in class tomorrow. Our presentations will take place tomorrow, so hopefully you guys all finished! We also got two sheets titled Electrical Forces and Coulomb's Law. Since we have our presentations taking place tomorrow we should be finished these sheets by Thursday! The scribe for tomorrow will be *KASIA*! okay have fun and have a good night everyone! =)


