On thursday's class we did a test.
Happy Holidays
Eric is the next scribe
Sunday, December 21, 2008
Wednesday, December 17, 2008
Saving the class from Eric's memory loss :P
Since "someone" forgot to blog again, I decided to blog for today. This class we went over our "Moving Charges" worksheet from question 11-16. We then went over the assigned questions from chapter 27's review worksheet. We also received a sheet with some short questions on the path of a charged particle in a magnetic field.
We can use the right hand rule to determine the path of the charged particle(positive or negative) with the following:
-For a positive charge, the palm of the hand will be pointing towards the magnetic force or away(on the paper) depending on the marking of the diagram(x is on to the page and o is away from the page). The thumb points in the direction of the moving particle and then wherever the fingers point, that is the direction the field will pull the charged particle to.
-On the other hand, if it is a negatively charged particle, the right hand rule will not apply so the negatively charged particle will be pulled in the opposite direction of the fingers.
Tomorrow is our unit test that will include everything(if Ms. K was telling the truth lol) we learned about electric and magnetic fields. Ms. K uploaded some of the derivations and equate smartboard notes for an extra study tool if anybody needs.
Richard blogs next.
We can use the right hand rule to determine the path of the charged particle(positive or negative) with the following:
-For a positive charge, the palm of the hand will be pointing towards the magnetic force or away(on the paper) depending on the marking of the diagram(x is on to the page and o is away from the page). The thumb points in the direction of the moving particle and then wherever the fingers point, that is the direction the field will pull the charged particle to.
-On the other hand, if it is a negatively charged particle, the right hand rule will not apply so the negatively charged particle will be pulled in the opposite direction of the fingers.
Tomorrow is our unit test that will include everything(if Ms. K was telling the truth lol) we learned about electric and magnetic fields. Ms. K uploaded some of the derivations and equate smartboard notes for an extra study tool if anybody needs.
Richard blogs next.
Tuesday, December 16, 2008
Moving Charges Worksheet
Scribe for Tuesday, December 16, 2008 since Eric was kidnapped by Santa Claus and his evil green elves and hasn't scribed since.
AGENDA:
* Corrected Moving Charges Worksheet #1-10
* Read Magnetic Formulae handout
* Chapter 27 Review #1-7 for homework
The Moving Charges Worksheet was difficult to do at first, which is why Ms.K suggested on doing the questions in reverse order. Then it’s just a matter of knowing how to use your formulas correctly. Also, visualizing the problem (i.e. drawing a picture) helps a lot too.
MOVING CHARGES WORKSHEET
10. An electron is moving east to west at 5.0 x 10^5 m/s through a magnetic field. A force of 4.0 x 10-6 N north is acting on the electron. What is the magnitude and direction of the magnetic field?
F = Bqv
(4.0 x 10^6) = B (1.6 x 10^-19) (5.0 x 10^8)
B = 5 x 10^7 (into page)
9. What is the magnitude and direction of a magnetic force on a proton moving horizontally to the north at 8.6 x 10^4 m/s, as it enters a magnetic field of 1.2 T pointing vertically upward?
F = Bqv = (1.2 N/(c*m/s)) (1.6 x 10^19 C) (8.6 x 10^4 m/s) = 1.7 x 10^014 [right]
8. An electron experiences the greatest force by a magnetic field as it travels at 2.1 x 10^5 m/s when it is moving south. The force is 5.6 x 10^-13 N and up. Find the magnitude and direction of the magnetic field.
F = Bqv
(5.6 x 10^-13)/ [(1.6 x 10^-29) (2.1 x 10^5)] = B
B = 16.7 T [left]
7. A 10.0 m long high tension power line carries a current of 20.0 A perpendicular to the earth’s magnetic field of 5.5 x 10^-5 T. What is the magnetic force experienced by the power line?
F = BIL = (5.5 x 10^-5 T) (20.0 A) (10.0 m) = 0.011 N
6. A wire connecting a taillight to a motorcycle battery is 0.50 m long and is perpendicular to the Earth’s magnetic field. It experiences a force of 6.0 x 10^-5 N when carrying a current of 1.5 A. What is the magnitude of the Earth’s magnetic field at that location?
F = BIL
6.0 x 10^-5 N = B (1.5 A) (0.50 m)
B = 8 x 10^-5 T
5. A 15 cm length of wire carries a current of 20.0 A. It is perpendicular to a uniform magnetic field. If it experiences a magnetic force of 0.40 N, what is the magnetic field intensity?
B = F/(IL) = 0.4 N / (20.0 A x 0.15 m) = 0.13 T
4. Using the Millikan apparatus, a sphere of mass 3.2 x 10^-15 kg is held motionless. The plates produce an electric field strength of 19 600 N/C.
a) What forces are acting on the sphere?
electric force, gravitational force
b) What type of charge must be on the sphere?
positive
c) Determine the amount of charge on the sphere.
F_g = F_e
mg = Eq
q = mg/E = (3.2 x 10^-15 kg) (9.8 N/kg) / (19 600 N/C) = 1.6 x 10^-18 C
d) How many electrons does this correspond to?
(1.6 x 10^-18 C) x (1 electron)/(1.6 x 10^-19 C) = 10 electrons
3. The electric field between these charged parallel plates is 23 800 N/C. The drop’s mass is 2.4 x 10^-14 kg. What force would be experienced by the drop if it had the charge of three excess electrons?
F_e = Eq = (23 800 N/C) (3 x 1.6 x 10^-19 C) = 1.14 x 10^-14 N [up]
F_g = mg = (2.4 x 10^-14 kg) (9.8 N/kg) = 2.35 x 10^-13 N [down]
F_net = 2.35 x 10^-13 N - 1.14 x 10^-14 N = 2.24 x 10^-13 [down]
2. The plates are 5.0 cm apart and produce an electric field strength of 1200 N/C in the region between them. An electron enters the electric field at 1.00 x 10^2 m/s. With what velocity does the electron strike the positive plate? (Ignore gravity.)
F_e = qE = (-1.6 x 10^-19 C) (1200 N/C) = 1.92 x 10^-15 N
ma = qE
a = qE/m = (1.92 x 10^-15 N) / (9.11 x 10^-3 kg) = 2.11 x 10^14 m/s^2
v_f^2 = v_i^2 + 2ad
v_f = [(1.00 x 10^2 m/s)^2 + 2 (2.11 x 10^14 m/s^2) (0.05 m)]^0.5 = 4.6 x 10^6 m/s
1. This oil drop of mass 4.5 x 10^-15 kg has had two electrons removed. The electric field strength is 3.0 x 10^4 N/C.
a) What will happen to the drop?
F_g = mg = (4.5 x 10^-15 kg) (9.8 N/kg) = 4.4 x 10^-14 N
F_e = Eq = (3.0 x 10^4 N/C) (2 x 1.6 x 10^-19 C) = 9.5 x 10^-15 N
F_g > F_e, so drop will fall
b) What is the total force experienced by the drop?
F_net = 4.4 x 10^-14 N + 9.5 x 10^-15 N = 3.45 x 10^-14 N [down]
Next scribe is DANA.
AGENDA:
* Corrected Moving Charges Worksheet #1-10
* Read Magnetic Formulae handout
* Chapter 27 Review #1-7 for homework
The Moving Charges Worksheet was difficult to do at first, which is why Ms.K suggested on doing the questions in reverse order. Then it’s just a matter of knowing how to use your formulas correctly. Also, visualizing the problem (i.e. drawing a picture) helps a lot too.
MOVING CHARGES WORKSHEET
10. An electron is moving east to west at 5.0 x 10^5 m/s through a magnetic field. A force of 4.0 x 10-6 N north is acting on the electron. What is the magnitude and direction of the magnetic field?
F = Bqv
(4.0 x 10^6) = B (1.6 x 10^-19) (5.0 x 10^8)
B = 5 x 10^7 (into page)
9. What is the magnitude and direction of a magnetic force on a proton moving horizontally to the north at 8.6 x 10^4 m/s, as it enters a magnetic field of 1.2 T pointing vertically upward?
F = Bqv = (1.2 N/(c*m/s)) (1.6 x 10^19 C) (8.6 x 10^4 m/s) = 1.7 x 10^014 [right]
8. An electron experiences the greatest force by a magnetic field as it travels at 2.1 x 10^5 m/s when it is moving south. The force is 5.6 x 10^-13 N and up. Find the magnitude and direction of the magnetic field.
F = Bqv
(5.6 x 10^-13)/ [(1.6 x 10^-29) (2.1 x 10^5)] = B
B = 16.7 T [left]
7. A 10.0 m long high tension power line carries a current of 20.0 A perpendicular to the earth’s magnetic field of 5.5 x 10^-5 T. What is the magnetic force experienced by the power line?
F = BIL = (5.5 x 10^-5 T) (20.0 A) (10.0 m) = 0.011 N
6. A wire connecting a taillight to a motorcycle battery is 0.50 m long and is perpendicular to the Earth’s magnetic field. It experiences a force of 6.0 x 10^-5 N when carrying a current of 1.5 A. What is the magnitude of the Earth’s magnetic field at that location?
F = BIL
6.0 x 10^-5 N = B (1.5 A) (0.50 m)
B = 8 x 10^-5 T
5. A 15 cm length of wire carries a current of 20.0 A. It is perpendicular to a uniform magnetic field. If it experiences a magnetic force of 0.40 N, what is the magnetic field intensity?
B = F/(IL) = 0.4 N / (20.0 A x 0.15 m) = 0.13 T
4. Using the Millikan apparatus, a sphere of mass 3.2 x 10^-15 kg is held motionless. The plates produce an electric field strength of 19 600 N/C.
a) What forces are acting on the sphere?
electric force, gravitational force
b) What type of charge must be on the sphere?
positive
c) Determine the amount of charge on the sphere.
F_g = F_e
mg = Eq
q = mg/E = (3.2 x 10^-15 kg) (9.8 N/kg) / (19 600 N/C) = 1.6 x 10^-18 C
d) How many electrons does this correspond to?
(1.6 x 10^-18 C) x (1 electron)/(1.6 x 10^-19 C) = 10 electrons
3. The electric field between these charged parallel plates is 23 800 N/C. The drop’s mass is 2.4 x 10^-14 kg. What force would be experienced by the drop if it had the charge of three excess electrons?
F_e = Eq = (23 800 N/C) (3 x 1.6 x 10^-19 C) = 1.14 x 10^-14 N [up]
F_g = mg = (2.4 x 10^-14 kg) (9.8 N/kg) = 2.35 x 10^-13 N [down]
F_net = 2.35 x 10^-13 N - 1.14 x 10^-14 N = 2.24 x 10^-13 [down]
2. The plates are 5.0 cm apart and produce an electric field strength of 1200 N/C in the region between them. An electron enters the electric field at 1.00 x 10^2 m/s. With what velocity does the electron strike the positive plate? (Ignore gravity.)
F_e = qE = (-1.6 x 10^-19 C) (1200 N/C) = 1.92 x 10^-15 N
ma = qE
a = qE/m = (1.92 x 10^-15 N) / (9.11 x 10^-3 kg) = 2.11 x 10^14 m/s^2
v_f^2 = v_i^2 + 2ad
v_f = [(1.00 x 10^2 m/s)^2 + 2 (2.11 x 10^14 m/s^2) (0.05 m)]^0.5 = 4.6 x 10^6 m/s
1. This oil drop of mass 4.5 x 10^-15 kg has had two electrons removed. The electric field strength is 3.0 x 10^4 N/C.
a) What will happen to the drop?
F_g = mg = (4.5 x 10^-15 kg) (9.8 N/kg) = 4.4 x 10^-14 N
F_e = Eq = (3.0 x 10^4 N/C) (2 x 1.6 x 10^-19 C) = 9.5 x 10^-15 N
F_g > F_e, so drop will fall
b) What is the total force experienced by the drop?
F_net = 4.4 x 10^-14 N + 9.5 x 10^-15 N = 3.45 x 10^-14 N [down]
Next scribe is DANA.
Monday, December 15, 2008
4 Days to go!!!
So today was simple and straight forward.
We were given out a handout for Electric and Magnetic Current.
We learned about the direction of a Force on a Current Carrying Wire, and read further into that. We learned again about the right hand rules.
Two things we were told to explicitly remember about the motion of a charged particle in a magnetic field were:
1. The charge must be moving. A Magnetic field will not influence the motion of a charged particle at rest.
2. The velocity of the moving charge must have a component that is perpendicular to the direction of the magnetic field.
We were then told to work on the Moving Charges Worksheet. Questions 1 - 10.
A Tip from Ms.K. If you get stuck at one, start at number 10 and work backwards.
The next scribe will be Eric!
We were given out a handout for Electric and Magnetic Current.
We learned about the direction of a Force on a Current Carrying Wire, and read further into that. We learned again about the right hand rules.
Two things we were told to explicitly remember about the motion of a charged particle in a magnetic field were:
1. The charge must be moving. A Magnetic field will not influence the motion of a charged particle at rest.
2. The velocity of the moving charge must have a component that is perpendicular to the direction of the magnetic field.
We were then told to work on the Moving Charges Worksheet. Questions 1 - 10.
A Tip from Ms.K. If you get stuck at one, start at number 10 and work backwards.
The next scribe will be Eric!
Sunday, December 14, 2008
December 12, 2008
Class today was simple and easy. Walking into class and hearing that I'm the scribe however wasn't as fun as I thought. Anyways, in class we did the following :
1. Finished reading "Grade 12 Physics: Gravitational Potential Energy."
2. Was handed an assignment called "Cathode Ray" but was handed in for marks.
3. Also was handed a sheet called "Appendix 2 - Fields."
So anyways, I'll see you soon and I won the bet against Mrs. Kozoriz that we had. How I was going to "forget" to scribe.
Lawrence can scribe for Monday
1. Finished reading "Grade 12 Physics: Gravitational Potential Energy."
2. Was handed an assignment called "Cathode Ray" but was handed in for marks.
3. Also was handed a sheet called "Appendix 2 - Fields."
So anyways, I'll see you soon and I won the bet against Mrs. Kozoriz that we had. How I was going to "forget" to scribe.
Lawrence can scribe for Monday
Thursday, December 11, 2008
Charges, Energy, Voltage Lab
Answers to 33-2 Electrical Fields and Potential are found on slides 1 and 2.
Answers to Electric Potential are found on slides 3 to 7.
After reviewing our answers to those worksheets, we did a lab entitled Charges, Energy, Voltage from Chapter 21 of the green book.
In the lab, we developed a model that showed the different amounts of charges at different energy levels. We erected a ruler in a Playdoh substance and made markings at 3 cm, 6 cm, 9 cm, and 12 cm. The 3 cm resembled 3 V or 3 J/C, the 6 cm resembled 6 V or 6 J/C, etc. We taped 4 pennies at the 3 cm marking, 3 pennies at the 6 cm marking, 2 pennies at the 9 cm marking, and a penny at the 12 cm marking. Each penny resembled a Coulomb.
A lab worksheet was handed out and has to be completed.
Next scribe is Vieteran.
Answers to Electric Potential are found on slides 3 to 7.
After reviewing our answers to those worksheets, we did a lab entitled Charges, Energy, Voltage from Chapter 21 of the green book.
In the lab, we developed a model that showed the different amounts of charges at different energy levels. We erected a ruler in a Playdoh substance and made markings at 3 cm, 6 cm, 9 cm, and 12 cm. The 3 cm resembled 3 V or 3 J/C, the 6 cm resembled 6 V or 6 J/C, etc. We taped 4 pennies at the 3 cm marking, 3 pennies at the 6 cm marking, 2 pennies at the 9 cm marking, and a penny at the 12 cm marking. Each penny resembled a Coulomb.
A lab worksheet was handed out and has to be completed.
Next scribe is Vieteran.
Wednesday, December 10, 2008
Nothing much, just more printouts
This will probably be one of the shortest post in the blog, since all we did was read out from a new printout that Ms. K gave us (Gravitational Potential Energy) and work on a fun worksheet (the one with that cartoons) and a not-so fun worksheet.
And finally the next scribe will be Zeph
And finally the next scribe will be Zeph
Tuesday, December 9, 2008
Happy Birthday Richard and Eric! (:
Hello everyone, Rojuane here for another scribe.
Today we went over the test questions that we finished the Electric Force worksheet.
The answers are:
1) FE= 3.6x10^10 N
2. a) FG=7.8x10^-47 N
b) FE=8.8x10^-8 N
3. Ne 2.5x10^19 N
4. Q=1.76x10^8 C
5. a) FE 1.76x10^-15 N [towards -2x10^-12 C charge]
b)a=2.6x10^11 m/s^2
Then we were given questions and pages to read from the green textbook. If you didn't catch the page numbers they are:
pg. 415-419, 1-5 p. 420 (answers from back of textbook)
and questions 1,9,10,16 (Sorry, you have to do those on your own)
DON'T FORGET to hand in those four questions for marks for tomorrow class!
Next Scribe will be, SHELLY :)
Today we went over the test questions that we finished the Electric Force worksheet.
The answers are:
1) FE= 3.6x10^10 N
2. a) FG=7.8x10^-47 N
b) FE=8.8x10^-8 N
3. Ne 2.5x10^19 N
4. Q=1.76x10^8 C
5. a) FE 1.76x10^-15 N [towards -2x10^-12 C charge]
b)a=2.6x10^11 m/s^2
Then we were given questions and pages to read from the green textbook. If you didn't catch the page numbers they are:
pg. 415-419, 1-5 p. 420 (answers from back of textbook)
and questions 1,9,10,16 (Sorry, you have to do those on your own)
DON'T FORGET to hand in those four questions for marks for tomorrow class!
Next Scribe will be, SHELLY :)
Monday, December 8, 2008
Magnetic and Electrical Fields
Today in class, we were introduced to our new unit, Magnetic and Electrical Fields. At first we wrote down a couple notes such as the differences between the two different energy fields [gravitational and electric fields-not ENERGY fields]. We also sketched out some examples on the attractive and repulsive forces between 2 different charges in an electrical field. We were also given some worksheets and notes to read from. The notes given include Grade 12 Physics: Electric and Magnetic Fields, Electric Fields and Coulomb's Law in One Dimension. Our worksheets given were Coulomb's Law, which was a brief and easy worksheet about Coulomb's Law. We were also given the Electrical Forces worksheet, which was a tougher worksheet, the questions were mainly centered around finding the electrical force of the given problems. We only worked up to question 3, and we should be done by tomorrow. Our answers so far were:
1) 3.6 x 1010 N
2)a) 7.8 x 10-47 N
b) 8.8 x 10-8 N
3) 2.5 x 1019 electrons
That was pretty much it for today, our next scribe will be Rojuane. :D
1) 3.6 x 1010 N
2)a) 7.8 x 10-47 N
b) 8.8 x 10-8 N
3) 2.5 x 1019 electrons
That was pretty much it for today, our next scribe will be Rojuane. :D
Friday, December 5, 2008
Post Gravitational Test
Okay we had a test today. How'd everyone do? Well the reason why I'm writing this is because I would like to know which part(s) of the unit did everyone have trouble with? Like what Mr K says, what were your muddiest parts? Clearest parts? Toughest? Easiest? You can post your thoughts in the comments section.
My muddiest part was combining the different formulas and equations. I can understand them individually but when they come together I take a bit longer working with them. My clearest part was the Kepler's 3rd law work, the formula with the ratio between the radii and period of the orbit.
So once again you can put your thoughts in the comment section and maybe we can help each other Past The Exam...
My muddiest part was combining the different formulas and equations. I can understand them individually but when they come together I take a bit longer working with them. My clearest part was the Kepler's 3rd law work, the formula with the ratio between the radii and period of the orbit.
So once again you can put your thoughts in the comment section and maybe we can help each other Past The Exam...
Thursday, December 4, 2008
Review Before Test
Hi everyone. I'm going to be posting the review for tomorrow's test on gravitation and orbits.
REVIEW
Kepler's 3 Laws
For the orbital of planets the Sun is one focus point.
The areas created between 2 equal intervals of the planet's orbit are equal
Kepler's Formula
Newton's law of Universal Gravitation
Gravitational Potential Energy
Escape Velocity & g of a Planet or Sun & Orbital Velocity
Thanks Richard!
Newton's Thought Experiment
To stay in Earth's Orbit the satellite must have a velocity between 8km/s to 11.2km/s
Orbital Period
Microgravity
What is it?
-Environment with very little gravity
How can it be achieved?
-By gaining altitudes rapidly then suddenly decreasing altitude rapidly like during the NASA Microgravity Simulating Planes
Problems of Space Exploration
-Hazards
-Expense
That is all we had for a review. I might add more stuff if I find something but for now I'm getting some rest. I gotta study. Good luck tomorrow and the next scribe will be Francis.
REVIEW
Kepler's 3 Laws
For the orbital of planets the Sun is one focus point.
The areas created between 2 equal intervals of the planet's orbit are equal
Kepler's Formula
Newton's law of Universal Gravitation
Gravitational Potential Energy
Escape Velocity & g of a Planet or Sun & Orbital Velocity
Thanks Richard!
Newton's Thought Experiment
To stay in Earth's Orbit the satellite must have a velocity between 8km/s to 11.2km/s
Orbital Period
Microgravity
What is it?
-Environment with very little gravity
How can it be achieved?
-By gaining altitudes rapidly then suddenly decreasing altitude rapidly like during the NASA Microgravity Simulating Planes
Problems of Space Exploration
-Hazards
-Expense
That is all we had for a review. I might add more stuff if I find something but for now I'm getting some rest. I gotta study. Good luck tomorrow and the next scribe will be Francis.
Wednesday Dec 3, 2008
Sorry everyone i forgot to scribe yesterday. Today is Thursday, but am actually scribing for Wednesday. Hopefully no one gets confused.
This is the day where Richard & Eric presented their space exploration(mars). They were supposed to presented on Monday, however their video did not work. Their project was taped and cleverly done. The video contained humor, and rich information of the planet Mars.
Afterward, Mrs Kozoriz asked us to answer several questions from the review sheet. Here they are:
1. The radius of earth is about 6400 km. What would be the earth's gravitational attraction on 75-kg astronaut in an orbit 6400 k above the earth's surface?
6.67x10^-11(5.98x10^24)(75)/( 6400+64000)x2
Fg= 182 N
2. The Mass of Mars is about 6.6x10^23 kg. and the acceleration due to gravity is 3.7 m/s(2). What is the radius of Mars?
3. The Earth's radius 6400 km. A 25-kg mass is taken 20l km above the earth's surface.
a. What is the object's mass at this height?
- Mass does not change
- It's still 25-kg
b. What is the weight of the object at this height?
4. A sphere of mass 85 kg is 12 m from a second sphere of mass 65 kg.
a. What is the gravitational force of attraction between them?
b. What is the acceleration of the first sphere towards the second?
6. How many years would it take a planet located four times as far from the sun as the Earth to orbit the sun? The Earth's distance from the sun is 1.5x10^11 m.
Next Mrs Kozoriz went over the Gravitational fields worksheet. She only went over question # 7 & 8, but question 8 was not fully done. so i wont be answering it.
7. The planet Jupiter has a mass of 1.9x10^27 kg and a radius of 7.2x10^7 m. Calculate the acceleration due to gravity on Jupiter.
We've also watched a video about "Dark Matter" that took half the class. Well that was all for Wednesday.
We have a Test tomorrow so everyone study hard. This unit aint easy. Well then i gotta go and do ma math homework ... Pce
I've already told benjamen, so he's next to scribe. He might even be scribing now, who knows.
This is the day where Richard & Eric presented their space exploration(mars). They were supposed to presented on Monday, however their video did not work. Their project was taped and cleverly done. The video contained humor, and rich information of the planet Mars.
Afterward, Mrs Kozoriz asked us to answer several questions from the review sheet. Here they are:
1. The radius of earth is about 6400 km. What would be the earth's gravitational attraction on 75-kg astronaut in an orbit 6400 k above the earth's surface?
6.67x10^-11(5.98x10^24)(75)/( 6400+64000)x2Fg= 182 N
2. The Mass of Mars is about 6.6x10^23 kg. and the acceleration due to gravity is 3.7 m/s(2). What is the radius of Mars?
3. The Earth's radius 6400 km. A 25-kg mass is taken 20l km above the earth's surface.
a. What is the object's mass at this height?
- Mass does not change
- It's still 25-kg
b. What is the weight of the object at this height?
4. A sphere of mass 85 kg is 12 m from a second sphere of mass 65 kg.
a. What is the gravitational force of attraction between them?
b. What is the acceleration of the first sphere towards the second?
6. How many years would it take a planet located four times as far from the sun as the Earth to orbit the sun? The Earth's distance from the sun is 1.5x10^11 m.
Next Mrs Kozoriz went over the Gravitational fields worksheet. She only went over question # 7 & 8, but question 8 was not fully done. so i wont be answering it.
7. The planet Jupiter has a mass of 1.9x10^27 kg and a radius of 7.2x10^7 m. Calculate the acceleration due to gravity on Jupiter.
We've also watched a video about "Dark Matter" that took half the class. Well that was all for Wednesday.We have a Test tomorrow so everyone study hard. This unit aint easy. Well then i gotta go and do ma math homework ... Pce
I've already told benjamen, so he's next to scribe. He might even be scribing now, who knows.
Monday, December 1, 2008
Monday dec 1st,2008
Hey everyone this is Jab....back on the class routine.
Today's class was fine and buzy.
First thing in today's class Mr. Joseph and Ms Shelly presented a presentation about the exploration of mars, soon as they finished Mr.Yonas and his partner (that's me) we took on the stage and presented about the biggest moon called Titan.
At last we watched a movie on weighlessness which we had to pick up seven points.
Mr. YONAS is next to describe.
Today's class was fine and buzy.
First thing in today's class Mr. Joseph and Ms Shelly presented a presentation about the exploration of mars, soon as they finished Mr.Yonas and his partner (that's me) we took on the stage and presented about the biggest moon called Titan.
At last we watched a movie on weighlessness which we had to pick up seven points.
Mr. YONAS is next to describe.
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