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.
Wednesday, November 26, 2008
Wednesday November 26 2008
Today in class the first thing we did was correct the questions that was assigned yesterday. On page 172 in the green book, Questions 9 to 12
Question 9 Answer : F = 8.0 x 10 -10 N
Question 10 Answer : F = 6.49 x 10 -8 N
Question 11a Answer : Fg = 491.3 N
Question 11b Answer : w = 490 n
Question 12 Answer : m = 9.01 x 10 -31 kg
The Next thing that we did was Watch some of the old Space Exploration Projects.
After that we then were taught to derive some formulas.
For number 1 you cancel out one of teh masses on each side.
For number 2 you cancel out one of the masses and radius on each side.
For number 3 you just cancel out one of the masses on each side.
She also gave us a forth one but she handed out a handout for this one.
The Next thing that we did was a review sheet for chapter 8.
And that was todays class the next scribe will be Kamil
Question 9 Answer : F = 8.0 x 10 -10 N
Question 10 Answer : F = 6.49 x 10 -8 N
Question 11a Answer : Fg = 491.3 N
Question 11b Answer : w = 490 n
Question 12 Answer : m = 9.01 x 10 -31 kg
The Next thing that we did was Watch some of the old Space Exploration Projects.
After that we then were taught to derive some formulas.
For number 1 you cancel out one of teh masses on each side.
For number 2 you cancel out one of the masses and radius on each side.
For number 3 you just cancel out one of the masses on each side.
She also gave us a forth one but she handed out a handout for this one.
The Next thing that we did was a review sheet for chapter 8.
And that was todays class the next scribe will be Kamil
Tuesday, November 25, 2008
November 25, 2008
To start things off we had gone over the questions on the Activity Sheet on Artificial Satellites and the Concept-Development practice page 14-1.
Here are the answers to the following:
Activity Sheet on Artificial Satellites
1) When the projectile is launched to a speed of 8km/sec
2) If it travels less than 8km/sec
3) around 8km/sec - 11.2km/sec
4)Yes, due to the atmosphere drag
5) about 850km - 35 800km
6) Yes
7) Yes, collisions with air particles slowly act to circularize the orbit and slows down the satellite. This causes it to drop to lower altitudes.
Concept-Development 14-1
1. d) Circle
e) Ellipse
2.d) Yes, the forces are the same strengh and are at equal distance
e)Yes, there is no acceleration
f) 90 degrees
g) No, "F" and "V" are perpendicular
h) No work is done
i) remains constant
j) remains constant
Right after we corrected the questions from the green book on page 172, and learned that the first 5 questions involves using this equation
(T1/T2)2 = (R1/R2)3
Answers
1) 12 years
2) was to be handed in to Ms. K (answer: 19.2 Re)
3) 3.0724 Re
4) 22.63 years
5) 0.35Tm
6) 6.11 x 10-9N
7) 4.17 x 1023N
8) 5.84 x 10-10N
Finally on the same page we are to work on next 5 (I think) questions.
Homework:
Questions 9 - 13 on page 172, Green book
Space projects are going to be due very soon!
Next scribe will be Richard
Here are the answers to the following:
Activity Sheet on Artificial Satellites
1) When the projectile is launched to a speed of 8km/sec
2) If it travels less than 8km/sec
3) around 8km/sec - 11.2km/sec
4)Yes, due to the atmosphere drag
5) about 850km - 35 800km
6) Yes
7) Yes, collisions with air particles slowly act to circularize the orbit and slows down the satellite. This causes it to drop to lower altitudes.
Concept-Development 14-1
1. d) Circle
e) Ellipse
2.d) Yes, the forces are the same strengh and are at equal distance
e)Yes, there is no acceleration
f) 90 degrees
g) No, "F" and "V" are perpendicular
h) No work is done
i) remains constant
j) remains constant
Right after we corrected the questions from the green book on page 172, and learned that the first 5 questions involves using this equation
(T1/T2)2 = (R1/R2)3
Answers
1) 12 years
2) was to be handed in to Ms. K (answer: 19.2 Re)
3) 3.0724 Re
4) 22.63 years
5) 0.35Tm
6) 6.11 x 10-9N
7) 4.17 x 1023N
8) 5.84 x 10-10N
Finally on the same page we are to work on next 5 (I think) questions.
Homework:
Questions 9 - 13 on page 172, Green book
Space projects are going to be due very soon!
Next scribe will be Richard
Monday, November 24, 2008
Physics November. 24, 2008
Today in class we watched a video about the "same same" teacher and the topic about the video was about the relationship of energy, gravity, and speed required to orbit and escape the earth, and the 's gravitational field, the vectors associated with the path of flight/orbit, and the shape of the orbits known as ellipses. It also explained why the paths of orbits were ellipses and how Kepler didn't see this visualization. Lastly, it went over the potential and kinetic energy relationships of the orbits.
Handouts on the "escape speed", "falling moon", "artificial satellites", and a practice sheet were given out.
Shelly blogs next..
Handouts on the "escape speed", "falling moon", "artificial satellites", and a practice sheet were given out.
Shelly blogs next..
Friday, November 21, 2008
I See Heavenly Bodies ;-)
Hello Hello,
This is Benofschool and for the record I didn't beg for this position as today's scribe. Well today we had a substitute and that person shall stay nameless because I do not have her permission to display his/her name online. Anyways we were assigned gravitational orbit and gravitational force questions from the Physics Green Textbooks. The questions are found on page 172 and questions #1-8 and including 8. For reference you can look at the past worksheet with planetary constants like the radii, period, and mass or you can look at page 159. I'd prefer the worksheet because it has all of the numbers required to answer the questions.
Sorry for the short scribe, but this is counted as a scribe. =P
Next scribe is Dana, but if he doesn't do it the next scribe will be Richard...
Okay I'm off to have a studying filled weekend =D
This is Benofschool and for the record I didn't beg for this position as today's scribe. Well today we had a substitute and that person shall stay nameless because I do not have her permission to display his/her name online. Anyways we were assigned gravitational orbit and gravitational force questions from the Physics Green Textbooks. The questions are found on page 172 and questions #1-8 and including 8. For reference you can look at the past worksheet with planetary constants like the radii, period, and mass or you can look at page 159. I'd prefer the worksheet because it has all of the numbers required to answer the questions.
Sorry for the short scribe, but this is counted as a scribe. =P
Next scribe is Dana, but if he doesn't do it the next scribe will be Richard...
Okay I'm off to have a studying filled weekend =D
Thursday, November 20, 2008
November 20, 2008
Walking into class, having a great day so far. Lawrence gave me a wonderful present. To scribe for today, what a nice guy huh? (Sarcasm). Anyways, let's get going.
The first thing we did today was the GRAVITATIONAL FIELDS WORKSHEET handed to us by the substitute teacher we had on November 19, 2008. (Mrs. Karras)
Next we were handed back our Transparency Worksheet (10)
I got the answers right so here they are.
1. 2M, 2M has the greatest force.
2. 2F.
3. Equal.
4. M1, M2, 2D = 1/4 F.
5. F=1.40 x 10 ^- 8 N.
6. 1/9 F.
7. 3F.
8. 3/4.
9. Remains the same.
Near towards the end of the class, we were handed a "Gravitational Potential Energy" booklet. If you don't have this booklet, get it because some very useful formulas are found at the back of this booklet.
Using this booklet, we were given a assignment (not due or anything) to finish in class. It's called "Gravitational Potential Energy Questions." Sorry I can't really give out the answers due to me leaving my Physics stuff in my locker or I might have lost it, I wouldn't know. Let's end this post by saying the next scribe is Benchman who begged me in class to blog for the weekend.
Wednesday, November 19, 2008
Wednesday, November 19, 2008
So by the blog, apparently Dana was supposed to be scribe, but upon my arrival to class, (Which was delayed due to my Phys Ed. Class, which took place at Sargent Pool) apparently Tony was chosen to be scribe, and since it was his birthday (which for some reason acts as a power of Veto), it was passed on to me. Honestly now.
Anyways, We basically went over the worksheets Ms. K assigned us the day prior. Mrs. Karras was also substituting.
Answers:
Study Guide 8
21) Cavendish used a rod with a ball on it, suspended from a thin wire. It also included a Large ball, a support, a mirror and a light source.
22) Force of attraction between the large and small balls caused the rod to rotate. It stopped rotating only when the force required to twist the wire equaled the gravitational forces between the balls.
23) F = G(M1*M2)/d^2 --> G = F*d^2/M1*m2 --> (6.67 *10^-11N)(1m)^2/(1kg)(1kg) = 6.67*10^-11 N*M^2/kg^2
Moving on, we went over the Chapter 11: Gravitational Interactions worksheet.
So since we know that if either mass in the equation for the law of universal gravitation is doubled, the force is doubled. Also we know that if the distance doubles, the force is one fourth of the original force.
1) If both masses are doubled, the force quadruples.
2) If the masses are not changed, but the distance of separation is reduced to 1/2 to the original distance, the force gets quadrupled.
3) If the masses are not changed, but the distance of separation is reduced to 1/4 the original distance, the force is sixteen times as much.
4) If both masses are doubled, and the distance of separation is doubled, the force remains the same.
5) If one of the masses is doubled, the other remains unchanged, and the distance of separation is tripled, the force is 2/9 of the original force.
6) The force remains the same (Force would be multiplied by 9, but because the distance is multiplied by 3, the force would be reduced to 1/9th of the original, keeping the force the same)
We then went over the Universal Law of Gravitation worksheet.
1) 7.41 * 10 ^-14 N
2) 81.82kg (In my case)
b)801.8 N (In my case)
c)180 lbs (In my case)
3)
a) Doubled
b) One quarter times
c) Four Times
d) Sixteen times
4) 1.62 * 10 ^ 6m
We were then handed a transparency sheet, which was later handed the same day. We were also given a Gravitational Fields worksheet, that would be due tomorrow I assume.
Next scribe is Tony.
Anyways, We basically went over the worksheets Ms. K assigned us the day prior. Mrs. Karras was also substituting.
Answers:
Study Guide 8
21) Cavendish used a rod with a ball on it, suspended from a thin wire. It also included a Large ball, a support, a mirror and a light source.22) Force of attraction between the large and small balls caused the rod to rotate. It stopped rotating only when the force required to twist the wire equaled the gravitational forces between the balls.
23) F = G(M1*M2)/d^2 --> G = F*d^2/M1*m2 --> (6.67 *10^-11N)(1m)^2/(1kg)(1kg) = 6.67*10^-11 N*M^2/kg^2
Moving on, we went over the Chapter 11: Gravitational Interactions worksheet.
So since we know that if either mass in the equation for the law of universal gravitation is doubled, the force is doubled. Also we know that if the distance doubles, the force is one fourth of the original force.
1) If both masses are doubled, the force quadruples.
2) If the masses are not changed, but the distance of separation is reduced to 1/2 to the original distance, the force gets quadrupled.
3) If the masses are not changed, but the distance of separation is reduced to 1/4 the original distance, the force is sixteen times as much.
4) If both masses are doubled, and the distance of separation is doubled, the force remains the same.
5) If one of the masses is doubled, the other remains unchanged, and the distance of separation is tripled, the force is 2/9 of the original force.
6) The force remains the same (Force would be multiplied by 9, but because the distance is multiplied by 3, the force would be reduced to 1/9th of the original, keeping the force the same)
We then went over the Universal Law of Gravitation worksheet.
1) 7.41 * 10 ^-14 N
2) 81.82kg (In my case)
b)801.8 N (In my case)
c)180 lbs (In my case)
3)
a) Doubled
b) One quarter times
c) Four Times
d) Sixteen times
4) 1.62 * 10 ^ 6m
We were then handed a transparency sheet, which was later handed the same day. We were also given a Gravitational Fields worksheet, that would be due tomorrow I assume.
Next scribe is Tony.
Tuesday, November 18, 2008
Tuesday, November 18th, 2008
Today in class, we got some examples of what our big project should look like and then we went over a couple of sheets.
Here are the answers for the worksheets we corrected:
Chapter 8 Study Guide (Universal Gravitation):
Inversely
Square
F=G(M1 x M2)/R²
Force
R
Constant
Doubled
Halved
1/4
Chapter 8 Study Guide (Newton's Use of His Law of Universal Gravitation):
Second
Kepler's
Chapter 8 Study Guide (Weighing Earth):
Henry Cavendish
Attraction
Law of U.G.
6.67 x 10-11 N x M²/Kg²
Chapter 8 Study Guide Section 8.1: Motions in the Heavens and on Earth:
1. True
2. True
3. False
4. True
5. False (Sun)
6. False (Area)
7. False (Ratios)
8. False (1st and 2nd)
9. Equal
10. 1
11. Rio -> need to calculate you need
Tio -> T + R of another of Jupiter's moons
12. C
13. A
14. D
15. B
We were assigned questions 16 - 23 on the "Chapter 8 Study Guide Section 8.1: Motions in the Heavens and on Earth" sheet and we also got 2 worksheets to do: "Universal Law of Gravitation" and "Chapter 11: Gravitational Interaction".
That's all folks, next scribe is Dana.
Here are the answers for the worksheets we corrected:
Chapter 8 Study Guide (Universal Gravitation):
Inversely
Square
F=G(M1 x M2)/R²
Force
R
Constant
Doubled
Halved
1/4
Chapter 8 Study Guide (Newton's Use of His Law of Universal Gravitation):
Second
Kepler's
Chapter 8 Study Guide (Weighing Earth):
Henry Cavendish
Attraction
Law of U.G.
6.67 x 10-11 N x M²/Kg²
Chapter 8 Study Guide Section 8.1: Motions in the Heavens and on Earth:
1. True
2. True
3. False
4. True
5. False (Sun)
6. False (Area)
7. False (Ratios)
8. False (1st and 2nd)
9. Equal
10. 1
11. Rio -> need to calculate you need
Tio -> T + R of another of Jupiter's moons
12. C
13. A
14. D
15. B
We were assigned questions 16 - 23 on the "Chapter 8 Study Guide Section 8.1: Motions in the Heavens and on Earth" sheet and we also got 2 worksheets to do: "Universal Law of Gravitation" and "Chapter 11: Gravitational Interaction".
That's all folks, next scribe is Dana.
Monday, November 17, 2008
Kepler's Three Laws Videotape
No one was selected for scribe in physics, so I'm assuming today's scribe post is up for grabs.
I am zeph. zeph is scribe. Today's scribe post has been zephed!
Answers to "8.1 Motion in the Heavens and on Earth: Kepler's Laws of Planetary Motion" and "Kepler's Three Laws Videotape" are given below.
---
8.1 MOTION IN THE HEAVENS AND ON EARTH
Kepler's Laws of Planetary Motion
Tycho Brahe studied the motion of the planets in order to be able to PREDICT astronomical events. He believed that EARTH was the centre of the universe. Johannes Kepler believed that THE SUN was the centre of the universe. He analyzed Brahe's data, and developed THREE laws of planetary motion. One law says that the paths of the planets are ELLIPSES with THE SUN located at one focus. Another law states that an imaginary line extending from the sun to a planet will sweep out equal AREAS in equal amounts of time. According to this law, planets move FASTEST when closest to the sun and move SLOWEST when farthest from the sun. Kepler's last law states that the ratio of the squares of the PERIODS of any two planets in orbit around the sun is equal to the ratio of the CUBES of their distances from THE SUN. This law can be states as an equation, (T_A/T_B)^2 = (R_A/R_B)^2. To use this law to calculate the period of a satellite, you must know the RADIUS of its orbit and the PERIOD and RADIUS of the orbit of another satellite.
---
KEPLER'S THREE LAWS VIDEOTAPE
Answer the following questions based on the videotape entitled "Kepler's Three Laws".
1. What planetary model did Johannes Kepler believe in?
Copernican
2. What planet's orbit did Kepler analyze?
Mars
3. How far off from a circle was the oribt of Mars?
8 min arc or 1/4th of the apparent width of the moon
4. What type of curve can explain the orbit of Mars?
ellipse
5. In an ellipse, is the total distance from 1 pin to the string and then to the other pin a constant?
yes
6. The name given to a single point in an ellipse is the __________.
focus point
7. Focus is the Latin word for __________.
fireplace
8. How would you describe Kepler's childhood?
poverty and illness
9. What journey did Kepler undertake in 1600 and for what purpose?
went to see Tycho Brahe in Denmark to learn more astronomy
10. What was a main interest of Tycho Brahe?
astronomy
11. How did Kepler obtain the careful observations that Brahe made during his lifetime?
stole it
12. Was the Copernican model of the universe easy to justify scientifically?
no
13. What were some of the problems that faced Kepler in discovering the secrets of the sky?
planets are in motion in its orbit and rotates on axis
14. How many pages of calculations did Kepler have?
about 900 pages
15. What does "in opposition" mean according to the positions of Mars, Earth, and sun?
Mars is in the same position as the Earth and the sun
16. Can any circle viewed obliquely be termed an ellispe?
yes
17. In ancient history, which group of people studied conic sections?
Greeks
18. Is a parabola a popular shape in our world today?
yes
19. In your own words, describe he term eccentricity?
how flat an ellipse is
20. State Kepler's 3 laws.
a) 1st: Each planet moves in an ellipse with the sun as its focus.
b) 2nd: A line from the sun to each planet sweeps out equal areas in equal times.
c) 3rd: T^2 directly proportional to R^3
---
HOUSEKEEPING
Homework:
* 8.1 Motion in the Heavens and on Earth Study Guide
* Exploration of Space Project due Thursday, November 27.
Next scribe is ERICT.
I am zeph. zeph is scribe. Today's scribe post has been zephed!
Answers to "8.1 Motion in the Heavens and on Earth: Kepler's Laws of Planetary Motion" and "Kepler's Three Laws Videotape" are given below.
---
8.1 MOTION IN THE HEAVENS AND ON EARTH
Kepler's Laws of Planetary Motion
Tycho Brahe studied the motion of the planets in order to be able to PREDICT astronomical events. He believed that EARTH was the centre of the universe. Johannes Kepler believed that THE SUN was the centre of the universe. He analyzed Brahe's data, and developed THREE laws of planetary motion. One law says that the paths of the planets are ELLIPSES with THE SUN located at one focus. Another law states that an imaginary line extending from the sun to a planet will sweep out equal AREAS in equal amounts of time. According to this law, planets move FASTEST when closest to the sun and move SLOWEST when farthest from the sun. Kepler's last law states that the ratio of the squares of the PERIODS of any two planets in orbit around the sun is equal to the ratio of the CUBES of their distances from THE SUN. This law can be states as an equation, (T_A/T_B)^2 = (R_A/R_B)^2. To use this law to calculate the period of a satellite, you must know the RADIUS of its orbit and the PERIOD and RADIUS of the orbit of another satellite.
---
KEPLER'S THREE LAWS VIDEOTAPE
Answer the following questions based on the videotape entitled "Kepler's Three Laws".
1. What planetary model did Johannes Kepler believe in?
Copernican
2. What planet's orbit did Kepler analyze?
Mars
3. How far off from a circle was the oribt of Mars?
8 min arc or 1/4th of the apparent width of the moon
4. What type of curve can explain the orbit of Mars?
ellipse
5. In an ellipse, is the total distance from 1 pin to the string and then to the other pin a constant?
yes
6. The name given to a single point in an ellipse is the __________.
focus point
7. Focus is the Latin word for __________.
fireplace
8. How would you describe Kepler's childhood?
poverty and illness
9. What journey did Kepler undertake in 1600 and for what purpose?
went to see Tycho Brahe in Denmark to learn more astronomy
10. What was a main interest of Tycho Brahe?
astronomy
11. How did Kepler obtain the careful observations that Brahe made during his lifetime?
stole it
12. Was the Copernican model of the universe easy to justify scientifically?
no
13. What were some of the problems that faced Kepler in discovering the secrets of the sky?
planets are in motion in its orbit and rotates on axis
14. How many pages of calculations did Kepler have?
about 900 pages
15. What does "in opposition" mean according to the positions of Mars, Earth, and sun?
Mars is in the same position as the Earth and the sun
16. Can any circle viewed obliquely be termed an ellispe?
yes
17. In ancient history, which group of people studied conic sections?
Greeks
18. Is a parabola a popular shape in our world today?
yes
19. In your own words, describe he term eccentricity?
how flat an ellipse is
20. State Kepler's 3 laws.
a) 1st: Each planet moves in an ellipse with the sun as its focus.
b) 2nd: A line from the sun to each planet sweeps out equal areas in equal times.
c) 3rd: T^2 directly proportional to R^3
---
HOUSEKEEPING
Homework:
* 8.1 Motion in the Heavens and on Earth Study Guide
* Exploration of Space Project due Thursday, November 27.
Next scribe is ERICT.
Saturday, November 15, 2008
Friday, November 14, 2008
Wednesday 12, 2008
On Wednesday we actually didn't do much.Mrs.K gave a review on circular motion and work and energy. Which took half of the class, afterwords Mrs.k gave us a brief article about space.Also we were given this assignment where you have to write about a historical technology or a particular planet, indicating whether its inhabitable or not, and such. The assignment is due on Nov 27, and a presentation is required.
Tuesday, November 11, 2008
Nov. 10
Today in physics we had a Remembrance day assembly, yeah.
Well, we also had physics class.
Today in physics we had a substitute teacher, named Mrs. Karras, she's pretty cool.
Anyways, we didn't do much at all in physics class, we just corrected the questions we got over the weekend. They were p.213-214 questions: 1-5, 9-12, 17-20 and p.237-238 questions: 1-4,12-14, 20-21.
The next scribe will be Yonas, cause he said we were absent.. and we fell for it. Life sucks.
Well, we also had physics class.
Today in physics we had a substitute teacher, named Mrs. Karras, she's pretty cool.
Anyways, we didn't do much at all in physics class, we just corrected the questions we got over the weekend. They were p.213-214 questions: 1-5, 9-12, 17-20 and p.237-238 questions: 1-4,12-14, 20-21.
The next scribe will be Yonas, cause he said we were absent.. and we fell for it. Life sucks.
Friday, November 7, 2008
Energy Transfer & Hooke's law
For today's class, we were given some time to work on the worksheet called "Energy Transfer" and and we also went over them. And for anyone who didnt get the answers here they are.
- Potential Energy and Kenetic Energy are the two types of energy that occurs in the spring- block system.
- PE is the energy position and KE is the energy due to motion.
- The kinetic energy is at a maximum when the distance = 0 , because thats when speed is the greatest.
- The potential energy is at a maximum when the speed = 0.
- As the potential energy increases the kinetic energy decreases. And this is an example of a closed system.
- The sum of the KE and PE stays constant. This demonstrate the Law of Conservation of Energy.
- The PE and KE equal to 4 times during one cycle.
- No, this cycle will not repeat forever because the displacement decreases.
- If the cycle stops the energy goes to the board.
- If the spring were vertical gravity would've affected the box which would result in an open system.
After going over the answers, we were given the rest of the time to work on our Lab from the class before. And we were also given question to do from the green book pages 213-214 questions number 1-5, 9-12, 17-20 ; pages 237-238 questions number 1-4, 12-14, and 20-21. the questions are DUE ON MONDAY.
AND DONT FORGET THAT THERE WILL BE A TEST ON THURSDAY, SO DONT FORGET TO STUDY !
And the next scribe is FRANCIS.
Thursday, November 6, 2008
Ugly Weather, Winter's Coming... Let's Scribe!
Hello :)
So today's class was very straight forward, we were given a lab to do related to Hooke's Law. We had to find the relationship between the extension in an elastic spring and the applied force. The materials that were needed for the apparatus were a stand, ring clamp, elastic metal spring, and a meter stick. We had to attach the spring to the ring clamp and adjust the spring so the lower end is on the zero point of the meter stick, or at a constant spot during the observations. Then we had to make a chart similar to the one on the paper like so:
Then we had to hang mass on the spring, and record the extension of the spring. We had to repeat the previous step for each of the mass given. Then we had to plot the graph (question 6 on the paper) and answer questions 7-11.
Don't forget to answer the Practice questions 1-5 shown on the back of the lab paper. The answers are given just show your work and how you got those answers.
Make sure to hand in yesterday Wind-up Toy Lab that was due today.
Today's lab 10.1: Hooke's Law is due TOMORROW!
No school Tuesday (Remembrance Day) ,
TEST on Wednesday so.. STUDY PEOPLE!
That's all for today's class,
your next scribe is christine d.
Bye :)
So today's class was very straight forward, we were given a lab to do related to Hooke's Law. We had to find the relationship between the extension in an elastic spring and the applied force. The materials that were needed for the apparatus were a stand, ring clamp, elastic metal spring, and a meter stick. We had to attach the spring to the ring clamp and adjust the spring so the lower end is on the zero point of the meter stick, or at a constant spot during the observations. Then we had to make a chart similar to the one on the paper like so:
Then we had to hang mass on the spring, and record the extension of the spring. We had to repeat the previous step for each of the mass given. Then we had to plot the graph (question 6 on the paper) and answer questions 7-11.
Don't forget to answer the Practice questions 1-5 shown on the back of the lab paper. The answers are given just show your work and how you got those answers.
Make sure to hand in yesterday Wind-up Toy Lab that was due today.
Today's lab 10.1: Hooke's Law is due TOMORROW!
No school Tuesday (Remembrance Day) ,
TEST on Wednesday so.. STUDY PEOPLE!
That's all for today's class,
your next scribe is christine d.
Bye :)
Wednesday, November 5, 2008
Scribe Time
Hi this is benofschool, and I'll be your scribe for today...
We started off correcting chapter 11.2 Study Guide. So here are the answers:
enter, leave, external, conservation of energy, form, constant, created, destroyed, mechanical, decrease, simple harmonic, potential, kinetic, potential, kinetic, increases , decreases , friction , thermal
Here are the answers for the second paragraph:
shape, potential, kinetic, all, conserved, some, other forms, conserved
Now we corrected a couple of worksheets from the beginning of the week.
Here are the answers for Concept- Development Practice Page: Chapter 8: Energy
1)
W= Fd
W= 200N (4m)
W= 800J
2)
P = W/t
P = [200N(4m)]/4s
P = 200 watts
3)
P=W/t
P= 6000 watts or 6kw
4) 16nore (IGNORE :P)
5) 10 m/s of all of the cases. Apply trajectory formula to answer this question.
6) Block A wil reach the bottom of the incline first because of the incline's shorter distance.
7) 100J, 75J, 25J
8) 10 m/s for Points B, D, and E. Max Speed at Point C
9) Yes there will be a change in wind speed because the windmills will take some of the wind's kinetic energy.
That was it for that work sheet. There was one more. Diagram Skills Section 5-3.
1a) 0 J
b) Ep = mgh
c) Ek = (1/2)mv2
d) E = mghA - mghB
2a) 0 m/s
b)v= sqrt(2Ek/m)
3) First Column: All 0
Second Column: All 19110J
Third Column: 9555J, 12740J, 15925J, 3185J, 6370J
Fourth Column: 9555J, 6370J, 3185J, 15925J, 12740J
Fifth Column: 17m/s, 19.8m/s, 22.1m/s, 9.9m/s, 14m/s
4)The values are equal.
After that we took a look at our Work and Energy Booklet and learned about spring energy and the potential energy of the spring otherwise known as HOOKE'S LAW.
The formula for Hooke's Law is F = kx where F is the force, k is the spring constant, and x is displacement of the spring if it is stretched using the spring at normal length as the reference point. So now we can calculate the potential energy of the spring we are given a new formula derived from Hooke's Law and our good old area of a triangle of a Force vs Displacement graph.
W = area of a triangle
W = (1/2) F x
W = (1/2) k x x ---Now we substitute F for Hooke's Law
W = (1/2) kx2
Now we have 2 Potential energy formulae:
Ep = mgh
W = (1/2)kx2
We started working on a lab that involved Wind Up toys. This lab is due tomorrow I believe. Other than that just study up because the test on Circular Motion and Work and Energy will be on Wednesday November 12, 2008. The next scribe will be my best friend Rojuane...
Bye Bye
We started off correcting chapter 11.2 Study Guide. So here are the answers:
enter, leave, external, conservation of energy, form, constant, created, destroyed, mechanical, decrease, simple harmonic, potential, kinetic, potential, kinetic, increases , decreases , friction , thermal
Here are the answers for the second paragraph:
shape, potential, kinetic, all, conserved, some, other forms, conserved
Now we corrected a couple of worksheets from the beginning of the week.
Here are the answers for Concept- Development Practice Page: Chapter 8: Energy
1)
W= Fd
W= 200N (4m)
W= 800J
2)
P = W/t
P = [200N(4m)]/4s
P = 200 watts
3)
P=W/t
P= 6000 watts or 6kw
4) 16nore (IGNORE :P)
5) 10 m/s of all of the cases. Apply trajectory formula to answer this question.
6) Block A wil reach the bottom of the incline first because of the incline's shorter distance.
7) 100J, 75J, 25J
8) 10 m/s for Points B, D, and E. Max Speed at Point C
9) Yes there will be a change in wind speed because the windmills will take some of the wind's kinetic energy.
That was it for that work sheet. There was one more. Diagram Skills Section 5-3.
1a) 0 J
b) Ep = mgh
c) Ek = (1/2)mv2
d) E = mghA - mghB
2a) 0 m/s
b)v= sqrt(2Ek/m)
3) First Column: All 0
Second Column: All 19110J
Third Column: 9555J, 12740J, 15925J, 3185J, 6370J
Fourth Column: 9555J, 6370J, 3185J, 15925J, 12740J
Fifth Column: 17m/s, 19.8m/s, 22.1m/s, 9.9m/s, 14m/s
4)The values are equal.
After that we took a look at our Work and Energy Booklet and learned about spring energy and the potential energy of the spring otherwise known as HOOKE'S LAW.
The formula for Hooke's Law is F = kx where F is the force, k is the spring constant, and x is displacement of the spring if it is stretched using the spring at normal length as the reference point. So now we can calculate the potential energy of the spring we are given a new formula derived from Hooke's Law and our good old area of a triangle of a Force vs Displacement graph.
W = area of a triangle
W = (1/2) F x
W = (1/2) k x x ---Now we substitute F for Hooke's Law
W = (1/2) kx2
Now we have 2 Potential energy formulae:
Ep = mgh
W = (1/2)kx2
We started working on a lab that involved Wind Up toys. This lab is due tomorrow I believe. Other than that just study up because the test on Circular Motion and Work and Energy will be on Wednesday November 12, 2008. The next scribe will be my best friend Rojuane...
Bye Bye
Monday, November 3, 2008
October 31, 2008
This Halloween in physics we did a Lab using a ramp, a cart, and a meter of floor, to measure potential energy and kinetic energy. With this lab we received a lab sheet which is due on Monday, We also received a chapter study guide which will be corrected in class.
The next to scribe is Benchmen.
The next to scribe is Benchmen.
Thursday, October 30, 2008
Work and Energy Study Guide
Previous scribe was abducted by aliens and hasn't scribed ever since, so I inherited his scribing powers under the "first come, first serve" rule.
The study guide is straightforward. Basic stuff. Grade 12 physics students can do this in their sleep.
10.1 WORK AND ENERGY STUDY GUIDE
Note: Capitalized words are the answers to fill-in-the-blanks.
Work
Work is the product of the FORCE exerted on an object and the DISTANCE the object moves in the DIRECTION of the force. The equation used to calculate work is W = Fd. In this equation, W stands for WORK, F stands for FORCE, and d stands for DISTANCE. Work has no direction, so it is a scalar quantity. The SI unit of work is the JOULE. When a force of one NEWTON moves an object a distance of one METRE, one JOULE of work is done. Work is done on an object only if the object MOVES. Work is done only if the FORCE and the DISTANCE are in the same direction.
Work and Direction of Force
If a force is exerted IN THE DIRECTION OF the motion, work is done. If a force is exerted PERPENDICULAR to the motion, no work is done. If a force is exerted at another angle to the motion, only the component of the force IN THE DIRECTION OF the motion does work. The magnitude of this component is found by multiplying the force applied by the COSINE of the angle between the force and the DIRECTION OF THE MOTION. When friction opposes motion, the work done by friction is NEGATIVE. When work is done on an object, ENERGY is transferred. Work is the transfer of energy as the result of MOTION. This transfer can be POSITIVE or NEGATIVE.
Power
Power is the RATE of doing work, or the RATE at which ENERGY is transferred. The equation used to calculate power is P = W/t. In this equation, P stands for POWER, W stands for WORK, and t stands for TIME. The unit of power is the WATT. One JOULE of energy transferred in one second equals one watt. This is a very small unit, so power is often measured in KILOWATTS.
1. Symbol for kinetic energy: K
2. Calculation of kinetic energy: mv^2/2
3. Symbol for work: W
4. Calculation of work: Fd
5. Statement that the work done on an object is equal to the object’s change in kinetic energy: Delta K = W
6. Equivalent to 1 kg*m^2/s^2: 1 J
7. Through the process of doing work, energy can move between the environment and the system as the result of FORCES.
8. If the environment does work on the system, the quantity of work is POSITIVE.
9. If the environment does work on the system, the energy of the system INCREASES.
10. If the system does work on the environment, the energy of the system DECREASES.
11. In the equation W = Fd, Fd holds only for CONSTANT forces exerted in the direction of displacement.
12. In the equation W = Fd cos theta, angle theta is the angle between F and the X-AXIS.
13. W > 0: B, E, F
14. W = 0: A, D
15. W < 0: C
16. What was the magnitude of the force acting on the crate? 30.0 N
17. How far did the crate move horizontally? 0.50 m
18. What does the area under the curve of this graph represent? work
19. How much work was done in moving the crate 0.1 m? 3.0 J
20. Rate of doing work: power
21. Unit of power: watt
22. Symbol for power: P
23. Calculation of power: W/t
24. 1000 watts: kW
Next scribe is KAMIL.
The study guide is straightforward. Basic stuff. Grade 12 physics students can do this in their sleep.
10.1 WORK AND ENERGY STUDY GUIDE
Note: Capitalized words are the answers to fill-in-the-blanks.
Work
Work is the product of the FORCE exerted on an object and the DISTANCE the object moves in the DIRECTION of the force. The equation used to calculate work is W = Fd. In this equation, W stands for WORK, F stands for FORCE, and d stands for DISTANCE. Work has no direction, so it is a scalar quantity. The SI unit of work is the JOULE. When a force of one NEWTON moves an object a distance of one METRE, one JOULE of work is done. Work is done on an object only if the object MOVES. Work is done only if the FORCE and the DISTANCE are in the same direction.
Work and Direction of Force
If a force is exerted IN THE DIRECTION OF the motion, work is done. If a force is exerted PERPENDICULAR to the motion, no work is done. If a force is exerted at another angle to the motion, only the component of the force IN THE DIRECTION OF the motion does work. The magnitude of this component is found by multiplying the force applied by the COSINE of the angle between the force and the DIRECTION OF THE MOTION. When friction opposes motion, the work done by friction is NEGATIVE. When work is done on an object, ENERGY is transferred. Work is the transfer of energy as the result of MOTION. This transfer can be POSITIVE or NEGATIVE.
Power
Power is the RATE of doing work, or the RATE at which ENERGY is transferred. The equation used to calculate power is P = W/t. In this equation, P stands for POWER, W stands for WORK, and t stands for TIME. The unit of power is the WATT. One JOULE of energy transferred in one second equals one watt. This is a very small unit, so power is often measured in KILOWATTS.
1. Symbol for kinetic energy: K
2. Calculation of kinetic energy: mv^2/2
3. Symbol for work: W
4. Calculation of work: Fd
5. Statement that the work done on an object is equal to the object’s change in kinetic energy: Delta K = W
6. Equivalent to 1 kg*m^2/s^2: 1 J
7. Through the process of doing work, energy can move between the environment and the system as the result of FORCES.
8. If the environment does work on the system, the quantity of work is POSITIVE.
9. If the environment does work on the system, the energy of the system INCREASES.
10. If the system does work on the environment, the energy of the system DECREASES.
11. In the equation W = Fd, Fd holds only for CONSTANT forces exerted in the direction of displacement.
12. In the equation W = Fd cos theta, angle theta is the angle between F and the X-AXIS.
13. W > 0: B, E, F
14. W = 0: A, D
15. W < 0: C
16. What was the magnitude of the force acting on the crate? 30.0 N
17. How far did the crate move horizontally? 0.50 m
18. What does the area under the curve of this graph represent? work
19. How much work was done in moving the crate 0.1 m? 3.0 J
20. Rate of doing work: power
21. Unit of power: watt
22. Symbol for power: P
23. Calculation of power: W/t
24. 1000 watts: kW
Next scribe is KAMIL.
Wednesday, October 29, 2008
October 29, 2008: Work and Power
In today's class we finished the rest of the question in Concept-Development Practice Page 9.2.
The answer to them are:
4.
a) normal
b) normal
5.
a)Yes
b)Provides centripetal force for circular motion.
After that, Mrs. K handed back our lab and assignment (questions 4 and 5), with it she gave us a small booklet on our last unit in mechanics, Work and Energy, which contains a small introduction to Calculus. We only read the first 4 pages then right after we started a new lab relating to the equations mentioned in the booklet.
Page 1
Page 2
Page 3
Page 4
At the end of the class we were given a chapter 10 study guide.
Finally the next scribe is Vieteran!!
The answer to them are:
4.
a) normal
b) normal
5.
a)Yes
b)Provides centripetal force for circular motion.
After that, Mrs. K handed back our lab and assignment (questions 4 and 5), with it she gave us a small booklet on our last unit in mechanics, Work and Energy, which contains a small introduction to Calculus. We only read the first 4 pages then right after we started a new lab relating to the equations mentioned in the booklet.
Page 1
Page 2
Page 3
Page 4
At the end of the class we were given a chapter 10 study guide.
Finally the next scribe is Vieteran!!
Tuesday, October 28, 2008
Tuesday, October 28th
Today in class we went over the Centripetal Acceleration assignment and the Centripetal Force assignment that was on the same sheet. Near the end of the class we also went over the Concept Development Practice Page (the one with the pictures).
Here are the answers for the Centripetal Force assignment:
1.A) 34.9 m/s
B) 24.4 m/s²
2. A) 18.85 m/s
B) 237 m/s²
3. 8 m/s²
4. 1.22 x 10-5 m/s2
5. A) 0.52 m/s
B) 1.8 m/s2
Answers for the Centripetal Force questions on the back of the same sheet:
1. 27 N
2. A) 46.3 m/s²
B) 4.63 N
3. A) 157.9 m/s²
B) 78.95 N
Answers for the Concept-Developtment Practice Page (Acceleration and Circular Motion side):
1. A) Backward
B) Forward
C) Forward
2. A) Forward
B) Backward
C) Backward
3. A) Outward
B) Inward
C) Inward
4. Opposite
5. A) Faster
B) Increases
6. A) More
B) Increases
Answers for the Concept-Development Practice Page (Centripetal Force):
1. A) T
B) T
2. A) Inward
B) Yes
C) Yes
D) Yes
3. A) Fn pointing up, Ff pointing towards the left, Fg pointing downward
B) Ff
C) Ff
D) Ac is not zero
4. A) Ff
B) Ff
Aaand that’s about all we did for class today! Next scribe is…..Shelly
Monday, October 27, 2008
Derivation for Circular Motion Formulas
Deriving a formula for velocity, acceleration, and the centripetal force when in a circular motion...
LEGEND
v = velocity (or speed, if direction is not indicated)
d = displacement (or distance, if direction is not indicated)
t = time (interval)
R = radius
a = acceleration
a_c = centripetal acceleration
F = force
F_net = net force
F_c = centripetal force
m = mass
1.
The distance an object moves in a circular motion is the circumference of the circular motion, which is equal to 2*pi*R. Using the definition of circumference (c=π*2*r) and the definition of velocity (v=d/t), we can derive this formula: v=2*π*R/t.
2.
Draw the circular motion, two radii, and two velocity vectors. Add the two radii vectors to get the net radius. Add the two velocity vectors to get the net velocity.
3.
Since the change in radii over the radius equals the change in velocities over a velocity (ΔR/R = Δv/v), using the definition of velocity (v=ΔR/Δt) and acceleration (a=Δv/Δt), we can derive this formula: a=v^2/R.
4.
Using the acceleration formula that we recently derived (a=v^2/R), we can use substitute velocity with v=2*π*R/t to get a more fancy-looking formula for centripetal acceleration (a=4*π^2*R/T).
5.
Using Newton's Second Law of Motion (F_net=m*a), we can substitute acceleration with a=v^2/R to get the formula for centripetal force (F_c=m*v^2/R), which can be derived even further using v=2*π*R/t to get F_c=4*π^2*R/t^2.
LEGEND
v = velocity (or speed, if direction is not indicated)
d = displacement (or distance, if direction is not indicated)
t = time (interval)
R = radius
a = acceleration
a_c = centripetal acceleration
F = force
F_net = net force
F_c = centripetal force
m = mass
1.
The distance an object moves in a circular motion is the circumference of the circular motion, which is equal to 2*pi*R. Using the definition of circumference (c=π*2*r) and the definition of velocity (v=d/t), we can derive this formula: v=2*π*R/t.2.
Draw the circular motion, two radii, and two velocity vectors. Add the two radii vectors to get the net radius. Add the two velocity vectors to get the net velocity.3.
Since the change in radii over the radius equals the change in velocities over a velocity (ΔR/R = Δv/v), using the definition of velocity (v=ΔR/Δt) and acceleration (a=Δv/Δt), we can derive this formula: a=v^2/R.4.
Using the acceleration formula that we recently derived (a=v^2/R), we can use substitute velocity with v=2*π*R/t to get a more fancy-looking formula for centripetal acceleration (a=4*π^2*R/T).
5.
Using Newton's Second Law of Motion (F_net=m*a), we can substitute acceleration with a=v^2/R to get the formula for centripetal force (F_c=m*v^2/R), which can be derived even further using v=2*π*R/t to get F_c=4*π^2*R/t^2.- Centripetal Force lab is due and was handed in today.
- Centripetal Acceleration and Centripetal Force assignments is due tomorrow.
- Next scribe is ERIC.
Friday, October 24, 2008
Centripetal Force Lab
Today in physics class we did a lab on centripetal force.
In this lab the equipment that we needed was
glass tube 10 to 20 cm long, and about 1 cm outside diameter (tube should be fire-polished at both ends)
Nylon cord (about 1.5 m) size 4 stopper, 2hole 25 iron washers (about 0.06n each)
Stopwatch or watch with second hand
Alligator clip
The objective of the lab was
During this investigation you will study circular motion experimentally. Use the graphical analysis of circular motion to study the relationships between the centripetal force and the velocity of an object moving in a circular path and the radius of the path.
In the lab we had to connect a rubber stopper to one end of the string and on the other end we had to add a bent paper clip with washers connected to it you also need to put the alligator clips somewhere near the middle of the string.
We also had to time how many seconds it would take to get 30 revolutions for 5 washers, 10 washers, 15 washers, 20 washers and 25 washers.
Don't forget to do the calculations and the Interpretation on the sheet. Well that's all the next scribe will be ZEPH.
In this lab the equipment that we needed was
glass tube 10 to 20 cm long, and about 1 cm outside diameter (tube should be fire-polished at both ends)
Nylon cord (about 1.5 m) size 4 stopper, 2hole 25 iron washers (about 0.06n each)
Stopwatch or watch with second hand
Alligator clip
The objective of the lab was
During this investigation you will study circular motion experimentally. Use the graphical analysis of circular motion to study the relationships between the centripetal force and the velocity of an object moving in a circular path and the radius of the path.
In the lab we had to connect a rubber stopper to one end of the string and on the other end we had to add a bent paper clip with washers connected to it you also need to put the alligator clips somewhere near the middle of the string.
We also had to time how many seconds it would take to get 30 revolutions for 5 washers, 10 washers, 15 washers, 20 washers and 25 washers.
Don't forget to do the calculations and the Interpretation on the sheet. Well that's all the next scribe will be ZEPH.
Thursday, October 23, 2008
Mole Day !
Hello physics lovers :)
Here's a summary of today's class!
First we corrected the Chapter 7 Study Guide (7.2 Periodic Motion Circular Motion) that was given the previous day. Here are the the answers:
If an object is moving with uniform circular motion, the speed is constant. Velocity is changing because the direction is changing. In uniform circular motion, the radius of the path is constant. The velocity is perpendicular to the radius and tangent to the circular path. Any vector that is tangent to the circle represents the instantaneous velocity at that point. All such vectors will have the same length, but different directions. The vector that represents acceleration is in the direction of the radius, pointing toward the center of the circle. Centripetal acceleration points toward the center of the circle. It is directly proportional to the square of the speed and inversely proportional to the radius of the circle. According to Newton's 2nd law, centripetal acceleration must be caused by a(n) force that acts toward the center of the circle. The force that causes centripetal acceleration is called centripetal force. If this force disappears, the object in uniform circular motion will travel on a path it is a(n) straight line tangent to the circle.
After that we corrected the Free-Body Exercise: Circular Motion,
here are the answers:
Then Ms. K handed out notes on Centripetal and Centrifugal Forces.
Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation. For example the car going around a curve there must be friction to provide the required centripetal force, if the force of friction is not great enough, skidding will occur.
Centrifugal force is the apparent force that is felt by an object moving in a curved path that acts outwardly away from the center of rotation like the the clothes being force against the wall of the washing machine.
We also got a hand out on Frames of Reference, that we read while Ms. K got the television for a video on Circular Motion.
Here are some videos I found on You Tube that I hope will help!
Look who I found, MR. SAME SAME !
It's Bill Nye the Science Guy!
PART ONE:
PART TWO:
PART THREE:
I hope this helped! Sorry for the late blog.
The next scribe for tomorrow is Richard,
you said to pick you.. you have all weekend :)
Goodnight Everyone!
Here's a summary of today's class!
First we corrected the Chapter 7 Study Guide (7.2 Periodic Motion Circular Motion) that was given the previous day. Here are the the answers:
If an object is moving with uniform circular motion, the speed is constant. Velocity is changing because the direction is changing. In uniform circular motion, the radius of the path is constant. The velocity is perpendicular to the radius and tangent to the circular path. Any vector that is tangent to the circle represents the instantaneous velocity at that point. All such vectors will have the same length, but different directions. The vector that represents acceleration is in the direction of the radius, pointing toward the center of the circle. Centripetal acceleration points toward the center of the circle. It is directly proportional to the square of the speed and inversely proportional to the radius of the circle. According to Newton's 2nd law, centripetal acceleration must be caused by a(n) force that acts toward the center of the circle. The force that causes centripetal acceleration is called centripetal force. If this force disappears, the object in uniform circular motion will travel on a path it is a(n) straight line tangent to the circle.
After that we corrected the Free-Body Exercise: Circular Motion,
here are the answers:
Then Ms. K handed out notes on Centripetal and Centrifugal Forces.
Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation. For example the car going around a curve there must be friction to provide the required centripetal force, if the force of friction is not great enough, skidding will occur.
Centrifugal force is the apparent force that is felt by an object moving in a curved path that acts outwardly away from the center of rotation like the the clothes being force against the wall of the washing machine.
We also got a hand out on Frames of Reference, that we read while Ms. K got the television for a video on Circular Motion.
Here are some videos I found on You Tube that I hope will help!
Look who I found, MR. SAME SAME !
It's Bill Nye the Science Guy!
PART ONE:
PART TWO:
PART THREE:
I hope this helped! Sorry for the late blog.
The next scribe for tomorrow is Richard,
you said to pick you.. you have all weekend :)
Goodnight Everyone!
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