Friday December 17

On Friday we received new notes on Circuits;
-Transparency 32: Circuits and Circuit Diagrams
-Transparency 33: Series Circuit/Parallel Circuit
-Study Guide 23: Simple Circuits

We also watched some videos on Parallel and Series Circuits as well as a demonstration
Here is a video similar to the demonstration we had:
http://freecircuitdiagram.com/2010/04/04/electricity-series-parallel-circuits-video-tutorial/

We also got our Electric and Magnetic Field tests back. Remember to do your corrections and hand them back in to Ms. Kozoriz.

December 16

Today we started the new unit by looking at Ohm's law and circuits. We also examined circuit diagrams and the symbols that can be found on them and were then assigned a worksheet called Circuit Symbols which we went over in class. A second worksheet, Resistivity Problems, was assigned. This assignment is homework if you didn't finish it in class.

Electric and Magnetic Fields

Hi Guys! today we started class with the questions given by Miss Kozoriz yesterday from the duck book.

75. d=5cm/100=0.05m
E=5.10^3 N/C
V=E.d=2.5.10^2 V

76.
a) F=E.q=5.0.10^-4N
b) Ek=EP
EP=qV
V=Ed
V=(50)(1.0)=50V
EP=(10^-5)(50)=5.0.10^-4J
c) m=2KE/v.v=1.6.10^-12 kg

77. delta E= E2-E1=k q1q2/R2-k q1q2/R1=-2.08.10^-19 J

79.
a) q=W/V=1.8.10^-3 C
# of electrons=1.10^16 electrons
b)v2= under root qV/m=6.6.10^-23

80.
a)a=3.0.10^10 m/s. square
b)v=1.2.10^-15 J

81. R=1.9.10^-14m

82.
a)delta dh=0.00025m
b)v=6.0.10^6m/s

90.
a) Fe=Fg
Fe=mg
E=v/d
q=Fe/E
=mg/v/d
=mgd/v
=4.2.10^-19C

91.
a)v= under root 2qv/m=1.2.10^7m/s
b)v=under root 2deltaEe/3m=7.26.10^6m/s

                               As we've test tomorrow so we reviewed all the topics about Magnetic and Electric Fields. We have test tomorrow on Electric and Magnetic Fields so be prepared and good luck

Magnetic and Electric Field

Hey Classmates!
                          Yesterday we went over the Chapter 27 Review worksheet and the answers are given below:
1. 1.4.10^-4 m
2. 1.7.10^6 m/s
5. 2.63.10^6 m
6.a) 4.3.10^6 m/s
b) 1290 N/C
7. 1.92.10^-25 kg
8. 15m
         We don't need to solve question 3, 4, 9 and 10. Miss Kozoriz gave us back our DARK MATTER and Chapter 21 Lab Worksheet. Then we read Cathode ray tube from duck book page#571 upto end of the chapter and solved questions 75, 76, 77, 79, 80, 81, 82, 90, and 91. Miss Kozoriz also announced that our test on MAGNETIC AND ELECTRIC FIELD will be on Wednesday.

Electric and Magnetic Fields

Hey Classmates we started from "Moving Charges Worksheet" 

and we went over problem 8 to 11 in the class.

8.v=2.1 . 10^5 m/s

   f=5.6 . 10^-13 N

   B=f/qv

   B=5.6 . 10^-13/(1.6 . 10^-19)(2.1 . 10^5)

   B=16.6T[left]

9.v=8.6 . 10^4m/s

    B=1.2T

    F=?

    F=Bqv

    F=(1.2 )(1.6 . 10^-19)(8.6 . 10^)

    F=1.7 . 10^-14N[E] 

10.v=5.0 . 10^5m/s

     F=4.0 . 10^-6N

     B=F/qv

     B=4.0 . 10^-6/(1.6 . 10^-19)(5.0 . 10^5)

     B=5. . 10^7T [out of the page]

11.v=2.0 . 10^6m/s

    B=5.0 . 10^-2T

    F=4.8 . 10^-14N

    q=?

    q=F/Bv

    q=4.0 . 10^-14/(5.0 . 10^-2)(2.0 . 10^6)

    q=4.8 . 10^-19C

Miss Kozoriz explain a bit The Mass of an Electron, we read the Green book from page 536-541 for some basic formulas.

We got home work for the Worksheet "Moving Charges" for problems 12-16 and Review Chapter 27. And we have test on Tuesday. Have a nice weekend and prepare for the test.

Magnetic Fields, B

HEY GUYS !

For today in class, first up, Ms. Kozoriz start the class off with a short video clip on Milikan's Oil Drop Experiment.

Then we went over the first 4 questions of the worksheet handed out yesterday, called MOVING CHARGES WORKSHEET. [Please see Amy's previous post for the scans if needed!]

The answers are:

After going over the 4 questions of the worksheet, we watched some few more video clips on magnetic fields, magnetism, magnets and 3 videos for each of the right hand rules.

We then received a worksheet called The Path of a Charged Particle in a Magnetic Field. We were given some time to work on it class, and went over it together after.

Last but not least, we were asked to do questions 8-11 on the Moving Charges Worksheet for tomorrow and we will be going over those questions in class !

Thanks for reading and have a good night ! (:

Wednesday, December 8, 2010

Hi hi guys! I am back with another scribe post for today. In class today, we first went over the two worksheets that were handed out on Monday called "Electric Potential" and "Physics 40S Electric and Magnetic Fields Assignment." Please click on the picture to see what the worksheets look like if you've missed it. You can ask Ms. Kozoriz for a copy if you didn't get one. If you want to see the steps on how to solve the problem, please come see me :). Otherwise, here are the answers for the worksheets if you can't read my writing:

Electric Potential

1.) 31 J/C (electric potential)

2.) a.) 3.6 x 10^-14 J (potential energy)

b.) 180 J/C (electric potential)

3.) 20,000,000 m/s (velocity)

4.) 3.0 x 10^-4 J (work)

5.) a.) 12 ev (work)

b.) 1.92 x 10^-18J ; 12 ev (energy)

c.) 2.05 x 10^6 m/s (speed)

Physics 40S Electric and Magnetic Fields Assignment

Part 1

Part 2

1.) a.) 3.2 x 10^-15 J (potential energy)

b.) 3.2 x 10^-15 J (kinetic energy)

c.) 8.38 x 10^7 m/s (velocity)

d.) 2.4 x 10^-10 s (time)

e.) 500,000 N/C (electric field)

f.) 8.78 x 10^16 m/s^2 (acceleration)

g.) 2.1 x 10^7 m/s (velocity)

h.) 2.5 x 10^-3 (displacement)

If you have any questions about the answers please come see me :).

Next in class, we were given 3 handouts. The handouts are "Moving Charges Worksheet," "Magnetic Formulae," and a cathode ray diagram that we previously got on Monday. Please click on the pictures to view the handouts.

Moving Charges Worksheet

Part 1

Part 2

Magnetic Formulae

Cathode Ray Diagram

It will be most likely that we will be correcting "Moving Charges Worksheet" tomorrow in class so for those who were in class please have it done and for the absentees, please see Ms. Kozoriz to get a copy.

Basically in today's class, we spent our time correcting the worksheets and were to work on the new ones that were handed out. New material was not taught in today's class so nothing important was missed. If you have any questions, please see Ms. Kozoriz or myself and we will be glad to help you.

Thanks for reading my post :)!

-Amy

Another scribe post FROM THE PAST (December 3)

By now I can't quite remember exactly what happened in class last friday, but that's why we have imaginations, right?

What I do remember however is going over question 5 from page 420 and questions 15, 16, and 19 from page 428, all from the green textbook. We went on to do a lab on charges, energy, and voltage. For homework we were assigned Electric Fields and Potential and the lab if it wasn't finished.

Hey classmates, today we started with Electric Potential 33-2 worksheet. We hand in our lab sheet and then Ms.Kozoriz explained Electron Volt and Equipotential line. We also discussed "Potential Energy and Kinetic Energy between two charged plates" worksheet. After  that we got 3 worksheet to work on; Appendix 2 -Fields (Illustrative examples), The physics of Cathode Ray Tube and Electric Potential. We'll hand these worksheets on wednesday. Thats all what we did in class.

  Have a great evening!

Scribe post FROM THE PAST (november 30)

Today, we started the unit on electric and magnetic fields by taking a look at Coulomb's Law. Coulomb's Law is simply a way to explain the force acting on one or more small charged objects called point charges by way of calculating the magnitude of the force (or forces) using the formula:



We were then assigned two sheets: one called Coulomb's Law for the sole purpose of better understanding the concepts of the new unit, and another titled Electrical Forces as practice applying this formula that we have definitely not seen in a similar form at any other point in our entire lives.

Wednesday, December 1, 2010

In today's class, we first went over the worksheets COULOMB'S LAW (the one where you had to underline the right answer) and ELECTRICAL FORCES.

For COULOMB'S LAW the answers are:
1-b. attraction
1-c. greater, less, 1/4 as much
2-a. doubled
2-b. quadrupled
2-c. quadrupled
2-d. 16 times

For ELECTRICAL FORCES the answers are:
1. -3.6 x 10^10 N (use Fe=kq1q2/R^2)
2-a. 7.80 x 10^-47 N (use Fg=GMm/R^2)
2-b. 8.80 x 10^-8 N (use Fe=kq1q2/R^2)
3. 2.5 x 10^19 electrons (use charge/e)
4. -1.76 x 10^8 C (use given mass/mass of electron)
5-a. 1.76 x 10^-15 N (add the force of each charged objects)
5-b. 2.63 x 10^11 m/s^2 (use a=Fnet/m)

Next we read through our first handout Coulomb's Law in One Dimension, Two Dimensions: Equilateral Triangle, and Two Dimensions: The Square.

To solve problems in one dimension, simply add the charges (like vectors). Don't forget that like charges repel and unlike charges attract.

To solve problems in two dimensions, first find the x and y components of each charge and add them together. Then combine the totaled x and y components by using the pythagorean theorem.

** Like vectors, electrical forces have direction.

Lastly, we were given question 5 on page 420 and questions 15, 16 and 19 on page 423 of the green books for homework!

Coulomb's Law

Coulomb's law

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Thursday November 25th, 2010

Today in class we watched a video called "Dark Matter", and received Worksheet 1 & Worksheet 2. Worksheet 1 contained multiple choice questions related to the video, and Worksheet 2 contained questions regarding the information shown on the video. We went over the answers for Worksheet 1, and #1 of Worksheet 2. We are expected to do the rest of the questions on our own.

We also received a worksheet called "Dark Matter within a Galaxy". This sheet, along with any other work you have not completed is tonight's homework.

Low Earth Orbit Simulation

Newton's Cannon

Last Friday November 19

Hey guys, sorry about the delay, I forgot

We worked on several sheets that were given to us including
- a review on Gravitational Potential Energy
-Gravitational Fields Worksheet
-Escape Speed

Sample Question from "Gravitational Potential Energy Questions"

A 500.0 kg communications satellite is initially at rest on Earth's surface before it is launched into space. What is its gravitational Potential Energy at this point?

Use Formula PEG= - Gm1m2/R
G= 6.67 x 10^-11
m1 = 50.0kg
m2= 5.98x10^24kg (mass of Earth)
R= 6.37x 10^6 (Radius of Earth)

- [(6.67x10^-11)(50.0 kg)(5.98x10^24) / 4.22x10^7m]

PEG= -3.13 x 10^10


If there are any questions or you need the sheets from Friday feel free to ask Mrs. Kozoriz

Sorry again for forgetting about it :P

The gravitational potential energy

Hi everyone,

On Thursday we learn some interesting facts about GPE that equation of PEG = -Gm1m2/R is always produces a negative value.As R increases,the masses get further apart,PEg increases by becoming less negative.As R increases to the point of approaching infinity R → infinity,the potential energy approaches zero,PEg → 0

The differences between

FGand EGPE

1. F inversely 1/R square E inversely 1/R

2. is a vector quantity is a scalar quantity

3. S.I unit measured in Newtons S.I unit measured in Joules

Hey grade 12's, in today's class we watched a film called "Gravity 1" where our task was to write down 5-7 facts that we thought that were interesting on a separate sheet of paper then hand it in. After that we were assigned questions 6-10 in the Green Textbook on page 172.

Homework for today was to complete those 3 sheets that were given to us yesterday, if you haven't received them yet you should have got them today. We will be going over those tomorrow since we didn't today.
- Universal Law of Gravitation
- Transparency 7-2: Gravitation
- Transparency 7-3: Kepler's Third Law

Have a good evening!

Janna's Scribe Post for November 16th 2010 !

Hey guys for today's class we got back our Energy in a Trebuchet worksheet & our questions 3, 4, and 5 that we were to do out of the Green Textbook.
Next we went over our Chapter 8 Study Guide and filled in the blanks in the Universal Gravitation, Newton's Use of His Law of Universal Gravitation & Weighing Earth paragraphs on the first page of the study guide. The answers can be found in the slide post directly below my scribe on the first page of Mrs. K's post. Then we went over the answers of the first three pages in the Chapter 8 Study Guide Motions in the Heavens and on Earth booklet. Again, these answers can be found in Mrs. K's slide post from page 2 on.
After going over the answers we picked up 3 worksheets: Universal Law of Gravitation, Chapter 7 Transparency 7-2 & Chapter 7 Transparency 7-3. If you were away make sure you pick up these worksheets ASAP and do them for we will be going over the answers in class tomorrow. =)

Homework:
- Universal Law of Gravitation
- Chapter 7 Transparency 7-2
- Chapter 7 Transparency 7-3

Happy Homeworking !! :)

Ch.8 Study Guides Answers

Ch.8 study guides answers

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Monday, November 15, 2010

Hello my fellow classmates! It's me Amy scribing again. In today's class, we went over the problems in the green textbook that was assigned to us on Friday. The page number is 172 for those who missed the chance to do those questions. Questions 1-5 were assigned for us as homework and we had gonve over the first two in class today. The other 3 were to be handed in at the end of today's class. I'm assuming those are for marks so hurry up and hand them in before Ms. Kozoriz returns the assignments back to the class! I would give out the answers for question 1 and 2 but sadly they are on the same paper with the 3 other questions that are to be handed in. In other words, I don't have them with me :\. Sorry!!!

So for half of the class, we had worked on those problems. Afterwards, we were given 4 hand-outs from Ms. Kozoriz!!! Those hand-outs included "Concept-Development Practice Page (Chapter 12: Universal Gravitation)," "Chapter 11: Gravitational Intructions," "Issues in Space Exploration: Benefits," and "Chapter 8 Study Guide." Please see Ms. Kozoriz if you missed/don't have any of these hand-outs!!!

We had went over the "Concept-Development Practice Page (Chapter 12: Universal Gravitation)" and "Chapter 11: Gravitational Instructions" worksheet in class.

Here are the answers and what the worksheets look like:

Concept-Development Practice Page (Chapter 12: Univeral Gravitation)

--Check out the cool experiment in question 3!! You'll be amazed, guaranteed :D.

Chapter 11: Gravitational Instructions (Side 1)

Chapter 11: Gravitational Instructions (Side 2)

If you're wondering what "Fold" means, it means the solved equation equals to the "old force." The old force is the force we use to calculate our "Fnew" ("new force") for what we're trying to find. This correlates to having constants (number in front of "Fold") or no constant at all to determine the new force out of Newton's Universal Law of Gravitation equation. For example, if there is "4Fold" means 4 times the "old force." By increasing the "old force" 4 times, a new force ("Fnew") is developed where the it is equivalent to 4 times the "old force." Please see me if you don't understand the concept :).

Along with the other two hand-outs, we were to read the "Issues in Space Exploration: Benefits" package and finish the first 3 pages of the "Chapter 8 Study Guide."

There was not much work to do in class today. So for those who were absent, you guys didn't miss much. That concludes the lession in today's class. If anyone has a question about the hand-outs, you can come see me about any questions and see Ms. Kozoriz for the ones that you didn't pick up. As for a heads up in tomorrow's class, you guys will be receiving an additional 2 new hand-outs called "Transparency 7-2" and "Universal Law of Gravitation." Thanks for reading my post :)!!!

November 12, 2010

Hello everyone!

In physics class that day we had a lab about ellipses, pick the sheet up from Ms.Kozoris, and if any trouble ask a friend or Ms.Kozoris herself. It's not very time consuming so that's a plus!

Reminder that it is due to hand in ASAP!

We also had practice questions to do on pg 172 in the green physics textbook.

November 10, 2010

Today, we got our Circular Motion, Work and Energy Test back that we did on Monday. We corrected it, and we started our new unit. Our new unit is about the Exploration of Space. To start this unit off, Ms.K read a sheet, " Why Explore Space?" . After that we watched a couple of videos of Kepler's three laws. We got the Chapter 8, Study Guide and we went over the first paragraph. We also picked up the sheet, Transparency 7-1. We answered questions on the back of the sheet.

Don't forget to get your Chapter 11 : Enrichment ; Energy in a Trebuchet in the folder.

Just to let you know, no school tomorrow.

Have a nice day off ~

Friday, November 5th, 2010

During Friday's class, we went over and answered questions on the following sheets:
- Kinetic Energy (Transparency 11-1)
- Potential Energy at Varying Locations (Transparency 11-2)
- Grade 12 Physics: Work and Energy: Springs

We also received the following sheets:
- Chapter 10 - Section 10-1 Quiz
- Chapter 11 - Section 11-1/Section 11-2 Quiz
- Chapter 11 - Enrichment

The answers for 'Chapter 10 - Section 10-1 Quiz' and 'Chapter 11 - Section 11-1/Section 11-2 Quiz' are already posted so you may make your corrections. 'Chapter 11 - Enrichment' is due on Wednesday. Also a reminder, the Circular Motion and Work/Energy test in on Monday.

Goodluck!

Chapter 11 Quiz Answers

Ch.11 quizanswers

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Chapter 10 Quiz Answers

10 1 quizanswers

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THURSDAY, NOVEMBER 04, 2010

Today in Physics class, we started off by getting back our Conservation of Energy lab and went through it quickly. Then we were given out 3 worksheets, and reminded to hand in our Power Lab if we have not done so [WHICH IS OVERDUE!] and the lab that we did on Wednesday, regarding the spring and weights [please see previous post to regard about the lab].

The three worksheets provided was the Chapter 11 Transparency 11-1, Chapter 11 Transparency 11-2, and Work and Energy: Springs. We were to complete them so that we could go over them on Friday.

Ms. Kozoriz then gave us the rest of the period to work on our labs and the worksheets.

Have an awesome day (:

Nov. 3. 2010

Hey,

So today in class we started with the Transparency 11-3 Worksheet. (the one about Potential Energy and Kinetic Energy)

answers:

1. What type of energy exists in the situation shown?
- Gravitational Potential Energy/ Kinetic Energy

2. How are these two types of energy defined?
-GPE: energy stored in a system as a result of the force of gravity between the object and Earth; Ep=mgh
-Ek: energy of motion; Ek=1/2mv^2

3. At what points is the kinetic energy of the ball at a maximum?
-Just as the ball is thrown up and when it returns to its lowest point, velocity is the greatest at these 2 points.

4. When is gravitational potential energy at a maximum?
-At the top of its trajectory; Gravitational Energy is proportional to the height of an object above Earth's surface

5. As the gravitaional potential energy increases, how does the kinetic energy change? Is this an example of an open or a closed system?
-Kinetic Energy decreases; closed system

6. How does the sum of the kinetic and gravitational potential energy change? What Principle does this demonstrate?
- Sum stays constant; Law of conservation of Energy

7. How many times during one rise and fall of the ball are the gravitaional potential and kinetic energy equal?
-twice


Next we moved onto the last part of the unit; Force and the Spring

One of the first things we need to use when working with springs is its deformation i.e. how much it stretches from its normal length when a force is added. This will be shown with X

This means that a graph of Force x X will be a straight line
http://www.pa.msu.edu/courses/1998spring/ISP209/homework/hmwk3/Hw03gr.gif

The slope of the graph, (K) can be calculated by k=rise/run=F/X

"K" is a constant called the spring constant and has the dimensions of force per unit length (N/m). "K" is also referred to as the stiffness because the larger the value for "k" the stiffer the spring

It is important to understand that, in the equation F=kx, the Force is coming from something else added to the spring. This means that when the spring is compressed, both X and F are negative (-). When the spring is stretched, bothe are positive (+)

This is explained in Hooke's Law ffor a spring: "The force applied to an ideal spring is given by F=kx where k is the spring constant and x is the displacement of the spring from its unstrained length"
................ for a diagram of the above, click on the link below
http://www.pa.msu.edu/courses/1998spring/ISP209/homework/hmwk3/Hw03gr.gif

Derivations for F=kx

W= area of triangle
W=1/2 Fx
W=1/2 (kx) (x)
=1/2 kx^2

Es=1/2kx^2


We also worked on a lab based on what we just learned, in which we put weights on a spring and measured the displacement and Force exerted. The lab is due tomorrow

If you have any questions, feel free to ask Mrs. Kozoriz

HAVE A GOOD EVENING!!!

Hello Everyone, today we went over the answers of chapter 10 Study Guide page 56 and 57 which are given below;

    

MATCHING:

1. f

2. c

3. a

4. b

5. d

6. e

MCQ'S:

7. a

8. b

9. b

10. a

11. a

12. a

  

For each of the situation shown, write the letter next to the number of the correct expression.

13. B,E,F

14. A,D

15. C

ANSWERS:

16. 30N

17. 0.5m

18. work done

19. 3J

MATCH THE FOLLOWING:

20. a 

21. c

22. e

23. b

24. d

         Those who were absent on Thursday did the lab and the other students were working on chapter 10 Transparency 10-1 (Force, Work and Distance). Ms.Kozoriz helped us solving those questions, the answers are;

1. 150N

2. 8.0m

3. 0-150N as force increases the object moves 3.0m. Between 3-6m a constant of 150N is applied to object. For last 4m force is upward from 150N to 250N.

4. w=f.d

5. w=f.d

w=150N.2m

w=300N.m(J)

          Then Ms.Kzoriz gave us a helpful lecture on Kinetic Energy and Gravitational Potentional Energy. After we performed an experiment on Conservation Of Energy with 3 balls of different sizes and a meterstick. We also get a chapter 11 Transparency11-3 Worksheet in which we have to answer 7 questions.

                      Thats all we did today and also we handed in our lab worksheet on work and power. Unfortunately, tomorrow we dont have a Physics period because its half day. 

                                                                                                              Have a great day,

                                                                                                              Nageena.

Hi everyone,

                      Today we did lab and before that Ms.Kozoriz discussed WORK  "which is defined as the integral of force over a distance of displacement." 

        Work=Force.Distance

         w=f.d

Its unit is joule. POWER "it is the time rate at which work is done"

           Power=Work done/Time

           p=w/t

Its unit is watt. Then we did lab on work and power. Also we got a chapter 10 study guide on Energy and Power. Lab worksheet on centripetal force and problem 4 and 5 were marked and were given back to us. Thats all we did today. Hope this post will help those who missed the class.

                                                        GOOD LUCK

Tuesday - October 26

In today's physics class, we went over Transparency 6-3 Worksheet (Uniform Circular Motion) , and both sides of the Centripetal Acceleration and Centripetal Force worksheet. Just a reminder, we had to hand in questions 4 and 5 of Centripetal Force worksheet for marks!

We also had to pick up one worksheet (Acceleration and Circular Motion) and two readings (Centripetal and Centrifugal Forces, and Motion Near the Earth's Surface). Please have the worksheet done for tomorrow's class and also read through both readings.

Also, if you still have your lab from Thursday, it is now overdue!

Enjoy the rest of your night!
-- Rainer

Applet for Circular Motion

Circular Motion Applet
Take a look at the direction of the force vector and the acceleration vector.
What is the relationship between the tangential velocity vector and the acceleration vector?
What is the relationship between the tangential velocity vector and the force vector?
In what direction do both the force & acceleration vectors point? Can you explain why this is so?

Centripetal Force

Today, we did the lab on Centripetal Force. We got into groups and tested how long a rubber stopper would go around if we put 5 washers and so on. We will do this lab again, since we needed more time and also because some people were away.

Also a reminder that the worksheet we got yesterday, Centripetal Acceleration is due by Monday.

Projectile Motion

Hi Dear Classmates,

                                      today Ms.Kozoriz returned our chapter 6 REINFORCEMENT. After that Ms.Kozoriz solved the problems 1-4 from chapter 6 REVIEW the solution is:

1. horizontal velocity vh=800m/s

horizontal distance dh=180m

time t=dh/vh

t=180m/800m/s

t=0.225s

vertical distance dv=1/2g.tsquare

dv=1/2.9.8m/secsquare.0.225s.0.225s

dv=0.25m

2.dv=8.0m

dh=10.0m

t=under root 2dv/g

t=under root 2.8.0m/9.8m/s

t=1.27s

vh=dh/t

vh=10m/1.27s

vh=7.8m/s

3.

    a.    vh=45m/s

theta=45 degrees

vh=cos45 degree.45m/s=31.8m/s

vv=sin45 degrees.45m/s=31.8m/s

t=2vv/g

t=2.31.8m/s/9.8m/secsquare

t=6.48s

   b.  dh=vh.t

dh=31.8m/s.6.48s

dh=206.06m

4.

    a.  v=8m/s

theta=30 degrees

vh=cos30 degrees.8m/s=6.928m/s

vv=sin30 degrees.8m/s=4m/s

t=2vv/g

t=2.4m/s/9.8m/secsquare=0.81s

   b.  dv=v2 square-v1 square/2g

dv=o-16m/s/2.-9.8

dv=-16m/s/-19.6

dv=0.82m

    c.  dh=vh.t

dh=6.928m/s.0.81s

dh=5.61m

                   the remaining 5 problems are related to our next topics so we'll do them in future.

We also solved the green book's 2-7 problems. The solutions are:

2.vh=8m/s

dv=78.4m

t=under root 2dv/g

t=4s

dh=vh.t

dh=8.4

dh=32m

3.

   a.    t=under root vv/g

t=under root 1.225/9.8

t=under root .25

t=.5 s

    b.  vh=dh/t

vh=0.40/.5

vh=0.800m/s

4. vh=2.8m/s

t=2.6s

dv=?

dh=vh.t

dh=2.8.2.6

dh=7.26m

dv=1/2gtsquare

dv=33.1m

5.

   a.  dv=1001m

vh=125km/hr=34.7m/s

t=under root 2 dv/g

t=14.3s

   b.dh=vh.t

dh=34.7.14.3

dh=496.21m

6.dv=61m

dh=23m

vh=3m/s

t=under root 2dv/g

t=3.52s

vh=dh/t

vh=6.51

7. vy=vyt+1/2gt square  since vy=o

t=under root 2y/g

t=under root 2.-0.32/-9.8

t=0.26

x=vxt=12.4m/s . 0.26s

=3.2m

          Then we revised the formulas for projectile quiz. We did quiz about projectile. Thats all we did today and TOMORROW WE'LL HAVE A TEST ON MOMENTUM and Projectile Motion so be prepared.

      

                        BEST OF LUCK

Hello everyone,

                        Today we started with chapter 6 study guide which is about motion in two dimension. The answers are 

VOCABULARY REVIEW

 1. centripetal force

 2. projectile

 3. uniform circular motion

 4. centripetal acceleration

 5. trajectory

                                         PROJECTILE  MOTION

1. To an observer at position A the ball would appear moving straight up and straight down.

2. To an observer at position B the ball would appear moving in straight line.

3. To an observer at position C the ball would appear as an arc (parabola).

4. Throughout its flight,a projectile is constantly being accelerated toward the ground. Vertical vector shrinks until it reaches the top of its flight and then it increases as it approaches the ground. The horizontal velocity vector remains constant throughout its flight. Vertical vector is under the influence of gravity and horizontal vector has no net external force acting on it.

5. Neither rock lands first. Both rocks hit the ground at same time. This is occuring because both are under the same external force that is gravity(g).

6. Time t = ?

     d =8500m

      g=9.8m/s

      t=  2d/g

      t=2.8500m/9.8m/s

      t=1734.69

      t=41.64s      answer 

7. Vertical velocity v=?

     a=vf-vi/t

     v=a.t

     v=9.8m/s(42s)

     v=410m/s

8. dh=?

    Horizontal velocity v=483km/hr=134m/s 

    d=v.t

    d=483m/s.42s

    d=5600m

       Ms.Kzoriz wrote the answers for transparency sheet 6-1 and 2 on whiteboard. The answers for 6-1 are:

1. The magnitude of the ball's velocity vector is smallest at its maximum height.

2. Equal in magnitude but opposite in direction.

3. Equal in magnitude and in direction.

4. The line for horizontal position is a straight line and for vertical position its an arc.

        The answers for 6-2 are:

1. Same vertical vector at each point.

2. Both balls fall at same acceleration.

3. Horizontal motion of red ball does not affect its vertical motion.

4. Vertical interval is greater near the bottom of diagram. Since the time interval between each two picture is same, vertical vector of ball must have increased.

5. Horizontal interval between each two pictures of ball are identical. Since the intervals represent identical time periods. Horizontal vectors of the ball must be constant.

            Ms.Kozoriz talked about upwardly launched projectiles and we started to solve the problems on page 11 PROJECTILE MOTION 2. While doing problems we enjoyed a song Harry the Human Cannonball. Thats all we did in class today. One other important thing is ourshort  momentum test will be on Tuesday not on Monday.

      Its my first blog so if you guys found any mistakes sorry about that.

Correction to yesterday's post

Horizontal velocity is independent of any force acting in the x direction (not time like you stated in your post). Vertical velocity is dependent on the acceleration due to gravity acting on the object.

Physics12Fall2010: Scribe List

Projectile motion,this is a non straight motion under gravity (g).

the horizontal component of velocity is constant i.e its independent of time,but the vertical component of the velocity (Vy)depends on time. it's 0 at the peak

october 12 scribe

October 12 2010

hi! today, we went over the Transparency worksheet.

Then miss K review a little about momentum (elastic, explosion, inelastic).

After that we have a lab with money :) and hand in after class.

Sorry!!!!!!! i don't know how to put answers or picture up. If you needed you can ask miss K

Oh yeah! we picked up study guide "Motion in Two Dimensions" and a graph. Take a look we'll go over it tomorrow...

October 7

On Thursday, we went over the answers to the Chapter 9 Review worksheet.

Here are the answers:

1: 1600 N

2: 0.0013 s

3: 6700 N

4: 0.05 m/s [W]

5: -8.8 m/s

6: 81 m/s

7: 5.7 x 10^-2 m/s

8: 6 kg m/s

We then went on to finish the momentum unit by looking at momentum in two dimensions. These kind of problems involve finding the missing side of a triangle using a combination of the Law of Conservation of Momentum and trig ratios.

For homework, we got the Chapter 7: Momentum and Transparency 9-3 worksheets do.

Momentum Notes

Momentum

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Pool Table Applet

Pool Table Applet

Janna's Scribe for October 4th, 2010

Heyy everyone,

to start off class we watched a video that explored the concepts on impulse and momentum.

After watching the short video we went over the answers to the Chapter 9 Study Guide. The answers are as follows:

Section 9.1: IMPULSE and MOMENTUM

1. B

2. B

3. B

4. A

5. D

6. B

7. D

8. B

Next we went over the IMPULSE AND MOMENTUM sheet

1. WILL BE POSTED LATER

2.

a) m=Fg/g

=900N/9.8N/kg

=92kg

F∆t = m∆v/ ∆t

= (92kg)(2.0m/s)/ 0.70s

= 260N

b) p=m∆v

= (92kg)(2.0m/s)

=184N

260N - 184N =76N

3.

a) Ft =mv

=(51kg)(20.0m/s)

= 1020kg . m/s

b) F = mv/t seatbelt

= 1020kg.m/s/ 0.12s

= 8500N

F = mv/t noseatbelt
= 1020kg.m/s/ 0.008s
= 128000N

ALWAYS WEAR A SEATBLET WHEN IN A VEHICLE !!

Don't need to do #4

Next we watched three short clips regarding the momentum of a ball, recoil and the conservation of momemtum.

KEY CONCEPTS:

Remember momentum before a collision is equal to the momentum after a collision occurs.

For every action there is an equal and opposite reaction.

Homework:

Transparency 9-1 and 9-2 are due on Wednesday in class.

Have an awesome afternoon off !!! :)

Friday - October 1

Friday's physics class was basically a chance for those who missed the momentum lab on Thursday to do and finish the lab.

For those of us who were in class on Thursday, our task was to finish and hand in the lab.

So, reminder to those who didn't finish or forgot to hand in their lab, it is now over due!!



Next thing on our to do list was to check the answers from Wednesday's questions from the duck book.

Basically, to find the momentum from a given force vs. time graph, we find the area under the given line.

If the line creates a triangle, its area can be found using the equation area = (1/5)(base)(height).

If the line creates a rectangle, its area can be found using the equation area = (base)(height).

If the line creates an irregular shape, simply cut up the shape into pieces of triangles and/or rectangles and find each area using the equations from above. (Don't forget to add up all the areas in the end).


The rest of the class should have been spent on doing questions 1 - 4 of the Impulse and Momentum worksheet.

If you did not get a chance to finish it in class, please do it for homework!



Enjoy the rest of your weekend!!!
-- RAINER

Momentum Lab

Today, we did a Momentum lab. The lab is from Chapter 9 (green text book). It was "The Explosion" lab. We had to calculate how far and fast the two carts went until they hit the bricks/books etc.

We had no time to finish it, due to the Pep Rally. I guess we will continue tomorrow with the lab.

Have a good evening, :D

September 29, 2010

Hello.

Today in class we finished up the video about Momentum and handed in 5-7 facts about it. Then we went over a few notes on deriving the impulse-momentum equation from Newton's 2nd law.



To go over it, the notes were;

There were also questions in the textbook;
pg287 question 6

pg308-309, questions 50-51


Check Your Answers.
On pg287: Find the impulse:
6. a) 125 N∙s [S]
b) 1875 N∙s [W]
c) 45 N∙s [W]

Pg208-309: Find the Impulse.
50. a) 1750 N∙s
b) 6∙10^-1 N∙s
c) -225 N∙s

51. 22.5 N∙s



*Sorry for the late post.

Dynamics Test

Dynamics test

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Momentum - September 28, 2010

Hey Grade 12's!
Today we started the class off by doing the "Dynamics" test corrections on the smart board which Ms. Kozoriz will post sometime today I believe. After that we began our new unit "Momentum" by watching a film about it, we were told to copy down 5-7 facts that we personally found interesting to us but since we didn't finish the whole film we are going to continue watching the rest of it tomorrow.
Enjoy the rest of your evening! :)

Dynamics Review

Hi Guys.
Last Friday in class we did some reviews in dynamics. We did the case questions 1,4,5.

We also did the question from the book in page 196 from 1 to 3.

1. A copy of Physics: Concepts and Connections leaves the printing press and slides down a 4.0 m long ramp into the arms of an eager physics student. If the ramp is inclined at an angle of 25 degrees to the horizontal ans has a coefficient of kinetic friction of 0.10, how long will it take the 2.0 kg textbook to reach the student?

2. The plastic case from your least-favorite CD recording is flung up a frictionless ramp, inclined at an angle of 20 degrees to the horizontal. If the case leaves your hand at a speed of 4.0 m/s, how long will it take before the case comes to rest?

3. A skateboarder slides down a frictionless ramp inclined at an angle of 30 degrees to the horizontal. He then slides across a frictionless horizontal floor and begins to slide up a second inclined at an angle of 25 degrees to the horizontal. The skateboarder starts at a distance of 10 m from the bottom of the first incline. hoe far up the second incline will he go if the coefficient of kinetic friction on the second incline is 0.10?

Lastly, we did a worksheet on transparency.

Hi Everyone!

Today we went over the "Dynamics Equlibrium Problems" worksheets. The answer is kinda blurred sorry guys i wrote it on pencil.

After we went over the worksheets, we did the 2 questions on pages 13 (#2 and 4) Dynamics booklet (the rest of the answer is on the booklet on page 22) the answer of those 2 questions are:

Lastly, we went over The Atwood's Machine on page 11 and she gave us a yellow booklet about that lessons and ask to answer some question on that.

If you don't have the yellow sheets , see Ms. Kozoriz tomorrow.

Our test will be on Monday on this unit.

September 22

Hey
So today in class we did a lab showing the forces working on a block moving down an inclined plane. We needed to find the FORCE OF GRAVITY (Fw), FORCE OF PARALLEL (Fgx), FORCE OF PERPENDICULAR (Fgy), the ANGLE, and COEFFICIENT OF FRICTION (µ).

The Objective of the lab was to find the value of the coefficient of sliding function using an inclined plane



To learn the steps involved read page 19 in the Dynamics Unit

Questions

1.) Calculate Fgx, Fgy, and µ

Fgx = (sin24)(o.75N) [24 being the angle at which the block moved, and .75 being the force of the block]

Fgx = 0.31 N

Fgy = cos24(o.75N)

Fgy = 0.69N

The formula for finding µ is: µ = Fgx/Fgy

µ = 0.31N/0.69 N

µ = o.45

2.) Show that µ = tanθ
tan24 = 0.45



Next we read pages 172-179, and 194- 198, and doing questions 1-4 on pg 96 of the textbook

Questions

1.) A book leaves the printing press and slides down a 4.0m long ramp into the arms of a student. If the ramp is inclined at an angle of 25 degrees to the horizontal and has a coefficient of kinetic friction of 0.10, how long will it take the 2.0kg textbook to reach the student?

To be honest I don't know how to get the answer, so here's a basic diagram

For better information please as Ms. K.

Tuesday, September 21, 2010

Hi,

Today we went straight into a worksheet that Ms K walked us through called the Transparency worksheet. As we worked through it, we learned how to find and use forces on an incline plane. For the sheet, see Ms K!

After that, we went back into our Dynamics Unit booklet and did the example question on page 7.

Problem

A 475 N trunk is resting on a plane inclined 40.0° above the horizontal. Find the components of weight force parallel and perpendicular to the plant.

To solve inclined plane problem, you could start by drawing a diagram and or a free-body diagram

(image not drawn to scale)

Firstly, we know Fg = 475N and angle FgFgy = 40°

To find Fgx, we multiply sin40° by 475N

Fgx = sin40°(475N) = 305N

To find Fgy, we multiply cos40° by 475N

Fgy = cos40°(475N) = 364N

And the answer is Fgx = 305N, Fgy = 364N


After Ms K went over this example question, she assigned practice problems in the booklet which are on page 18. The answers to these questions are in the back. She also assigned a worksheet called Equilibrium Problems. If you don't have it, see her as soon as possible tomorrow. Both of these assignments are to be done for tomorrow.

Monday, September 20, 2010

Grade 12 Physics: Dynamics Unit

Hi everyone.

Today in class we went over the 'Free-Body Diagrams' on pages 12 and 13 of the Dynamics booklet. Unfortunately I don't have a scanner. So if anyone has some images of those diagrams, please stick them here:



After that we went over the 'Static Equilibrium Problems' on page 16 and 17. The answers to those questions are:

1. Fg = 42N

2. FT = 980N

3. FT = 346N

4. FT = 281N

5. FT = 57N


On page 15 you'll find the 'Net Force' questions which we also did in class. To do these questions you must find the net force of the forces acting on a box.

Ex: 3N <---O----> 4N
Fnet = 3N, left + 4N, right
Fnet = 1N right

Hint: use the Pythagorean Theorem (c² = a² + b²) for triangular vectors.

Answers:
1. 0N
2. 1N, right
3. 5N, 53°
4. 5N, 217°
5. 5N, 53°
6. 5N, 217°
7. 10N, 45°
8. 10N 45°
9. 0N
10. 7N, 270°

Lastly we were assigned some problems on page 5 as homework.

Friday, September 17th 2010

Grade 12 Mechanics Unit: Acceleration Test (Answers):

1) Step 1: Square both sides of the equation
V2² = V1² + 2V1at + a²t²
Step 2: Factor out 2a from the last two terms
V2² = V1² + 2a(V1t + ½at²)
Step 3: Substitute a term in the equation given in step 2
V2² = V1² + 2ad
Step 4: The equation is _____
V2² = V1² + 2ad

2) V1 = 40.0km/h ÷ 3.6 = 11.11m/s
a = -2.3m/s²
t = 2.7s
d = ?
V2 = ?

2a) d = V1t + ½at²
d = (11.11)(2.7) + ½(-2.3)(2.7)²
d = 21.6m

2b) V2 = V1 + at
V2 = 11.11 + (-2.3)(2.7)
V2 = 4.89m/s

3) V1 = 0m/s
a = 0.300m/s²
d = 25.0m
V2 = ?

V2² = V1² + 2ad
V2² = 0² + 2(0.300)(25.0)
V2² = 15
V2 = 3.87m/s

4) motorist = 80.0km/h(W)
police car = 95.0 km/h(W)

4a) motorist - police car
80.0 - 95.0 = 15km/h(E)

4b) police car - motorist
95.0 - 80.0 = 15km/h(W)

5) airplane's airspeed = 2.0 x 10²km/h(E)
approching destination = 15°(N of E)

5a) tan15° = 200 ÷ wind speed
wind speed = 200 ÷ tan15°
wind speed = 53.6km/h

5b) c² = a² + b²
speed approaching destination = (2.0 x 10²)² + (53.6)²
speed approaching destination = 207km/h

6) V1 = 25m/s
t = 1.45s
a = -8.5m/s²
V2 = 0m/s
d = ?

reacting to child and applying the brakes:
d = Vt
d = (25)(0.45)
d = 11.25m
after applying the brakes:
V2² = V1² + 2ad
d = V2² - V1² ÷ 2a
d = (0)² - (25)² ÷ 2(-8.5)
d = 36.76m
total distance = 11.25m + 36.76m = 48.0m

_____________________________________________________

Equilibrium: When an object has zero acceleration, meaning that the object is at rest or moving at constant velocity
Free Body Diagram: A drawing that represents the object and the forces acting on it

Steps to Drawing a Free Body Diagram:

1) Focus on the object you will be studying. If there is more than one object in the drawing, do each seperately.
2) Begin drawing the free body diagram by looking at only the desired object. The drawing will show where forces are applied to the object, so draw those forces on your free body diagram in exactly the same direction thats shown on the drawing.
3) Draw x and y axes on your free body diagram so you will be able to analyse the diagram. Try to draw your axes so that as many forces as possible are close or directly on the axes. It will help when calculating.
4)When the situation is equilibrium, make sure that the x and y components of the forces will be equal to zero when added.
5) Finally, solve for the unknown quantities.

_____________________________________________________

In class we received the 'Grade 12 Physics: Dynamics Unit' booklet, some of us recieved this booklet after finishing the Acceleration Test on Thursday. We read pg.1, completed pg.14, and completed #1 on pg.16. For homework we were assigned pg.12 & 13 (Free Body Diagrams) and #2-5 on pg.16 & 17 (Static Equilibrium Problems).

Test Day

Test on Acceleration. How did you do?

Relative Motion Problems

okay, so today we didn't really learn something new. we just went over the questions on page 10 of the first booklet we were given: Acceleration & Relative Velocity

I don't know about you guys but questions 4-7 gave me a hard time solving them, until Ms. Kozoriz gave me some really important tips on how to solve them.

so here are some steps of how you can solve them.

1) draw your x-axis and y-axis

2) draw the first information that you are give on the graph

3) draw the second information that you are given, but draw it from the end tip of the first information. i will show you what i mean by this

4) now you use this equations to solve for y and x.

y = sin(angle) * given distance or velocity

x = cos (angle) * given didstance or velocity

5) now you make sure that you have the total x or y

6) use the x and y to find the resultant velocity if you are asked to find it.

7) after the above steps you now have enough information to solve for tan (angle)

okay let us put the above steps into actions

question: A dog's owner walks 20 meters due west and then turns and walks 45 degrees south of west a distance of 150. where is he with the respect to where he started?

y = sin45 (150) = -106

x = cos45 (150) = -106

the total value for x is -126

this is because we had -20m which we were given in the question, plus the -106 that we just solved for.

total value for y is -106

* okay now you might be thinking why this values are negative, and that is becuase if you look on the graph the side that they are drawn on is negative for both y and x axis

now to find the resultant distance you use pythegorean theorem.

r = 165

now the last thing is find the angle.

tan (angle) = 106/126

tan (angle) = 40s degrees south of west

so now this is what you do when you get a question like that. just solve it step by step.

Cosine Law

Cosine law example

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Scribe List

This is The Scribe List. Every possible scribe in our class is listed here. This list will be updated every day. If you see someone's name crossed off on this list then you CANNOT choose them as the scribe for the next class.

jad YASSIN(2) marinhel Alex mark e

Justin Aruni(3) sakhone(3)

shahani katrina(3) chemutai(3)

This post can be quickly accessed from the [Links] list over there on the right hand sidebar. Check here before you choose a scribe for tomorrow's class when it is your turn to do so.

Monday, September 13, 2010

First we began our class with handing in the assignment from Friday, titled "Grade 12 Physics: Acceleration Assignment" (which if you haven't, reminder that you should!) and copied and checked answers 8 - 11 on the practice worksheet titled "Grade 12 Physics: Acceleration II."

The answers are as followed :
(Note: Thanks Amy, and I hope you like the colours added.)

8b. = 121m
8c. = 391m
8d. = 946m
9. = 15m
10. = 1350m
11a. = 15.4s
11b. = 427m
11c. = 200km/h

For full, see a friend or Ms.K.

Once finished, we went back to the first booklet we received titled, "Grade 12 Physics Acceleration & Relative Velocity" and went over page 6 - 9, doing the questions along the way. Don't worry if you weren't there, all the answers are stapled along with the booklet on page 11.

Page 6 extending to 7 is the lesson about relative velocity.

The concept of relative velocity can be a bit confusing, but with a little more understanding it can be easy.

It's the relationship between 3 different velocities, that can be different depending on the question. In the example included in the booklet, there are the velocities of a passenger on a train, the train itself, and someone standing on the ground watching the train.

To the passenger, who is walking to the front of the train, their velocity is +2.0 m/s.
Note: The + sign indicates direction, so if the passenger was walking towards the rear, the velocity would be -2.0 m/s.

The train's velocity, to the person on the ground, is +9.0 m/s.

We can calculate the velocity that the person the ground can see, or any of the three as long as we have any 2 velocities.

The formula that can be used is:

Vpg = Vpt + Vtg

Vpg = (2.0 m/s) + (9.0 m/s) = 11 m/s

p = passenger, t = train, g = ground.

Each velocity has letters in the subscript that are in relation to the objects in the question. Look at the bottom of page 6 for more details.

At the end of page 7 there is a small chart which can be filled out to find the answer.

So on the next page, 8 , there are two diagrams with a matching sheet of questions. The questions are pretty straight forward, but it's never bad to take a second check on the answer key.

For homework, we were assigned two things:

1) Page 10 of the booklet, for practice. (Answers included in the key)

2) A relative velocity sheet with 7 questions which you can get from Ms.K, if you don't already have it, which we will correct in class.

Well that concludes our day, and I will come back to fill in those answers, sorry everyone! Reminder, TEST ON THURSDAY! So study, study!

-Sakhone P