SCRIBE POST Thursday/Friday

I don't see anyone doing it, so I guess I'll take initiative, because I don't want to keep wishing and waiting.

From what I recall, unless I mixed up the days, Ms. K wasn't there but, Ms. K was there! Don't question that... it does, technically make sense. xD

But yes, I do believe we watched another Same-Same guy video on electromagnetic induction and had to write eight points on it and handed it in. Funny though, I could've sworn that the video was REALLY familiar. I have a feeling I watched it last sem during gr. 11 physics.

On Friday's class, we spent the first couple of minutes and filled in more blanks in the study guide and finished the Chapter 37: Electromagnetic Induction

The correct circled answers were:

1. related/

the motion of electric charges/

the motion of a magnet/

magnetic field in the loop/

electromagnetic induction

2. doubles/

doubles/

electric field/

magnetic field [although baseball is acceptable xD haa]/

electromagnetic waves.

Then afterwards... we were assigned to read an information sheet regarding EMF, and to have answered most of the worksheet questions on Electromagnetic Induction by Monday. Then we sort of... dispersed into a foreign land-- the quad and had a lovely class time. We should do it again **hint hint** it's quite lovely outside. nyar.

Physics Class May 27, 2009

GREETINGS FELLOW PHYSICISTS.
On May 27th we continued our work on electric circuits.

We went over the following:

• Resistivity Problems.
• Chapter 25 Study Guide.
• Electric Circuit Problems.

We received the following:

• Cross-Word Puzzle.
• Notes on Electromagnetic Induction.

Here is a summary of the notes:

Induced current

A current set up in a closed circuit located in a magnetic field whenever the total magnetic flux linking the circuit is changing.

Necessary conditions for an electric current to be produced with a magnetic field

1) must have a closed circuit
2) must have a changing magnetic flux

Factors that determine the value of the induced EMF

1) length of the wire
2) flux density of the field
3) speed at which conductor moves though the field

EMF = Blv

• Some of us also rewrote the test on Electric and Magnetic Fields.

Thank you Ms. Kozoriz for uploading the worksheets.

~Katherine~

Electric Circuit Problems

http://www.box.net/shared/nyrb1ghezv

Ch. 25 Study Guide

http://www.box.net/shared/9hronbgcrq

Resistivity Problems

Resistivity Problems

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Niwatori-san's corner (Ohm's Law and Circuits sounds electrifying LOL)

Annyong-hi jumushyossoyo?
( Korean for Did you sleep well? / Good morning phrase in Korean)


Back to physics, sorry I couldn't make it early for my blog but I was reformatting my computer and it took long than I had anticipated. (Don't ask plz -___- )

Anywho Quoting Pacifico's last post just in the morning also, I want to say that the answers for the textbook questions from 2 days back are on Pacifico's scribe post so say thanks to Pacifico for his contribution.

Other than that a new set of new info was given by the title "2.0 Resistance, Ohm's Law and Resistivity"


It is an elaboration to Ohm's Law V / I = R

Adding Resistivity to the formulae of the Resistance R = p*L / A

where "p" is the proportionality resistivity constant, "L" is the Length of the resistance and "A" is Amperes or current flowing through the resistance.



For now ill close out Annyong-hi kashipshio ( Good bye in Korean =) )


Next scribe is Moofatt like pacifico said...
*rolls into a ball and summersaults to class*

Early Morning Unemotional Scribe Post.

What happened today:

• We went over the circuit questions from the "duck book" (p.670, #35, a. - c. if I recall correctly) that Miss Kozoriz assigned us to do. Then added in question number 35.d.
• And gave us handouts.

Because I don't have a scan/drawings of the circuit questions, I will just post the answer like how it was shown on the board.

p.670(?), #35 a. - d.

a.

	Volts (V)	Current/I (A)	Resistance (Ω)
Resistor1	24	38	8
R2	9	38	3
R3	21	38	7
RTotal	54	38	18

b.

	Volts (V)	Current/I (A)	Resistance (Ω)
Resistor1	9	2	4.5
R2	9	2	4.5
RTotal	9	4	2.25

c.

	Volts (V)	Current/I (A)	Resistance (Ω)
Resistor1	24.2	0.92	25
R2	2402	1.61	15
R3	25.8	2.58	10
RTotal	50	2.58	19.4

d.

	Volts (V)	Current/I (A)	Resistance (Ω)
Resistor1	75	3	25
R2	20	2	10
R3	5	1	5
R4	1	1	15
RTotal	9	3	21.67

For the Handouts, if you were absent, be sure to ask:

• 2.0 Resistance, Ohm's Law and Resistivity

• Grade 12 Electric Circuits: Resistivity Problems

• Table 14.4: Factors that Affect Resistance

• Fundamentals of Physics: An Introductory Course (Factors Affecting Resistance)

ANSWER: Grade 12 Electric Circuits: Resistivity Problems. We might go over them in class today.

Next scribe is:

Ryan MOOffatt

Scribe!!

Hey everyone it's me Aldrin B. scribing for yesterday's class...

At the start of the class Ms. K gave back our test papers..
woooo... scary!!XD.. She also gave us the answers....

refer to slide 5/25/09 posted by Ms K.

after that, we were asked to answer the problems on the handout she gave last friday.

---

Simple Circuits (#'s 1 & 2 a-f)

1. Three 10Ω resistors are connected in series to a 12-V battery.

a. Draw a diagram of the circuit.
(-----series circuit diagram----)to be posted

b. What is the total resistance of the circuit.
RT = 10Ω + 10Ω + 10Ω
= 30Ω

c.What is the total current in the circuit?
I = V/R
= 12V/30Ω
=0.4A

d. what is the potential drop across each resistor?
V= IR
=(0.4A)(10Ω)
=4V

e. What is the current in each resistor?
I=V/R
=12V/30Ω
=0.4A

f. What is the power dissipated by each resistor?
P=IV
=0.4A(4V)
=1.6W

2. Three 10Ω resistors are connected in parallel to a 12-V battery.

a. Draw a diagram of the circuit.
(-----parallel circuit diagram----)to be posted

b. What is the total resistance of the circuit?
1/R = 1/10 + 1/10 + 1/10
= 3/10
R/1 = 10/3 = 3.3Ω

c. What is the total current in the circuit?
I=V/R
=12V/3.3Ω
=3.6A

d. What is the potential drop across each resistor?
(-----------------------------------------)

e. What is the current in each resistor?
I=V/R
= 12V/10Ω
=1.2A

f. What is the power dissipated by each resistor?
P=IV
=1.2A(12V)
=14.4W

---

We then answered problems regarding compound circuits.??

well, thats all I remember. i'll put more information as soon as I get home.

*** i think we have an assignment about the problems on the big book***:)?

see you all in class!!!

Electric and Magnetic Fields Test

E And B Test

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Hi... ^^ uhm,, I forgot to scribe last Thursday.. I'm sorry ,, my bad.. oh well,, last Thursday... we did a lab about circuits... Basically, the purpose of that lab is for us to know the difference between the series cicuits and Parallel circuits.. We spent the whole class doing the lab and answering questions regarding it..

For Friday,, Ms. K. gave the answers for the study guide... (I'll post the answers tomorrow... ^^)
Also she gave us some worksheets to do as homework... Yeheyyy!!! (??) and we answered sheet 34-1 - 35-2 .... here it is..

that's it... have a wonderful weekend... see you guys on Monday...

next scribe is ALDRIN B. ^^,

hi...!!!!

In today's class:

-we watched a video (electrical current)

- jot down 6-7 info....

- received the following worksheets

-Direction of the electrical current

- transparency worksheet and..

- chapter 23 study guide

The following were the answers in the Transparency Worksheet

1.Which symbol represents a battery?Draw it in the space provided.

2. Draw the symbol thjat represent and instrument that measures current.

3. How would you show that an electric connection exist

4. if you illustrated a circuit that included a lamp dimmer ,what symbol you use for the resistor.

5. Draw a circuit that includes the following :a switch ,a battery, and a lamp

6.If you wanted to measure the voltage in the circuit in question 5,what symbol would you add?

That's what we all did for today's class......the next scribe will be arjeannelle..

Corrections...

hello....

Today in class Mrs. K corrected the last 6 questions (11-16) from the worksheet called " Moving Charges" and the corrections are on Mrs. K slide .

Also we corrected The path of a Charged Particle in a Magnetic Field and Review worksheets



Review...

1) R = mv/Bq
= (9.11x10^-31 )(2.0x10⁶ )/(8.0x10^-2)(1.6x10^-19)
= 1.42x10⁴ m

2) v = E/B
= 5.0x10^-4/3.0x10^-2
= 1.67x10⁶ m/s

# 3, 4, 9 and 10 you don't have to do it

5) v = BqR/m
= (5.0x10^-4)(1.6x10^-19)(0.03)/9.11x10^-31
= 2.63x10⁶ m/s

6) a.
v = BqR/m
= (3.0x10^-2)(1.6x10^-19)(1.5x10^-2)/1.67x10^-27
= 4.3x10⁴ m/s

b. E = V.B
= 4.3x10⁴/3.0x10^-2
= 1.3x10³ N/C

7. m = BqR/v
= (0.28)(1.6x10^-19)(0.30)/7.0x10⁵
= 1.92x10^-26 kg

8) for this question calculate the velocity using the formula below

v = BqR/m
= (0.030)(1.6x10^-19)(8.0x10^-3)/9.11x10^-31
= 4.4x10⁷ m/s

then substitute the velocity to the formula below to find the radius

R = mv/Bq
= (1.6x10^-27)(4.4x10⁷)/(0.03)(1.6x10^-19)
= 14.66 m

Reminder:
-test on tuesday... Good Luck!!

the next scribe will be KimC
have a great long weekend everyone :D

Moving Charges Questions 11-16

Moving Charges 11-16

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Moving Charges Questions 11-16

Moving Charges 11-16

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Just a random question

Hi ms. K ! I'm just wondering, is the research paper supposed to be one page double space, or one page single space?

Today. ;D

This is Ann here and I'm your scribe for today. Awesome.

Today we didn't do very much. Ms. K corrected a worksheet. It was called 'Moving Charges' or at least that's what it says on the slides that Ms. K posted up. Check your work and do corrections if you havent done them already.

Ms. K also gave out 2 worksheets today. Pick them up if you haven't yet.

Our research paper is due tomorrow. Hand it in, it's work lots of marks !

Reminder that our test on this current unit is on Tuesday. You have the whole long weekend to study physics. :D

The next scribe will be raminaR18. Have a nice time scribing tomorrow ;D

Moving Charges Question 5-10

Moving Charges 5-10

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Scribe Time!!

HEYYYYYYY!!!!!! It's me, Dion and I'm here to scribe!!!!!

First off in class we got back our lab sheets from, i believe it was tuesday we did them?....anywho.. here are the answers for it!

----------------------------------------------------------------------

Observations and Data

Level Charge (C) Voltage(J/C) Energy (J)

3v 4 3 12

6v 3 6 18

9v 2 9 18

12v 1 12 12

(You need to multiply charge by the voltage so the C's cancel out and your left with J.)

Analysis

1.  

2. 18J

3. Because there is more than one charge in that voltage level.

4. 6J because each charge falling to the 6V level gives off 3J each.

Applications

1. A larger 12V battery holds more charges than 2 small 9V batteries.

----------------------------------------------------------------------

Well that's it for that sheet we then went over the Moving Charges Worksheet. Here are the answers for that sheet!!

----------------------------------------------------------------------

1a. Because the drop has had 2 electrons removed it becomes a positively charged. Because of gravity and that this charged particle will move towards the negative plate, the drop will move downwards.

1b. There are 2 forces acting on this little drop, the electric field force and the force of gravity. To solve for it, we need to figure out each of these forces and because they are going in the same direction we simply add them together. 

Fg=mg=4.4x10^-14N Fe=Eq=9.5x10^-15

Fnet=Fg+Fe=5.35x10^-14N [down]

2. For this question, we need to solve for its final velocity. To do this we need to first find the force of the electric field. with that, we can then solve for acceleration of the electron. Once we have acceleration we can find the final velocity.

Fe=Eq=1.92x10^-16N a=Fe/m=2.1x10^14m/s^2

v2^2=v1^2+2ad=4.58x10^6m/s

3. For this question, we first need to figure out what the charge of 3 electrons is. Once we know this we can solve for the electric force and gravitational force acting upon it. Then just add them together and there is the force! Do not forget that the 2 forces act in different directions!!

q=(3)(1.6x10^-19)=4.8x10^-19

Fe=Eq=1.14x10^-14N Fg=mg=2.35x10^-13N

Fnet=Fg+(-Fe)=2.24x10^-13N[down]

The force is downwards because the force of gravity is the stronger force in this case.

4a. There are 2 forces acting on the sphere, gravity and electric field.

4b. Because the particle is stationary, and the force of gravity acts downwards, the sphere must be positively charged so it will go against the force of gravity causing it to remain motionless.

4c. To figure out the charge, we know both forces of gravity and electric forces are equal so we can rearrange the equations to make it do work for us. 

Fg=Fe mg=Eq

q=mg/E=1.6x10^-18C

4d. To solve for the number of electrons, simply divide the charge by the constant for the charge of one electron.

N=q/e=10 electrons

----------------------------------------------------------------------

We then received one more sheet called Magnetic Formulae! This sheet contains numerous equations and is basically a review from what we did in grade 11 physics. 

That's all we did for today so I'm off!! I choose the next scribe to be....Pacifico!

Moving Charges Questions 1-4

Moving Charges 1- 4

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yesterday we hand in our lab from monday. we also went over the Charged Particles in an Electric Field work sheet.

We had a new lesson yesterday which we went over the formulas before doing the worksheets.





next scribe is..................
i choose you ryan moffatt

Charged Particle in an Electric Field

Charged Particle In An Electric Field

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5/11/09

Hey guys!

So today, we finished correcting the worksheet "Electric Potential"

1) V = PE/Q
5.0x10^-18/1.6x10^-19
= 31V

2a) ( 9x10⁹)(2X10^-16)(2X10^-10)
= 3.6X10^-12
(3.6X10^-12)(0.01)
= 3.6X10^-14 JOULES
2b) V=PE/q
= 3.6x10^-14/2.0x10^-16
= 180V

3)Ek= qV
(1.6x10^-19)(1000 V)
= 1.6 x 10^-16J
Ek= 1/2 mv²
v= sqrt( 2(1.6x10^-16/9.11x10^-31)
= 1.9x10⁷ m/s
4) V= kQq/r
=9x10⁹ (3.0)(1.6X10^-19)/ 0.80
= 5.4 x10^-9
V= kQq/r
= 9x10⁹ (3.0)(1.6X10^-19)/ 0.5
= 8.64x10^-9

W= V2-V1
= 8.64x10^-9 - 5.4 x10^-9
= 3.24x10^-9 J

5a) W= qV
= (1.6x10^-19)(12)
= 1.92x10^-19J
OR
= (1.92x10^-18J/ 1.6x10^-19 )
= 12eV electronvolt
b) same amount = 1.92x10^-18J or 12eV

c) Ek= 1/2mv²
= sqrt( 2(1.92x10^-18/ 9.11x10^-31)
v= 2.05x10⁶ m/s

after we corrected the worksheet, we did a lab:

Chapter 21: Charges, Energy, Voltage

which is due tomorrow, so don't forget to hand it in.
Also don't forget to do your research paper, its due Friday !

George will scribe tomorrow, since he didn't do it last week =P

* correct my mistakes when you see some , please and thanks*

=)

Electric and Magnetic Fields

hi there, it me, lorenz. i am the scribe for wednesday and for some reason George still hasn't post his blog for tuesday...

Anyways, so for the people who were not there yesterday, we handed in out "Concept Map for Radiation" and we answered 3 questions for a "Health Physics Reflection". She gave us some notes on "Electric and Magnetic Fields" , "Coulomb's Law in One Dimension" and a "Chapter 32: Electrostatics Coulomb's Law Worksheet". I asked ms. Kozoriz to post the notes that she wrote on the smart board and i hope that will help you more.

Don't forget to hand in your Health Physics Research on Friday, May 15.

the next scribe will be Daureen Joy Nacario.

see you guys later!!!

=)

Electric Fields

Electric Fields

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Effects of Radiation on Humans ...

Hi guys, this is Nelsa, (finally) scribing for Friday's class. Whoot, everybody rejoice, hahaha.

ANYWAYS.


On Friday, we read chapter 3 in our Health Physics duotang, so.. you should probably do that if you haven't already. We answered the questions on pages 36, 39 and 41, which Ms. K posted the answers to on the previous post.

We read about a new formula on Friday too, I(d) = Ioe^-ud, which can be used to mathematically determince the amount of gamma radiation a substance will absorb. In the equation, Io stands for the original number of gamma rays (or incident intensity), I(d) is the number of gamma rays that don't pass through the substance, d represents the thickness of the substance and the coefficient u depends on the amount of energy the gamma ray carries. The higher the energy, the smaller the absorption coefficient.

Aaaand yeah, that's it. Don't forget to do your 'Sort and Predict' sheet, I think it's due today.


Next scribe will be.. George.

Cool.

'Kay bye.

Health Physics Problems: Chapter 3

Activity #3 Problems

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