May 31

Today we were given our exam reviews and also started our new unit on electricity.

We preceded the unit with a lab; using a light bulb, wire and a 'big' battery, we were to find a few ways on how to light the bulb, and a few ways where we couldn't. We were to illustrate the circuit diagrams in our hand-out booklet on the back page of the lab.

Within the booklet, we were to read pages 5 to 12 and were to do worksheets 44,45 & 46.

The current, I, is measured in C/s and is represented by the unit A, amperes.

Potential Difference, V, is measured in J/C.

Power, P, is measured in J/s and is represent by the Watt, W.

I=q/t V=E/q P=IV E/t=IV

P=(q/t)(E/q)=E/t

P=(1 C/s)(1 J/C)=1 J/s

Please feel free to correct.

May 30 - Today's Class

In the beginning of class, we went over the questions from the "duck" book on page 585. The answers are below.

82.

a)

q= + 2e-

M = 6.69x10^(-27)kg

V1 Vertical = 0 m/s

V1 Horozontal = 6.0x10^(6)m

v = 500V

a = Fe/m

= qE / m

= qV/mdv

= (2(1.6x10^(-19)c)(500N))/((6.696x10^(.27)kg)(3.0x10^(-2)m))

= 7.97x10^11 m/s ^squared

dV = 0.03m

dH = 0.15m

t = dH/vH

= (0.15m)/(6.0x10^(-8))

= 2.5x10^-8 s

Therefore...............

change in dH = 1/2at^squared

= 1/2(7.97x10^(11)m/s^SQAURED)(2.5x10^(-8) s)^SQUARED

= 2.5x10^-4m or 0.025cm

3.0cm-0.025cm = 2.975cm from negative plate

*NOTE THIS QUESTION WILL NOT BE ON THE TEST*

b)

V2V=V1V+at

= 0 + (7.97x10^(11)m/s^SQUARED))(2.5x10^-8)s)

= 2.0x10^(4)m/s

V =_/(6.00x10^(6)m/s)+(2.0x10^(4)m/s)^SQAURED

= 6.0x10^6 m/s

83)

V = 39(N/C)(m)

D = 0.050m

V = Ed

E = V/d

= 39(N/C)/(0.050m)

= 7.8x10^2 N/C

85)

a)

Fg = Fe

Mg=Eq

mg/q = E

(4)(m)(g)/(2)(q) = E

E = (2(1.67x10^(-27))(9.8m/s^SQAURED))/(1.6x10^(-19)c)

= 2.04x10^-7 N/C

b)

V = (d)(E)

= (.03m)(2.04x10^(-7)N/C)

= 6.12x10^-9 V

86)

E = (v)/(d)

= (90V)/(0.12m)

= 766.67 N/C

87)

d = v/E

= (1200)/(450N/C)

= 0.27m

90)

a)

((m)(q))=((q)(E))

E = (V)/(d)

q = ((2.2x10^(-15)kg)(.0055m))/(280N)

= 4.2x10^-19C

b)

N = c/q

= (4.2x10^(-19))/(1.6x10^(-10))

= 2.6e-

= 3 e-

Today we also got a review for the exams. Questions like 83 and 85 will most likely be on the test this wednesday. STUDY!!

may 25, 2011 :D

today we did the work sheet on moving charges numbers 5-16

answers:

5) 1.33 x 10 ^-3 T

6) 8.0 x 10 ^-5 T

7) 0.011 N

8) 16.667 T [<--]

9) 1.7 x 10 ^-14 N [east]

10) 5.0 x 10^7 T [into page]

11) 4.8 x 10^-19 C

12) a. 5.12 x 10^-14 N

b. 2.8 x 10^-4 m

13) 4.39 x 10^7 m/s

14) 5.3 x10^5 m/s

15) 1.3 x10^8 m/s

16) 360 V

also read pages 31-36

and answer pages 51-52

on the pink booklet(aka electric and magnetic fields)

:D

May 20, 2011

At the beginning of the class on Friday, we went over the answers for the questions on pages 45-47 as well as those on pages 49 and 50.

Here are the answers to those questions:

Pages 45-47:
















Pages 49 and 50:

















As well, don't forget to finish the yellow worksheet: Moving Charges Worksheet.
The next scribe is Arveen.

May 18th, 2011

Hi class today Mrs. Kozoriz was away and we had a sub. The instructions Mrs. K left for us were that we were to read pages 36-40 in our pink Electric and Magnetic Fields booklet and complete the problems on pages 45 and 47 as well as complete pages 49-50. Also if you did not yet finish the Charges, Energy and Voltage Lab you were to also work on it in class.

May 17th, 2011

Hi Class, today we went over pages 42 and 43 in the pink Electric and Magnetic Fields booklet. The answers on page 42 are as follows:


1. PE decreases, KE increases.


2. Similarly, a force pushes the charge closer to the charged sphere. The work done in moving the test charge is the product of the average FORCE and the DISTANCE moved. W = F x d. This work INCREASES the PE of the test charge. If the test charge is released, it will be repelled and fly past the starting point. Its gain in KE at this point is EQUAL to its decrease in PE.


3. Electric PE/CHarge has the special name Electric POTENTIAL.


Since it is measured in volts it is commonly called VOLTAGE.


4. 1 Volt


5. 12 Joules


6. 5000 Volts


7. 5000 Volts


8. 5000 volts


9. 0.005 Joules


10. CHARGE

We also got and worked on a lab called Charges, Energy , Voltage. If you weren't there for the lab please go see Ms. Kozoriz for the lab sheet and complete the lab. The lab setup is on page 430 in the Green Textbook.

MAY 16 - Today's Class

At the begginning of class, we were asked to open the new pink bookleet titled Electric and Magnetic Fields and turn to page 13 which we went over the question. The answer for this question is:

F1 on B = (k)(qA)(qB)/R^2
= (9x10^9)(4.5x10^-6)(-8.2x10^6)/(0.040m^2)
= 208 N {L}

DISTANCE BETWEEN THE OTHER TWO CHARGES IS:

_/ means sqaure root

C = _/(0.040m^2)+(0.030m^2)
= 0.050m

" means inverse

Oi = tan"~0~(0.030m/0.040m)
= 37 degrees

F2 Fc on B = (k)(qC)(qB)/(R^2)
= (9x10^9)(8.2x10^-6 c)(6.0x10^-6 c)/(0.040m^2)
= 177 N at 217 degrees from positive x-axis

~0~ means theta
dg means degrees

COMPONENTS OF F1 are:

F2 x = (F2)(cos~0~)
= 177 N(Cos~0)(~217dg)
= 142 N {L}

F2 y = (F2)(Cos~0~)(Sin~0~)
= (177 N)(Sin~0~217dg)
= 106 N {Down}

COMPONENTS OF NET (Resultant Force) FORCES ARE

Fnet x = -208 N - 142 N
= -350 N or 350 N {L}

Fnet y = 106 N

Then...............

Fnet = _/(350 N^2)+(106 N^2)
= 366 N or 350 N {L}

~0~ 2 = Tan" (106 N)/(350 N)
= 17 dg below the negative x-axis

Fnet = (3.7x10^2) at 197dg from positive x-axis

Then in class we were asked to read pages 14, 15, 17, 18, 19, 20. The things that was important were the things in bold. The pages also went on a previous lessons about work. Remeber that potentional energy increases if the height of an object increases creating a bigger fall. Then we were asked to do questions in the booklet pages 42, 43 but we didn't go over them. Tommorow we will I think.

MAY 4 :D

HI GRADE 12s! :D

so today, we just take over the questions on pages 27, and 36-37

~on page 27, the answers are already given.

Here the formulas used for each question to find the answer:

#1 & 3 K = R^3 / T^2 -> grey booklet page 13

#2 (Ta/Tb)^2 = (ra/rb)^3 -> grey booklet page 7

#4-7 F=G(m1m2/r^2) ->grey booklet page 11

#8 & 9 T=2pi ­(square root of r^3/Gm) -> blue booklet page 5

~answers on page 36-37

1.) 180N

2.) 3.5x10^6m

3.a.) m=25kg

b.) 230N

4.a.) 546N

b.) 490N

5.a.)2.6x10^-9N

b.) 3.9x10^-11 m/s2

6.) 3.9x10^-11m/s2

7.a.) 1470 N

b.) 160 N

c.) 4500m/s

and we are going to take over pages 10 and 11 tomorrow, make sure you've done it.. :D see yeah later peeps! :D

Monday, May 02

Hey, grade 12's. If you were not in class today, this is what we do:

First, we watched a video about Satellite Motion. We took down 5-7 facts about the movie. Also we handed our facts to Ms.K

After the movie, we went over the questions on page 32 in our grey booklet - Grade 12 Physics: Fields Exploration of Space. Here are the answers:

1. PE= -3.13x10^10 J

2. ET=EK+EP

=-2.36x10^9 J

3.∆Etotal=∆Etotal(in orbit)-∆Etotal(on earth)

=-2.36x10^9 J - (-3.13x10^10 J )

= 2.89x10^10 J

4. Etotal(in orbit) =-2.36x10^10 J ; you would need to put in 2.36x10^9 J of energy to escape earth's orbit

5. * R must include Re+ Rsatellite

PE= -1.18x10^11 J

Although there would be no class tomorrow due to a staff meeting, we would be going over the questions on REVIEW: CHAPTER 8 found on page 36-37 in the grey booklet.