June 13, 2011

In today's class we did pages 58, 43, 52 and 55

Page 58

1. true
2. false, thickness
3. false, closes
4. true
5. false, parallel
6. false
7. A

Page 43

1. 40 ohms
2. 3 A
3. no, the current 3A is less than 10A fuse
4. 40 ohms
5. 3A
6. No, 3A < 10A
7. 9.375 ohms
8. 12.8A
9. Yes
10. Increase value of one of the resistors

Page 52 -53

1. 2.5 ohms
2. 
   
3. 
   

b.)1 ohms ; 4W; 2A; 2V

2 ohms; 8W; 2A; 4V

c.)6 ohms; 6 W; 1 A; 6V

3 ohms, 12W, 2A, 6V

d.) 2 ohms; 4.5 W; 1.5A, 3V

2 ohms; 4.5 W; 1.5 A; 3 V

1 ohm; 9V; 3A; 3V

Page 55

1. b
2. c
3. d
4. a
5. a
6. b
7. b

June 9

Answers for circuits booklet

Page 48
1. 3A
2. 6ohms,0.5A
3. 12 ohms
4a. b
4b. a
4c. 4,6
4d. 5
4e. 4,6
4f. b
4g. b

Page 49
1a. 3A
1b. 6A
1c. All 3A EXCEPT for the two bottommost which the answer is 6A
2b. d
3a. 6V
3b. 3A,3A,6A
3b. 12A
3c. 0.5 ohms

Page 50
1.



2.



We also need to answer page 37,38 and 41

JUNE 8, 2011

In today's class we did a lab experiment from the green booklet about Circuits.
The materials that you need are:

• batteries
• alligator clip.
• light bulbs
• sockets

We did two kinds circuits. Thes are Series and Parallel Ciruits.

• This is one of the kinds of the circuits that we did;
  This one is called series circuit.
• With this kind of circuit if you loosen one of the bulbs espescially the one in between, all of the lights, turned off.
  
  
• Another type of series that we did is this one.
• It is called,parallel circuit.
• With this type of circuit, when you loosen up one of the bulbs. the current will still flow and the light did not turn off.
  
  
  
  
  The next blogger will be SAI.
  
  
  
  
  
  
  
  
  

june 6 2011

READ PAGES 15-20!!!

DO PAGES 48-50!!!

finish page 14

Answer keys:

page 29

1) P=v^2/R

= (3)^2/30

= 0.3 W

2) T=v/R

=3/30

=0.1 A

3)power varies inversely with resistance. as R of bulb increases the amount of power of the bulb increases

4) E(or W)=Pt

= (.3)(3600)

= 1000J

page 31

1. resistance- property of an object that determines how much current will flow through that object

factors- length of the wire cross sectional area temperature material the object is made of.

2. voltage is a potential difference in other words voltage is PE/ charge. current is the movement of charge due to potential difference. thus, voltage is nothing more than a difference in potential and does not necessarily create a current. given the proper conditions, voltage causes charge to move and when they do a current is created

3. R=v/I

since v is a constant and R is increasing, the circuit must be decreasing

4. they allow complex circuits to be graphically illustrated clearly and concisely

Electric Current Booklet!

Answers for pages 44, 45, 46

p. 44

1a. volts
b. voltage
c. current

2a. 1
b. 15
c. 1
d. narrow pipe,thin wire

p. 45

1. 1.5x10^-3 A
2. 4 A
3. 1.2 A
4. 4A
5. 1.5 x 10^-2 A -> 3.0 x 10^-2 A
6. 1000 Ω
7. 10 Ω
8. 100 V
9. Yes the resistance change, case 1: 240 Ω case2: 50 Ω

p. 46

1. R=pL/A, A=pL/R, A= (1.72 x 10^-8 Ω·m )(20 m) / (.10 Ω)
A= 3.44 x 10^-6 m^2

d= sqr(4A/π) d= 2.09 x10^-3 m

2. A= (0.25 x 10^-3 m)(1 x 10^-3m) A=2.5 x 10^-7 m^2

L= RA/p L=(1.5Ω)(2.7 x 10^-7 m^2) / (100 x 10^-8 Ω·m )
L= 0.38m

3. p=RA/L p= (0.04 Ω)(2 x 10^-6 m^2) / 5m
p= 1.6 x 10 ^-8 Ω·m

4. L= RA/p L= (3 Ω)(1.5 x10 ^-6 m^2) / (100 x 10^-8 Ω·m )
L= 4.5m

5. A = pL/R A= ( 55 x 10^-8 Ω·m )(.2m) / .10 Ω
A= 1.1 x 10^-7 m^2

d= sqr(4A/π) d = sqr((4 x 1.1 x 10^-7 m^2) / π)
d= 3.74 x 10 ^-4 m

NOTE:
Nichrome resistivity is 100 x 10^-8 Ω·m
Tungsten resistivity is 55 x 10^-8 Ω·m


We also need to answer the 2 questions on the bottom of page 14 on separate sheet of paper

The next one to scribe is ctc

June 2nd

Worked on Pages 44,45,46, and 48
Pages 5- 12

Notes:

An electric current is a flow of charged particles.
A conventional current is just a flow of a positive charge.
Charges can not be created or destroyed, but they can be seperated.
A resistor is a device designed to have a specific resistance.
Resistance is known as the ratio of potential difference to the current.
Temperature can increase resistance since a moving charge is affected by molecules, greater molecular motion would thus increase the resistance.

Formulas:

I = q/t
I is the conventional current measured in Amperes (A)
q is the charge measured in Coulombs (C)
t is the time measured in seconds (s)

P = IV
P is power measured in Watts (W)
V is the Potential Difference also reffered to as Voltage, it is measured in Volts (V)

R = V/I
R is the resistance, measured in ohms, the symbol omega (Ω) can be used to represent it.

R = pL/A
p is a constant(the resistivity of the metal) (Ω·m)
A is the crossectional Area (m^2)
L is the length of the material (m)

Useful Ratios to take note of:

R1/R2 = L1/L2
R1/R2 = A2/A1
R1/R2 = p1/p2

For P46

Nichrome wire resitivity at room temperature varies from 1.10 × 10−6 Ω*m to 1.5 × 10−6 Ω·m
Tungston wire resistivity at room temperature is 5.28 × 10−8 Ω·m

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.

Thursday, April 28

Hey mates(haha lame)... if you were away during the class, here is what you need to do:

• read page 28 - page 31 (The Exploration of Space booklet)
• Answer the question on pages 25 and 26 (answers at the back of the booklet)
• Answer page 32 (no answers at the back)

Ms. K also gave back the Activities 1-5 on the green book, we also derive an equations, and give us notes (check bellow).

Hey... last thing... if i could have Katrina to do the next scribe. thank you!

Law of Universal Gravitation

HEY!

for those who weren't in the class.. here's what we did. (''.)

STUDY GUIDE

8.1

Kepler's Law of Planetary Motion

• predict
• Earth
• sun
• 3
• ellipses
• sun
• areas
• fastest
• slowest
• periods
• cubes
• sun
• (TA/TB)^2 = (RA/RB)^3
• radius
• period
• radius

Universal Gravitation

• inversely
• square
• FG= GM1M2/ d^2
• force
• d
• constant
• doubled
• halved
• 1/4

Newton's Use of His Law of Universal Gravitation

• 2nd
• Kepler's

Weighing Earth

• Henry Cavindish
• Attraction
• Universal Gravitation
• 6.7x10^-11 N.m^2/kg^2

8.2

Motion of Planets and Satellites

• parabolic
• vertical
• horizontal
• horizontal
• Earth
• orbit
• Uniform Circular Motion

Weight and Weightlessness

• acceleration
• inverse
• 2nd
• decreases
• freefall
• upward
• weightless

The Gravitational Field

• gravitational field
• gravity
• inversely
• square

Einstein's Theory of Gravity

• force
• space
• curved
• accelerated
• relativity
• sun
• black hole
• mass

8.1 page 17

1. true
2. true
3. false, mathematical
4. true
5. false, sun
6. false, area
7. false, ratio
8. false, first and second only
   
9. they are equal
10. point 1
11. period and radius of another moon
12. c.
13. a.
14. d.
15. b.
16. 2F
17. 2F
18. 4F
19. 1/4F
20. 4F

We also need to answer pags 19&26!

-patrickd.

Tuesday April 26 2011

Today, we study the new formula F=Gm1m2/d^2.
And also we go over pg11, 12, 23, 24 in grey booklet.
If you are not finish the Study Guide, so go ahead in pg15, 16

Monday, April 25, 2011

Hi guys! If you miss today's class here are the things that we did.
In the beginning of class Ms. K let us watch videos explaining Kepler's Law of Planetary Motion. If you're unfamiliar with Kepler's three laws, refer to your gray booklet in pages 6&7.

After we watched the videos, we went over Transparency 7-1 in page 21-22. Here are the answers to the questions:
1. gravitational force
2. elliptical
3. at one focus of Earth's elliptical orbit
4. at the point closest to the sun
5. at the point farthest from the sun
6. area(1)=area(2)
7. T(1)=T(2). Since the Earth's velocity varies, the time it takes to sweep out equal areas along its orbit is the same.

Lastly, for tomorrow's class we are to finish pages 15-18.

Hello grade 12's!
During Thursday's class, there were instructions given to us on the board and they were:
1. Finish the orbit lab on page 20 in your grey booklets. (It was due on Thursday at the end of class.)
2. Finish page 22 in the grey booklet.
3. Do problems 1-4 on page 160 in the green textbook. (Answers are at the back of the textbook)
4. Do problems 1-5 on page 172 in the green textbook. HAND IN ON MONDAY!

Here are the questions that were assigned from the textbook:
Questions on page 160:

1. An asteroid revolves around the sun with a mean (average) orbital radius twice that of Earth's. Predict the period of the asteroid in Earth years.
Answer: 2.8 years

2. From Table 8-1, you can calculate that, on the average, Mars is 1.52 times as far from the sun as is Earth. Predict the time required for Mars to circle the sun in Earth days.
Answer: 684 days

3. The moon has a period of 27.3 d and has a mean distance of 3.90 x 10^5 km from the centre of Earth. Find the period of an artificial satellite that is 6.70 x 10^3 km from the centre of Earth.
Answer: 88.5 minutes

4. From the data on the period and radius of revolution of the moon in Practice Problem 3, find the mean distance from Earth's centre to an artificial satellite that has a period of 1.00 d.
Answer: 4.30 x 10^4 km

Questions 1-5 on page 172 (Don't forget that these questions are for marks!)

1. Jupiter is 5.2 times farther than Earth is from the sun. Find Jupiter's orbital period in Earth years.

2. Uranus requires 84 years to circle the sun. Find Uranus' orbital radius as a multiple of Earth's orbital radius.

3.Venus has a period of revolution of 225 Earth days. Find the distance between the sun and Venus, as a multiple of Earth's orbital radius.

4. If a small planet were located 8.0 times as far from the sun as Earth, how many years would it take the planet to orbit the sun?

5. A satellite is placed in an orbit with a radius that is half the radius of the moon's orbit. Find its period in units of the period of the moon.

Enjoy!