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!

Friday, April 15th 2011

Today, we did the lab in Physics class. The lab is in pink booklet page 37.
That's everything we did on Friday.

Wednesday, April 13th 2011

Heyy class for Wednesday's class we went over the answers on Page 24 CHapter 11 Study Guide in the Pink Booklet. The answers are as follows: FORMS OF ENERGY: itself environment kinetic work position shape form DOING WORK TO CHANGE KINETIC ENERGY: mass velocity 1/2mv2 joule increases net change same increases opposite decreases POTENTIAL ENERGY: potential kinetic kinetic potential energy kinetic potential zero kinetic potential inceases potential kinetic mgh (formula) height reference level constant After we went over those answers we went over the answers for the questions on page 237 in the Green Book. After that we did questions 20. b) and c) , 21, 22 and 23 and handed them in. That's everything that we did on Wednesday. Thank you. --janna k
Wei to scribe for Friday's class.

Friday, April 8

In class we looked over the Review from yesterday as well as the the questions on p 213 in the green text book.

Questions 1-5, and 17 -19

1.
W= F*D
W= 80N * 10m
= 800J

2. d= 8m
F = mg
= 150kg * 9.8m/s2
=1470N
W=FD
=1470N*8m
=11760J

3.
w=176J
d= .300m
F=W/D
= 176/.3
= 586.67

m = 586.67/ 9.8
= 59.9kg

4.
W=FD
= (2.25 x 10 3N)(7.50)
= 1.60 x 10 4 J

m=dv
=( 2.00 x10 3 kg/m3)(1.15m3)
m=2.30 x 10 3 kg
Fg=mg = ( 2.3 x 10 3 kg)(9.8m/s2)
= 2.25 x10 3 N
totally Fg = 2.25 x 10 4 N + 2.25 x 103 N
= 2.48 x10 4 N

5.
W=FD
=(3N)(.16m)
= .48J
total .080J

W= 1/2 FD
= 1/2( 4N)(.10m)
=.32J

answers for 17-19

17
a. W= 5527.7J
b. 0 work is done.
c. W= -5527.2J
d. No, doesn't exert the force
e. W = 2352.0N
P = 2.2 x 10 3 watts

18.
a. W= 9 x 10 3 J
b. p .3 x 10 3 W

19.
a. P= 3.44 watts
b. P= 696 watts

lab was also due today.

04/07/11

stair walking lab due on friday april 8 :D

answer problems 1-5,17-20 (p 213-215) on green text book

and do study guide of page 10 of pink booklet (work and energy)

Friday-March 25 Working on Circular Motion Questions

In today's class, we went over the "Momentum and Projectile Motion Test". The answers were posted in the next post. Then, we had the whole class time to do the "green booklet" page 21 and questions 1-6 in the big text book, page 206.




Page 21



* We will be having a test after spring break*

Have a nice break.

Momentum and Projectile Motion Test

Momentum test

View more documents from ekozoriz

Thursday, March 24 -Working on Centripetal Force Lab

For today's physics class we continued on working on the Centripetal Force Lab. It seems as though most people finished gathering their data and started to answer This lab shows the relations between centripetal force and speed. After writing down the data from the lab procedures, we begin to answer the questions stated in the booklet. The accuracy of the data might produce a variety of different results for this lab, however they should all be relatively close. It is imperative to keep close attention to the lab beforehand so that answering the questions will be easier.The data of the graphs provided us with the information to conclude that there is a power relation between force, speed an speed squared. Even though the shape of the two graphs are different they both illustrate that there is a power relation between their variables.

Question 1 and 3 of the booklet tells us to graph our data, with force as our independent variable and speed or speed squared as our dependent variable.

Questions 2 and 4 asks us what the relationship between the two variables are. It is shown that speed and speed squared have a power relation to force.

Question asks how if the formula Fc = mv^2/r supports the data we have gathered from the experiment. Since both mass and radius does not change it is clear that the formula is supposed to support the gathered data.

Also we were supposed to work on page 17 and 18, with the answers posted on page 25
and pages 21 and 22. Finishing all of these would be good practice and also would be good so that there will be lessened homework during spring break. That is all I have to say for this scribe.

Wednesday March 23 2011 - Centripetal Force

In today's class, we were instructed to carry out our Centripetal Force labs. I'm unable to post any data which my partner and I received during the lab as we're suppose to perform the procedure on our own and also because the way the lab is performed may cause differences in data from group to group. I highly suggest that any of those who were unable to make it to class and could not do the lab, do their best to finish this particular one because it's fairly straight forward and easy to record and also very enjoyable. Another suggestion is that before going ahead with the lab, read it through and record the measurements you're using as they are important for calculations concerning time and speed. Also remember to look at the "Interpretation" questions on the next page as they will require the plotting of a graph and comprehension of the obtained data.

*Note
When carrying out the lab be sure to keep a constant distance with the stopper attached to the string. This is indicated by the alligator clip. Use that specific measurement for each time you go and change the number of washer. Be sure to record this measurement.

There's not much more I can say for this scribe but to just get the lab done and to finish the homework in the green booklets. It's almost spring break and I think everyone would like little or no homework for the break so do as much as possible.

Circular motion

During our class we red pages 141 - 142 on the green book about circular motion and answered page 14 on the green circular booklet. The answers on the fill in the blanks activity are in the list bellow as ordered.

• Constant
• Changing
• Changing
• Constant
• Perpendicular
• Tangent
• Instantaneous velocity
• Length
• Direction
• Radius
• Center
• Centripetal
• Directly
• Inversely
• Second
• Force
• Centripetal force
• Straight
• Tangent

Next we answered page 7 Free-Body Exercise: Circular Motion (Attached), and the back of the book page 28 (Attached)

Answers are at the back of the booklet. (Attached)

Answers page 27

Page 28

After all the answering (and guessing) we red pages 3 - 5 (Describing Circular Motion and Centripetal Acceleration). On page 5 we derive a equation...

There is no lessons tomorrow because of the dance, so we are expected to answer pages 15, 16 and 20. Also we are going to talk about circular force on Wednesday this week and as Ms.K promise us we are watching the amusement park video also on Wednesday.

Jaryl to scribe for Tuesday

Projectile Motion continued

During today's class we went over the ivory booklet pages 10, 15,17,18,23, and 24.
On page 10: Tossed Ball Problem
The answers are as follows:
20 m/s
10 m/s
5 m/s
5 m/s
11.2 m/s
20.6 m/s
30.3 m/s

On page 15: Transparency 6-1 Worksheet
The Trajectory of a Projectile
1. the highest point of the projectile has the smallest magnitude because ate this point vertical velocity equals zero.
2. their the same in magnitude but different in directions.
3. their the same in magnitude and directions.
4.


On page 17: Transparency 6-2 Worksheet
Vertical and Horizontal Projectiles
1. they are the same
2. they are the same
3. the horizontal motion does not affect the vertical motion
4. it is because of the vertical spacing between the ball's motion is changing
5. it's because the horizontal distance between the red ball's motion are the same

On page 18:
Projectile Motion I


On page 23:
Projectile Motion II


On page 24:
Chapter 7 Review


Don't forget that the "On Target" lab is due tomorrow.

projectile motion

For those who were not in class today.. we have discussed about Projectile Motion.

for your reference, just refer to the pics posted. :]

Test on Friday! :]

..next to scribe is Marinhel S.

Physics Lab Worksheet ; On Target

For those who were not in class today we did our 'On Target' lab which can be found in your ivory booklet Projectile Motion page 19. We got into our groups and built a target launcher that would be able to fling a ping-pong ball a distance of 1 meter. We were able to use anything we could think of that would help us with this lab. Once we built our target launcher, we started to fill in our data tables. Although, it was kinda difficult, most of us groups got our data tables filled.

I think the next step would probably answer the 'Analyze' and 'Conclude and Apply'.

patrickd to scribe next.

Momentum

Sorry for the late post.

If you missed class last Thursday, we just went over the questions in the momemtum booklet. We went over on questions on page 9, 17, 21, and 22.

Page 9
1. 6.3 m/s
2. -249 m/s
3. -0.884 m/s
4. 2.5 m/s
5. 540 N

Page 15
15. C
16. C
17. D
18. B
19. A
20. C

Page 21
1. 1600N
2. 1.3x10-3 s
3. 6.7x103 N
4. 0.005 m/s in direction of 65kg skater
5. 8.8 m/s westward

Page 22
6. 81 m/s
7. 0.058 m/s
8. ?
9. ?

Answers are in sig. fig.

 A reminder to hand in test corrections.

MOMENTUM

at the beginning of the class Ms. Kozoriz show us a video and then we answered the questions in the booklet.

After that, we read p. 189-191 of the greem book and the problems 13,14,15

PROBLEMS:

13. A 1325 kg car moving north at 27.0 m/s collides with a 2165 kg car moving east at 17.0 m/s. They stick together. draw a vector diagram of the collision.

14. A 6.0 kg object, A, moving at velocity 3.0 m/s collides with a 6.0 kg object, B, at rest. After collision, A moves off in a direction 40.0º to the left of its original direction. B moves off in a direction of 50.0ºto the right of A's original direction.

a. draw a vector diagram the momenta of the object A and B after collision.

b. what is the velocity of each object after the collision?

15. A stationary billiard ball, mass 0.17 kg is struck by an identical ball moving at 4.0 m/s. After the collision, the second ball moves of at 60.0º to the left of its original direction.The stationary ball moves off at 30.0º to the right of the second ball's original direction . What is the velocity of each ball after the collision?

Impulse and Momentum

At the beginning of the class, we went over the answers for the worksheets assigned yesterday.
Here are the answers for those worksheets:

Page 6
1. twice
2. twice
3. the same as
4. a) the same
b) the same
c) the same
d) bug
e) the bug of course!

Page 15















Later in the class, we took down notes on how to derive the impulse-momentum equation from both the 2nd law and a graph.










This equation can be used to solve problems such as finding time and the force exerted.











Unit for impulse is N x S
Change in p is measured in
kg x m/s







We also took down notes on the different types of collisions.

















*Remember, in an elastic collision, the objects do not stick together after the collision, but bounce off each other. In an inelastic collision, the objects stick together and move as one after the collision.

Remember to finish page 7, 8 and 16 in the booklet.

The next scribe is Sarika.

Dynamics Worksheet

Today, we did answer the dynamics worksheet. here are the answers for the dynamic worksheet and the questions itself.

The dynamics test will be on Monday, February 28, 2011

The next to scribe is goldenfalcon

Frictional Forces

ms. K gave us this sheet, titled Frictional Forces, we did answer the questions on it and the third and fourth images are the answer key...

Inclined Plane

Today, four groups answered the questions on page 196, 1 number per group. But only number 3 was solved on the white board. (Second image)
Ms. K also gave an example of a diagram about Forces acting on an inclined plane. (Third image)
We also answered the questions on page 18 of the booklet for Dynamics Unit. The answers are on the back of the booklet.








**We will do Lab experiment about inclined plane tomorrow

The next person to scribe is kun

(continuation) static equilibrium problems

Today, we read lessons regarding frictional force, normal force and inclined planes from the textbook. There's an assignment on p. 196 numbers 1-4. Ms. K said that the we will a do lab on Tuesday about inclined planes.

We also answered the static equilibrium problems # 3-5.

**If you're given the mass of the object, solve first for the weight.

**Draw your triangle with its corresponding angles, then solve for the unknown side which is Ft(tensional force) in this case. Solve it by using trigonometric identies.

The next scribe is Sai

Physics Lesson for 2/17/11