Monday, July 6, 2009

Thank you

thanks ms.kozoris for such a splendid semester!!..yeah i know school ended awhile ago.

Wednesday, June 10, 2009

6/10/09

Hello everyone =) in today's class, we went over the physics exam review "MECHANICS".
Things to know:
- question #8 will not be on the exam=)
- when solving question #13, you must know vertical / horizontal velocity
- use newtons 2nd law for question #14
- both question 15 & 16 are momentum questions
- remember for question 17 , when force and distance are the same direction, no work is done
- know work , force, distance for question 18
- question 19 is energy over time
- question #20 , add the areas together
- question #21 is potential energy
- question #22 is kinetic energy
- use newtons second law for question 23
- for question #24 , use hooks law
- question 25 is a spring potential energy question

In the past few days this week, we corrected our last test. You will find the corrections on the blog. We also went over some of the exam review questions.

Since a lot of people will be away tomorrow, we will go over the fields question on Friday =)
have fun studying=)

Oh and Jasmin will scribe tomorrow =)

Thursday, June 4, 2009

June 4 2009

Hello everybody.
This is Hao. : )
We talk about the answers of the phyz book ( green book) today.

P532
1. EMF=BLV=(4*10^-2 T)(31m)(15m/s)=18.6V

2. EMF=BLV=(0.5 T)( 20m)(4m/s)=40V

3. EMF=BLV=(4.5*10^5 T)(75m)(950000m/3600s)=0.89V

4. a. EMF=BLV=(0.3T)(0.75m)(16m/s)=3.6V
b. I=V/R=3.6V/11Ω=0.33A

5. EMF=BLV=(0.32T)(0.4m)(1.3m/s)=0.17V
I=V/R=0.17V/10Ω=0.017A

14. a. X/150 = 625V/120V
X= 781.25 turns
b. X/150 = 35V/120V
X= 43.75 turns
c. X/150 = 6V/120V
X= 7.5 turns

15. a. X/120v = 1200/80
X=1800V
b. Pp = Ps = (I s)(V s) = (2A)(18000V)=3600w
Ip=Pp/Vp =(3600w)/(120v)=30A
c. Pp=Ps =(Is)(Vs)=(2A)(1800V)=3600W

16. a. X/475 = 9V/120V
X=35.625
b. Ps=(Vs)(Is)=(9v)(0.125A)=1.125W=Pp
Pp= (Vp)(Ip)
Ip=Pp/Vp=1.125w/120v=9.375*10^-3A

P533
20. a. Pp=Ps=150W= (Is)(Vs)=(5A)(Vs)
Vs=150W/5A=30V Bigger than 9V
So, step up
b. 30V/9V = 10/3

21. Pp=(Ip)(Vp)=(3A)(12V)=36W=Ps
Vs=Ps/Is=36W/0.75A=48W
Tp/Ts = Vp/ Vs
Tp= (Vp/Vs)(Ts)=(12V/48V)(1200)=300 turns

we will have unit test tomorrow. good luck everybody......

Electromagnetic Induction Problems

June 3 2009

I was so ready to go to sleep.. not only did it slip my mind, but it flew out of it, crashed into a garbage can, met a homeless person, made friends with a mime, took a taxi but, got hit by a train, broke a leg, joined the paraolympics, won silver losing to a Japanese woman, retired, got married to a platypus, discovered a world in a wardrobe, and finally found its way back to my head, but had a fight, argued for days, went out for dinner, asked to come back home, ate a lobster and finally decided to finish this scribe post together.

Answers to Chapter 26 Review

1.EMF=Blv
EMF=(4x10^-3T)(0.05m)(2.5m/s)
EMF= 5x10^-4 V

2.EMF=Blv
EMF=(4x10^-5T)(2m/s)(0.07m)
EMF=1.12x10^-5 V

l=v/R
l=
1.12x10^-5 V/0.5
l=2.2x10^-5 A

3.CC is down EC is up

6. South ^ ^ ^ North

8.a)V=P/l
V=1500 Watts/175 A
V= 8.57

9.a)
Np/Ns=Np/Vs
1/2500=1200/Vs
Vs=300 000

b)Pp=Ps
Pp=IpVp
Pp=15 A x 120 V
Pp=18 000 W

Ps=IpVp
18 000=(Ip)(300 000)
Ip=0.06 A

c)p=IV
p=(300 000)(0.06)
p=18 000 W
-----------------------------------------------------

"A transformer does not create or destroy energy, a transformer only converts voltage and current"
- Ms.Kozoriz

We completed Chapter 37: Electromagnetic Induction Transformers, Ms.Kozoriz has posted the answers on the blog rendering it useless to give you the answers myself.

She presented to us a copper coiled magnetic flashlight (CCMF for short) using the power of inserting a magnet into a copper coil.. that`s as far as i can explain.


If you must study for the Physics Test, a list of problems are provided in the green book to help prepare you for the test:

Ch.25

Reviewing Concepts: 1, 2, 3, 10, 14 p.530
Applying Concepts: 13 p.531
Problems 1-5, 14-16, 20, 21 p.532-533