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.

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.