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 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.