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Iowa State - CS - 611
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Iowa State - CS - 611
%!PS-Adobe-2.0 %Creator: dvipsk 5.58f Copyright 1986, 1994 Radical Eye Software %Title: paper.dvi %Pages: 18 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: Helvetica-Bold Times-Italic Times-Roman Helvetica %EndComments %DVIPSCommandLine
Iowa State - CS - 611
%!PS-Adobe-2.0 %Creator: dvips(k) 5.86 Copyright 1999 Radical Eye Software %Title: layers-tr.dvi %Pages: 43 %PageOrder: Ascend %BoundingBox: 0 0 612 792 %DocumentFonts: Times-Roman Times-Italic Times-Bold Courier %+ Times-BoldItalic Helvetica %Docume
Iowa State - CS - 611
%!PS-Adobe-2.0 %Creator: dvips(k) 5.92b Copyright 2002 Radical Eye Software %Title: disc.dvi %Pages: 15 %PageOrder: Ascend %BoundingBox: 0 0 596 842 %DocumentFonts: CMBX12 CMBXTI10 CMR10 CMR7 CMMI7 CMR6 CMR9 CMSY9 CMTT9 %+ CMBX9 CMTI9 CMTI10 CMMI10 C
Virginia Tech - PHYS - 2306
35.8: a)For the number of antinodes we have: m mc m(3.00 108 m s) sin = = = = 0.2317 m, so, setting = 90, d df (12.0 m)(1.079 108 Hz) the maximum integer value is four. The angles are 13.4, 27.6, 44.0, and 67.9 for m = 0, 1, 2, 3, 4. (m +
George Mason - ISA - 774
GMU ISA 774 Intrusion Detection Spring `06 Homework 1 Due date March 7, 2006, via email to me and the TA. Subject line of email should start "ISA 774 HW1" 1. Briefly describe the differences between anomaly detection and misuse detection 2. Briefly
Virginia Tech - PHYS - 2306
35.12: The width of a bright fringe can be defined to be the distance between its two adjacent destructive minima. Assuming the small angle formula for destructive interference (m + 1 ) 2 ym = R , d the distance between any two successive minima is
Virginia Tech - PHYS - 2306
35.16: (a)Dark fringe implies destructive interference. 1 d sin = 2 624 10 9 m d= = = 1.64 10 6 m 2 sin 2 sin 11.0 (b)Bright fringes: d sin max = mmax 1.6410 m The largest that can be is 90, so mmax = d / = 624109 m = 2.6 Since m is an int
Virginia Tech - PHYS - 2306
35.23:a)The distance from the central maximum to the first minimum is half the distance to the first maximum, so: R (0.700 m) (6.60 10 7 m) y= = = 8.88 10 4 m. 4 2d 2(2.60 10 m) b)The intensity is half that of the maximum intensity when you are ha
Virginia Tech - PHYS - 2306
35.24:a) =c 3.00 108 m/s = = 2.50 m, and we have: f 1.20 108 Hz 2 2 (r1 r2 ) = (1.8 m) = 4.52 rad. = 2.50 m 2 2 4.52 rad = 0.404 I 0 . b) I = I 0 cos = I 0 cos 2 2
Virginia Tech - PHYS - 2306
35.40:Destructive interference occurs 1.7 m from the centerline. r1 = (12.0 m) 2 + (6.2 m) 2 =13.51 m r2 = (12.0 m) 2 + ( 2.8 m) 2 =12.32 m For destructive interference, r1 r2 = / 2 = 1.19 m and = 2.4 m. The wavelength we have calculated is the
Virginia Tech - PHYS - 2306
35.53: a)There is a half-cycle phase change at the glass, so for constructive interference: 1 x 2d x = 2 h + x = m + 2 2 2 21 x 2 + 4h 2 x = m + . 2 Similarly for destructive interference: x 2 + 4h 2 x = m.b)The longest wavelengt
Virginia Tech - PHYS - 2306
21.2:current 20 ,000 C s and t 100 s 10 Q = It = 2.00 C Q ne 1.25 10 19. 19 1.60 10 C4s
Virginia Tech - PHYS - 2306
21.36: a) x b) vv01 2 at 2at2(0.0160 m) (1.67 10 27 kg) 1 eE 2 t E 2 mp (1.60 10 19 C) (1.50 10 6 s) 2 eE t 2.13 10 4 m s . mp148 N C .
Virginia Tech - PHYS - 2306
21.37: a) tan c) tan111.35 0 , r 2^ jb) tan112 .2 ^ ,r 42^ i 22^ j 22.6 1.10^ 1.97 radians 112.9 , r^ 0.39i0.92 ^ (Second quadrant). j
Virginia Tech - PHYS - 2306
21.43: a) Point charge q1 (2.00 nC) is at the origin and q 2 ( 5.00 nC ) is at x 0.800 m. k | q1 | k | q2 | 575 N C right. i) At x 0.200 m, E 2 (0.200 m) (0.600 m) 2 k | q2 | k | q1 | 269 N C left. ii) At x 1.20 m, E 2 (0.400 m) (1.20 m) 2 k | q1 | k
Virginia Tech - PHYS - 2306
21.58: a) Since field lines pass from positive charges and toward negative charges, we can deduce that the top charge is positive, middle is negative, and bottom is positive. b) The electric field is the smallest on the horizontal line through the mi
Virginia Tech - PHYS - 2306
21.60: a) d p q (8.9 10 30 C m) (1.6 10 19 C) 5.56 10 11 m. b) m ax pE (8.9 10 30 C m)( 6.0 10 5 N C) 5.34 10 24 N m. Maximum torque:
Virginia Tech - PHYS - 2306
21.68: a) F13 F131 q1q3 ^ cosi 2 40 r131 q1q3 sin ^ j 2 40 r13 F13 Similarly for the force from the other charge: 1 q2 q3 ^ 1 (2.00 nC)(6.00 nC) ^ F23 j j (1.20 10 4 N ) ^ j 2 2 40 r23 40 (0.0300 m) Therefore the two force components are: Fx
Virginia Tech - PHYS - 2306
21.70: Examining the forces: Somg sin cos Fxd 2LT sin Fe0 andFyT cos mg0.Fekq 2 d2But tan d32 kq 2 L mgdq2L 1 3 2 0 mg.
Virginia Tech - PHYS - 2306
21.71: a)q2b) Using the same force analysis as in problem 21.70, we find: 40 d 2 mg tan and2 (1.2)sin256dq40 (2 (1.2) sin 25 )2 tan 25 (0.015 kg)(9.80 m s )kq2 d22q2.79 10C.c) From Problem 21.70, mg tan .d kq2 (8.99 109
Virginia Tech - PHYS - 2306
21.85: E 12 .0 N Ck (16 .0 nC) (3.00 m) 2k (12 .0 nC) (8.00 m) 2kq (5.00 m) 2q12 25.0 m k21.60 10 8 C 9.0 m21.20 10 8 C 64.00 m27.31 10 8 C73.1 nC.
Virginia Tech - PHYS - 2306
21.94: The electric field in the x-direction cancels the left and right halves of the semicircle. The remaining y-component points in the negative y-direction. The charge per unit length of the semicircle is:Q and dE a2k a0k dl a2 2k d but
Virginia Tech - PHYS - 2306
21.95: By symmetry, E xE y . For E y , compared to problem 21.94, the integral over theangle is halved but the charge density doublesgiving the same result. Thus,ExEy2k a2kQ . a2
Virginia Tech - PHYS - 2306
21.97: a) b) 1429qE 10 mgm qE 10g1.4 105 N C 10(9.8 m s 2 )1429 kg C.kg 1 mol 6.02 1023 carbons 1.6 10 19 C carbons . . . 1.15 1010 . 3 C 12 10 kg mol excess e excess e
Virginia Tech - PHYS - 2306
21.98: a) E xE y , and Ex2 Elength of wire a , charge Q21 Q 40 x x 2a 2 2, wherea 2Q 2Q Ex , Ey . 2 2 0a 0a 2 b) If all edges of the square had equal charge, the electric fields would cancel by symmetry at the center of the square. x
Virginia Tech - PHYS - 2306
21.101:By inspection the fields in the different regions are as shown below:EI EIII E ^ ^ ( i k ), 2 0 EII ^ ^ ( i k) 2 0 ^ ^ ( i k) 2 0 ^ ^ ( i k ), EIV 2 0 ( 2 0 x ^ i x z ^ k ). z
Virginia Tech - PHYS - 2306
38.2: a) Pt = (0.600 W) (20.0 10 3 s) = 0.0120 J = 7.49 1016 eV. hc = 3.05 10 19 J = 1.90 eV. b) hf = Pt = 3.94 1016. c) hf
Virginia Tech - PHYS - 2306
38.10: K max = hf Use the information given for = 400 nm to find : (6.626 10 34 J s) (2.998 108 m s) = hf K max = (1.10 eV) (1.602 10 19 J eV) 400 10 9 m = 3.204 10 19 J Now calculate K max for = 300 nm : (6.626 10 34 J s) (2.998 108
Virginia Tech - PHYS - 2306
38.31: E =hc (6.63 10 34 J s)(3.00 108 m s) = = 1.88 10 20 J. 5 1.06 10 mTotal energy in 1 second from the laser E = Pt = 7.50 10 3 J. So the number of E 7.50 10 3 J = = 4.00 1017. photons emitted per second is 20 E 1.88 10 J
Virginia Tech - PHYS - 2306
38.41: The derivation of Eq. (38.23) is explicitly shown in Equations (38.24) through (38.27) with the final substitution of h p = h and p = h yielding = (1 cos ). mc
Virginia Tech - PHYS - 2306
38.44: From Eq. (38.30), 2.898 10 3 m K c m = = 0.966 mm, and f = = 3.10 1011 Hz. a) 3.00 K m Note that a more precise value of the Wien displacement law constant has been used. b) A factor of 100 increase in the temperature lowers m by a factor o
Virginia Tech - PHYS - 2306
E ( mole) 1.00 10 5 J = = 38.51: a)Energy to dissociate an AgBr molecule is just E = 1 mole 6.02 10 23 1.66 10 19 J = 1.04 eV. (1.60 10 19 J eV) b) = c) f = hc (6.63 10 34 J s)(3.00 108 m s) = = 1.20 10 6 m. 19 E 1.66 10 J c 3.00 108 m/s =
Virginia Tech - PHYS - 2306
38.55: a)Plot: Below is the graph of frequency versus stopping potential.Threshold frequency is when the stopping potential is zero. 1.89 f th = Hz = 4.60 1014 Hz 4.11 10 15 c (3.00 108 m s) th = = = 6.52 10 7 m. b)Threshold wavelength is 14
Virginia Tech - PHYS - 2306
38.62: a)The levels are E 4 = 1.0 eV, E3 = 5.0 eV, E 2 = 8.0 eV, and E1 = 10.0 eV.b)We can go from 4-3(4 eV), 4-2(7 eV), and 4-1(9 eV) directly, but also 3-2(3 eV), 3-1(5 eV), and 2-1(2 eV) after starting from 4.