轉學考的解答
※ [本文轉錄自 Transfer 看板 #1CG3XV25 ]
作者: hsnuyi (Tsubasa) 看板: Transfer
標題: [其他] 台大99年普物A解答
時間: Fri Jul 16 18:45:49 2010
1.
0 GMmr GMm 1
- S ------dr = ----- = ---mv^2 --> v = (GM/R)^(1/2)
R R^3 2R 2
GMmr ..
------ = -m r --> 令 r = e^lt --> l = +-i(GM/R^3)^(1/2)
R^3
r = Ae^it(GM/R^3)^(1/2) + Be^-it(GM/R^3)^(1/2) 由 t = 0 時 r = R 及 t = 0 時 v = 0
A = B = R/2
r = Rcos[t(GM/R^3)^(1/2)] , r = 0 時 t = (pi/2)(R^3/GM)^(1/2)
..
或: 由 m r + kr = 0 , w = (k/m)^(1/2) 得 w = (GM/R^3)^(1/2)
t = T/4 = (pi/2)(R^3/GM)^(1/2)
2.
MgL ML^2
----- = ------w^2 --> v = Lw = (3gL)^(1/2)
2 2*3
3.
dQ1 = m1c1dT , dQ2 = m2c2dT
T m1c1 T m2c2
S = S ------dT + S ------dT = m1c1ln(T/TH) + m2c2ln(T/TL)
TH T TL T
4.
令原狀態為 P,V,n,T 則:
2V nRT
Isothermal : Wi = S -----dV = nRTln2 = 0.69nRT
V V
Adiabatic : TV^(r-1) = T2(2V)^(r-1) , r = 5/3 , T2 = 0.63T
Wa = -U = -1.5nR(0.63T - T) = 0.56nRT
由 T > 0 , 0.69nRT > 0.56nRT > 0
5.
(a) 1/C = 2/{[8.85*10^(-12)*3.2*10^(-4)]/[1.5*10^(-3)]} +
1/{[8.85*10^(-12)*2.5*3.2*10^(-4)]/[2*10^(-3)]}
C = 7.74*10^(-13) F
(b) Eb Ca 7.74*10^(-13)
---- = ---- = ----------------------------------------- = 1.37
Ea Cb [8.85*10^(-12)*3.2*10^(-4)]/[5*10^(-3)]
6.
(a) W/2 uID uI W
B = S ---------------dx = ------tan^(-1)[----] , 向正x方向
-W/2 2piW(x^2+D^2) piWD 2D
(b) QwuIr^2 W
tau = --------tan^(-1)[----]
2piWD 2D
7.
(a) 2*1.2*500 = m*波長 --> m = 2 時 波長 = 600nm
(b) 2*1,2*500 = (m+0.5)*波長 --> m = 2 時 波長 = 480nm
8.
(a) I = 100/4pi = 7.96 , 7.96 = [n*6.626*10^(-34)*3*10^8]/[589*10^(-9)] --> n = 2.36*10^19
(b) eV = [6.626*10^(-34)*3*10^8]/[1.60*10^(-19)*589*10^(-9)] - 1.8 = 0.31
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※ 編輯: hsnuyi 來自: 59.117.4.132 (05/03 01:19)