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UBND TH X BM SN
UBND TH X BM SN
K THI OLIMPIC THPT NM 2010 THI CHNH THC MN: VT L LP 11
Thi gian lm bi: 180 pht
thi gm 01 trang. SBD......
Cu 1 (4 im): mp bn A ca mt chic bn chiu cao h =1 m, t mt qu cu ng cht, bn knh R = 1 cm , ( hnh 1). y nh cho tm O ca qu cu lch khi ng thng ng i qua A, qu cu ri xung t. Hi qu cu ri xung t cch mp bn theo phng nm ngang mt on bao nhiu?Ly g = 10 m/s.
Cu 2 (4 im): Cho mch in nh (hnh 2), bit R0 = 3, r2 = 2r1. Khi k1 ng k2 m hay khi k1 m k2 ng, cng sut ca mch ngoi khng i v s ch ca vn k(c in tr rt ln) u bng 0,5V. (b qua in tr cc dy ni v cc kho K)a.Tnh r1, r2 v E1, E2b.So snh cng sut ni trn vi cng sut ln nht m b ngun in c th cung cp cho mch ngoi?
Cu 3 (4 im): Thanh kim loi CD chiu di , khi lng m = 100g t vung gc vi hai thanh ray song song nm ngang v ni vi ngun in nh (hnh 3).
b. Nng hai u A, B ca ray ln ray hp vi mt phng nm ngang gc ( = 300. Tm hng v gia tc chuyn ng ca thanh bit thanh bt u chuyn ng khng vn tc u.
Cu 4 ( 4 im):
Cu 5 ( 4 im): Trn th pV i vi mt lng kh l tng no , gm hai qu trnh ng nhit ct hai qu trnh ng p ti cc im 1,2,3,4 (hnh 5 ). Hy xc nh t s nhit ca cht kh ti trng thi 3 v 1, nu bit t s th tch . Bit th tch kh ti trng thi 2 v 4 bng nhau.
H T
H NG DN V P N CHM VT L LP 11K THI OLIMPIC THPT NM 2010
Cu 1 (4 im):Ni dungim
- Ban u tm o ca qu cu chuyn ng trn quanh tm A.0,25
- Khi bn knh R = AO lm vi ng thng ng gc no (Tm 0 chuyn ng trn cung 00) th qu cu bt u ri khi mp bn vi vn tc , ch cn trng lc mg tc dng ln qu cu , Hnh v :
0,25
- Phn tch ti lc bt u ri khi mp bn: Thnh phn php tuyn cung trn l lc hng tm= mg.cos ( 1 )0,5
- p dng nh lutBTCN (chn w= 0): mgR = mg.Rcos+ (2 )0,5
- Thay ( 1 ) vo ( 2 ) c: g.R = v+
EMBED Equation.3 v = 0,26 m/s0,5
- Thay v vo ( 1 ) cos =
0,5
- thi im ny qu cu c. nm xin xung vi vn tc ban uhpvi mt phng ngang gc . 0,25
- V R rt nh so vi h nn ban u coi 0 gc A .Trong h to xAy c cc p/t chuyn ng: x =v.cos.t =0,26.. t = 0,17.t ( 3 ) 0,25
y = v.sin .t + gt= 0,19.t + 5t ( 4 )0,25
- Ta c h = y =1 m. Thi gian chm t l: t ( 4 )5t+0,19.t 1 = 0
Loi nghim m. Ta c t 0,43 s . 0,5
- T (3 )V tr c vt ri xung t cch mp bn theo phng ngang
l: x = CH = 0,17.0,43 = 0,073 m0,25
Cu 2 (4 im):
aKhi k1 ng k2 m: in tr mch ngoi R1 = 1,5R0
0,5
Khi k1 m k2 ng: in tr mch ngoi
0,5
T gi thit: P1 = P2 nn t (1) v (2) ta c:
R1.R2 = 9r12 thay R1 v R2 ta tnh c R02 =9r12Hay 3r1 = R0 t r1 = 1, r2 = 20,5
Vy
S ch ca vnk l 0,5V: UBC = Ir2 - E2 = 0,50,5
V I2 > I1 nn: gii h trn ta c E1 = 5V, E2 = 2,5V0,5
bThay vo (1) ta c: P1 = P2 = 4,5W
Cng sut cc i ca b ngun:
theo Cosi:
Vi
Pmax = 4,6875W0,50,5
0,5
Cu 3 (4 im):
Ni dungim
p n
a) Dng in qua CD
Cc lc tc dng ln thanh CD:
- Trng lc thng ng hng xung.
- Lc n hi ca ray vung gc vi ray.
- Lc t , vung gc vi v CD.
- Lc ma st , hng dc theo ray, ngc chiu chuyn ng.
Thanh CD trt sang tri lc t s hng sang tri. Theo quy quy tc bn tay tri, dng in I qua thanh s c chiu t C n D.
p dng nh lut II Niutn ta c: + + + =
(1)
Chiu phng trnh ln hng chuyn ng: F Fms = ma
(2)
Chiu phng trnh ln phng thng ng: - P + N = 0
Fms = kN = k.P = kmg
(2)
I . . B - kmg = ma.
Suy ra:
b) Do Psin( > Ft nn AB s trt dc theo ray i xung p dng nh lut II Niutn: + + + = (1)Chiu (1) ln Oy: N Py Fy = 0
N = Py + Fy =
ln ca lc ma st trt: Fmst = kN = 0,05 + 0,02 (N)
Chiu (1) ln Ox: Px Fx Fms = ma
( c hng theo chiu ca Ox)0,50,50,50,50,50,5
Cu 4 (4 im):
Ni dung yu cuim
a.(2 im)
Xt OIJ cn ti O r1 = r2 Suy ra c i1 = i2 -----------------------
- p dng nh l hm s sin trong OJF:
EMBED Equation.3 -----------------
v i1