南京市2012屆高三學(xué)情調(diào)研物理試卷講評(píng).pptVIP

南京市2012屆高三學(xué)情調(diào)研 物理試卷講評(píng),3 題 C 9,17,E=kQ/r2 r2,1、定義：穿過(guò)某一面積的磁場(chǎng) （即穿過(guò)某一面積磁感線的條數(shù)）,適用條件：勻強(qiáng)磁場(chǎng) SB,一般計(jì)算式：,磁通量可理解為穿過(guò)一個(gè)面的磁感線的條數(shù)。如圖：,磁通量,2、定義式：=BS,=BSsin 或： =BS,最大,較小,=0,4題 A 3,12,.,U,S,E,F=eE,f=evB,v,F=eE f=evB,10題,9 8 7 6 5 4 3 2,18 7 23 8 2 0 2 0,6 9 11 7 11 5 4 3,5題 D 34,45,O,1,2,3,4,5,12, 23, 34,3=0, 45,F1,F1,F1 F1,v,順時(shí)針,順時(shí)針,逆時(shí)針,逆時(shí)針,6 題 BC 0:20,35; 2:10,6,GmM/r 2=m a=mg,a=g 1/r2,GmM/r 2=m v 2/r,v 2 1/ r,GMm/r2=m(2 /T)2r,T 2 r3,=2 /T,2 1/T 2 1/r3,r月=60R地,r同=6R地,7題 AC 0:3,12; 2:2,4,v0,v0,v0,H,WG=mgH,t上 t平 t下,機(jī)械能守恒定律： mv2/2=mv02/2+mgH,8題 AB 0:9,12; 2:0,1,U1=220v,U2=n2U1/n1=110v,I2=U2/R=110/55=2A,P1=P2=U2I2=220W,T=0.02s,f=1/T=50HZ,9題 ABC 0:3,8; 2:25,36,M,N,a,b,c,d,Ed+,Ed-,Eb+ EO+,Eb,Ed Eo 用極限分析法,O,Ed EO Eb,Uab = Ubc,EO Eb,EO+,EO-,Eb- EO-,d,b,a,c,10題 (1) a=3.2m/s2,.,.,.,.,.,.,.,.,.,cm,6.19,7.72,7.21,6.70,x12-x01=a1(2t)2,0,1,3,2,4,t,t,t,t,保留二位有效數(shù)字,x23-x12=a2(2t)2,x34-x23=a3(2t)2,=(x34-x01)/3(2t)2,x24-x02=a(4t)2,10題 (1) a=3.2m/s2,.,.,.,.,.,.,.,.,.,cm,6.19,7.72,7.21,6.70,x24-x02=a(4t)2,0,1,3,2,4,t,t,t,t,(7.21+7.72)-(6.19+6.70)10-2=a(40.02)2,a/ms-2,1/m/kg-1,0,2.00,1.00,4.00,3.00,2.00,1.00,110,1101,10,保留一位有效數(shù)字,保留二位有效數(shù)字,m/kg,11題 (1),9 8 7 6 5 4 3 2,28 13 2 9 4 0 2 2,7 12 5 16 8 5 2 1,11題 (3) C,I,I,I,I,U,U,U,U,I=U/R,k=1/R,U,T,（BD）,12題 B (選修3-4）,(1) B D(4分）按多項(xiàng)選擇題解題策略,x ,(2) 負(fù)(2分） T=0.2S v= /T=8/0.2=40m/s (2分）,(3),30,i=60,60,C=45,i=60 C,i=30,i=30 C,r,Sinr/sin30=n,r=45,12 11 10 9 8 7 6 5 4 3,27 9 10 5 6 2 0 0 1 0,22 6 8 2 13 1 1 2 0 1,（AD）,12題 C (選修3-5）,(1) A D,(2) W=h 0=hc/ 0,h =W+EKmax,EKmax= h-W=hc/-hc/0,(3) m Av A=mBvB+mAvA,1.52=90.5+1.5vA,vA=-1m/s,12 10 8 7 6 4 3,23 5 22 1 5 4 0,8 7 26 1 7 5 2,14題,M,O,N,A,(1) 2R=gt2/2,(2) 4R=vMt,vM,vN,(3) N M由機(jī)械能守恒定律： mvN2/2=mvM2/2+mg2R,mvN2=mvM2+4mgR=8mgR,mg,F,在N點(diǎn)：,F-mg=mvM2/R=8mg,F=9mg,由牛頓第三定律,小球?qū)壍赖膲毫?F=F=9mg,2分,1分,2分,2分,3分,3分,1分,1分,1分,方向豎直向下,1615131110 9 87 65310,28 19 1 3 5 1 1 2 0 0 0 0 0,16 6 1 2 9 0 0 12 3 3 3 0 1,161413121110 9 87 65210,3 2 0 1 47 0 0 0 1 4 0 0 0 1,0 0 1 0 24 1 1 1 0 6 3 1 2 16,15題,v,F安=ILB,mg,a,(1) E=BLv,I= E/(R1+R2)=BLv/(R1+R2),F安=ILB,=B2L2v/(R1+R2),當(dāng)a=0時(shí) v=vmax,mgsin =F安m=ImLB,=B2L2vmax/(R1+R2),vmax= mgsin (R1+R2)/ B2L2=7.5m/s,(2) vmax= mgsin (R1+R2)/ B2L2=3m/s,mgvmaxsin =I2(R1+R2)=2I2R2,I2R2=mgvmaxsin /2=0.075（W),1分,1分,1分,1分,2分,2分,1分,2分,I= BLvmax/(R1+R2) =0.05A 1分,P2=I2R2=0.075W,2分,mgsin =ILB,I=mgsin /LB,15題,v,F安=ILB,mg,a,(3),mgsin ILB=ma,I=q/t,q=CE=CBLv,=CBLv/ t=CBLv/t=CBLa,mgsin CL2B2a=ma,a=mgsin /(m+CL2B2),v=at=mgsin t /(m+C