湖北省高中名校联盟2024届高三第四次联合测评物理试卷参考答案与详解—、选择题：本题共10小题，每小题4分，共40分。35题号24678910-CAABDD答案BBOBCAD1.【答案)A【解析】打AI+扞He--罚P甘n,罚P--节S叶�e;人工转变过程，电荷数守恒，质讯数守恒；过狱的接受射线辐射，会危及身体健康。选项A正确。2.【答案】CPVC【斛析】由一-=C有V=-T,从a-b,圆弧的斜率在变化，不是等斥变化；b-+-c,温度升高，体积膨胀，Tp巾6U=W+Q,可知该过程气体吸热；c一d,体积膨胀，气体对外界做正功；d一a,温度降低，气体分子平均动能变小。故选项C正确。3.【答案】AQ===【韶析】在等屈异种电荷连线和中垂线上，由E=k了知EAEc>EuEo;由电场线知cpucpD>=vb,易知回路中感应电流为逆时针。a的速度为v时，由动拭守恒定律，2mv4==—====7B九为2m--u+3m-ub,·ui,-v;此时EDLuDL--u,EIR.F1,DIL3m叭，所以uL;f设464636mRBL(父,-xi,)=共，=共==6Rmv后稳定时2mv5mv对b,BLq3mv-O,q，所以X•-Xi,------,。选项D正确。R5B九8.【答案】BD220T/1220111=＝=220【斛析】—=—,n11300匝；若711不变，̍，所以b.112156匝；若llz不变，—－＝6.6393339+t::.11233n,—t::.n,=，所以细，1040匝。选项BO正确。399.【答案】BCRmv,qBR【俏析】如图，到达C点的离子的速度最小，由—=—－得，v=一—-;到达圆形2qB2m3Rmv,3qBR区域速度最大的离子刚好到达D占八、，由一＝㜍得，v=；设能到达0点2qB2m2的离子轨迹半径为r'圆心为E,与圆形区域的交点为P,由勾股定理有(2R-r)=矿＋产，解得r=3mv,3qBR-R,由r=—得，v=。选项BC正确。qB4m10.【答案】AD===【解析】开始时，对Bmg,kx,,A在Q点时，对C,2mgkx2,A从P到Q点，易知x,十X28L'所=3mg=16==以k—一，x2-L。A从P到Q,巾能批守恒定律，mgl2Lmg8L+t::.Ep所以�,4mgL,因8L3m—为弹簧原先有形变狱，所以EPQ#-4mgL。D从Q到P,由能从守恒定律tiEp十mg8L=g12L+2趴，所以肛=6mgL。选项AD正矿0。11.【答案】(2)M>2m(2分）(3)0.102(3分）(4)偏大(2分）【解析】(2)由动滑轮规律，欲使钩码下降重物上升，盆耍M>2m;VA—2—12——12—1(3)理论上，若满足机械能守恒，有Mg�mgh=上M(卫)+mvM(—)—m让，代入数2222222Vv;,2据整理可得h=—-—，所以k=0.102s•mI一;gg物理试卷参考答案与详解第2页（共5页）{#{QQABAYCAggAoQIIAABhCEwWSCAKQkACCAAoGBBAIIAAASRFABAA=}#}(4)若实际实验过程中阻力不能忽略，觅力势能减少址大于系统动能增址，图线斜率会增大。12.【答案】(1)862(2分）22(3)0.80士0.02(2分）7.8Xl0~8.2Xl0(2分）(4)偏小(2分）偏小(2分）【解析】(1)电阻的测狱采用的是替代法，1178n+R。=2040n,可得R。=862!1;(3)根据图像截距可得E=O.80土0.02V;根据图像斜率可得内阻为r=820土20!1;(4)测讥误差主要来源于电斥表分流，故电动势和内阻的测狱值均偏小。464R13.【答案]Cl)一(2)315c【解析】如图所示，1°°(1){3=—2(180-试卷参考答案与详解第3页（共5页）{#{QQABAYCAggAoQIIAABhCEwWSCAKQkACCAAoGBBAIIAAASRFABAA=}#}笥根据抛体运动规律，沿垂直电场方向（垂直MN方向）v=vin0·······································(J),ps=m2gcos0m2a.Y·································®y=三．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．．＠2ayH=y+R+Rcos0......…...…...………...@)aALB解得H=O.7776m······························@评分参考：第(1)间7分，Q)@@式各2分，＠式1分；第(2)问3分，©式2分，＠式1分；第(3)间5分，(])@®®@式各1分。15.【答案】0)5m/s,10m/s(2)5.625m(3)7.5护(m)【解析]Cl)规定水平向左为正方向，根据动批守恒定律可得O=mAvA+muvu······································································································CD根据能杖守恒11E=—mv�+-mv�=J.................................................................................CZ)p2A2u37.5可解v,,=5m/s............................................................................................................®=v8-10m/s.........................................................................................................®(2)弹簧之后弹开，对两球受力分析可得=Fm11a···············································································································@A向左匀速运动，B先向右匀减速，再向左匀加速直到两者第一次碰搅，此时l。=v1t。十ati·································································································®VA/1了碰撞时B球的速度为1=。.........................................................................................................v8vB+at(J)第一次碰撞，根据动址守恒定律可得，=mAvA+mllvllmAvA,+mllvlll····················································································®根据能让守恒21112=