山东省济宁市全市高考模拟考试（济宁三模）物理+答案

物理答案第1页（共3页）物理答案2026年济宁市高考模拟考试物理试题答案2026.05题号123456789答案CCDBABDAADCDBCBC13.（1）C（2）1.84（3）g（每空2分）14.（12）bkR0（3）无影响（每空2分）1_k1_k15.（7分）解析：（1）气缸内气体温度升高过程中做等压变化，有（2分）解得d=H··················································································（1分）5（2）设气缸内气体压强为p，对气缸由平衡条件得G+pS=p0S·············（1分）气缸内气体温度升高过程中，外界对气体做功W=_pSd·······················（1分）由热力学第一定律ΔU=W+Q··························································（1分）解得ΔU=Qp0SH····································································（1分）16.（9分）解析：（1）设粒子做匀速圆周运动的半径为r，A点速度方向与x轴正方向夹角为θ。2由牛顿第二定律qvB=mv·····························································（1分）r解得rL由几何关系sinθ=L·······································································（1分）r解得θ=60o·················································································（1分）yA=r_rcosθ···············································································（1分）物理答案第2页（共3页）物理答案解得yAL··············································································（1分）（2）粒子在O点和A点的速度大小相等，所以OA连线为等势线，电场强度与y轴正方向所成夹角为“=30o··········································（1分）设粒子从O点到A点运动时间为t。沿OA连线方向vcos“.t·······················································（1分）沿y轴方向··························································（1分）解得E··············································································（1分）17.（14分）解析：（1）若小滑块恰好到达圆管轨道最高点E，则vE=0从D点到E点，有一mgx2Rmv·············································（1分）在D点有FNmg=m·································································（1分）解得FN=10N···············································································（1分）解得h=1.8m·················································································（1分）（2）①若只经过C点一次，则有mgh一μmgL一μ1mgx············或mghμmgLμ1mgx3x·······················································（1分）=1········································································（1分）3②当滑块第二次到达C点且和传送带速度相等时，有mghμmgLμ1mg2smv··························································（1分）解得μ1=0.225················································································（1分）≥0.225时，滑块第二次到达C点后再从传送带离开时，速度大小不变。解得μ1tn5,7,9)此时μ1<0.225，不符合要求。··························································（1分）物理答案第3页（共3页）物理答案当μ1<0.225时，物体第三次回到C点时速度为v0。此后运动过程根据动能定理有_μ1mg.nmv·····························（1分）解得n=1,3,5…)································································（1分）18.（16分）解析：（1）对金属棒a、b组成的系统，由动量守恒定律得mv0=mmv1······（1分）解得v··················································································（1分）由能量守恒定律得mvmvQ总·······························（1分）金属棒b上产生的热量QbQ总······················································（1分）解得Qbmv············································································（1分）（2）对金属棒b由动量定理得LΔt=Σ2mΔv···············（1分）即=2mv1·····································································（1分）解得xax0·······································································（1分）对金属棒a、b由动量守恒定律得Σmv0Δt=ΣmvaΔt+Σ2mvbΔt··············（1分）即mv0t=mxa+2mx0·······································································（1分）解得t········································································（1分）（3）设金属棒b进入MN右侧后，整个系统达到稳定时金属棒b、c的速度大小为v。对金属棒b由动量定理得_ΣBIbLΔt=Σ2mΔvb即_BLqb=2m·····································································（1分）对金属棒c由动量定理得ΣBIcLΔt=ΣmΔvc即BLqc=mv·················································································（1分）对电容器C有q=CBLv····································