2027届湖南省邵阳市高三上学期第一次学情检测物理试题及答案

邵阳市2027届高三第一次联考试题卷物理本试卷共6页，15个小题。满分100分。考试时间75分钟。2．作答选择题时，选出每小题答案后，用2B铅笔黑；如需改动，用橡皮擦干净后，再选涂其他答案。答案不能答3．非选择题必须用黑色字迹的钢笔或签字笔作答，应位置上；如需改动，先划掉原来的答案，然后再写上新答案；不准使用铅笔和涂改液。不持秒表在站台的尽头测得列车匀加速出站时通过他用时20.0s，小华在该列车上测得列车的加速度为始终相对字典静止。则该过程中，橡皮擦对字螺线管右侧有竖直悬挂的铜环，匀强磁场垂直于圆盘平面向上，从上向下看，圆盘顺时针转动。下述说法脱钩，货车司机保持原牵引力前进L的距离后才发现，便立即撤去牵引力。已知车头与拖车的质量M、m，两车均停止运动后，两车之间的距离为7．如图，在大小、方向均未知的匀强电场中，边长为0.75m的正方形ABCD处于某一竖直6N/CB6V向右侧直管中沿管壁缓慢注入适量的水银，忽略环境温度的变化，U形管导热性能良好。下列说法正确的是面内放置有记录纸。当弹簧振子沿竖直方向上下自由振动时，以速率在纸上留下如图所示的余弦型函数曲线印迹，图中的y1、y2、x0、2x0、3x0为记录纸上印迹的位置坐标值，P、Q分别是印迹上纵坐标为y1和y2的两个点。若空气阻力、记录笔的质量及其与纸之间的作用力均A．余弦型函数曲线为波的图像且波长为2x010．如图，在xOy平面内有一以O点为中心的正五边形，顶点到O点的距离为R。在正五边形的顶点上顺116分）小赞同学在实验室进行电表的改装实验。将一电流表（满偏电流为1mA，内阻为500Ω)改装定值电阻R2的阻值为kΩ;1210分）同学们利用如图甲所示的装置探究加速度Hz，打点计时器打下计数点C时小车的速度为m/s，运动过程中小车的加速度大小为图像如图丙所示，导致该图像没有过坐标原点的原因是，导致该图像后部分发生弯曲的原因是________C．木板与水平面倾角过大，平衡摩擦力过量D．木板与水平面倾角过小，平衡摩擦力不足1310分）道威棱镜是医用内窥镜的重要组成部分（如图甲它被广泛用于图形已知小物块每次经过y轴时速度方向均与y轴垂直。重力加速度为g，忽略空气阻力。（2）求小物块从点P到第三次经过y轴所用的时间；成相互垂直的复合场。小物块仍以初速度v0从点P沿y轴正方向开始运动，求运动过程中小物块速率最小面上以v0=8m/s的速度垂直于传送带的方向匀速运动。现将弹簧解除锁定，弹性势能释放后，物传送带，随后经传送带右端水平抛出，打在木板上表面。已知弹簧锁定时的弹性势能为EP=2J，小物块邵阳市2027届高三第一次联考试题参考答案与评分标准物理123456789CCBBADAACACD1310分1）见解析解析1）棱镜全反射临界角sinC=····························································解得C=45°···························折射定律························································解得折射角r=30°·····································································故光线在BC面上发生全反射。····················································L·················································································光在棱镜传播速度：v=c····························n5m2qBE5 1分2由洛伦兹力提供向心力有qv0B=·······················解得v0=·····················································象限做半个圆周运动再次进入第一象限，即为第三次经过y轴，小物块在磁场中做圆周运动的周期T=································小物块在电场中的加速度a=·······························m则小物块在电场中的时间tE=·································3τm2BLBE2qBE将物块在P点速度v0分解为沿y轴的两个分速度和，其中分速度对应的洛伦兹力f1=Bq与电场力平衡，物块的运动可分解为速度为的匀速圆周运动和沿y轴方向速度为的匀速直线运动的由Bq=···························· 1516分1）2m/s（2）1.5s（3）52m/s·········································································解得v1=2m/s·····························又·····················································································································（3）设物块碰撞前、后瞬间竖直方向的速度分别为vz、v′z，则v=2gh··································物块刚落到木板上时，水平面内相对木板的速度关竖直方向ΣFNΔt=mv′z−····················x轴方向FNΔtsinθ=mvx−mv······································