资源简介 A.图甲磁电式电表铝框骨架能起到电磁驱动的作用2023~2024学年第二学期高二年级期中学业诊断B.图乙电流表运输时,把正负接线柱用导线连在一起,利用的是电磁阻尼原理C.图丙阻尼摆片摆动中产生感应电流,很快停下来物理试卷D.图丁变压器硅钢片相互绝缘叠加能减小涡流4.如图所示,绝缘圆环a与金属圆环b同心共面放置,α带负电并绕中心0点顺时针转动,下列(考试时间:下午4:30—6:00)说明:本试卷为闭卷笔答,答题时间90分钟,满分100分。说法正确的是A.当α顺时针加速转动时,b产生顺时针方向电流并有扩张趋势题号二三四总分0B.当a顺时针减速转动时,b产生逆时针方向电流并有扩张趋势得分C.当a顺时针匀加速转动时,b产生逆时针方向电流并有扩张趋势D.当a顺时针匀速转动时,b不产生电流但有收缩趋势一、单项选择题:本题共10小题,每小题3分,共30分。请将正确选项前字母标号填入下表内5.如图所示电路中,线圈L电阻不计,D为理想二极管,下列说法正确的是相应位置。A.S闭合时,A不亮,B立即变亮题号3456910B.S闭合时,A立即变亮,B逐渐变亮答案C.S断开时,A不亮,B立即熄灭1.关于交变电流,下列说法正确的是D.S断开时,A立即变亮,然后逐渐变暗至熄灭A.手机电池提供的电流是交变电流6.如图所示为甲、乙弹簧振子的振动图像,下列说法正确的是B.交变电流可能是按正余弦规律变化的个x/cmC.交变电流的大小一定随时间做周期性变化A.两弹簧振子的初相位不同D.家用照明电压的大小随时间做周期性变化,方向不变B.甲系统的机械能比乙系统的大2.关于简谐运动,下列说法正确的是C.甲、乙两弹簧振子加速度最大值之比为2:1-2A.弹簧振子的回复力在半个周期内的功一定为0D.1=2s时,甲具有负方向最大速度,乙具有正方向最大位移B.弹簧振子的回复力在半个周期内的冲量一定为0C.若物体做简谐运动,合外力与位移的关系符合F=-:7,男子蹦床比赛中,运动员某次下落从最高点到着网点用时0.8s,从着网点到最低点用时D.物体所受合力大小与时间满足一次函数关系,物体一定做简谐运动0.4s。运动员的质量为60kg,该过程中网对运动员的平均作用力约为3.关于电磁感应,下列说法不正确的是A.600NB.1200NC.1800ND.2400N摆片mA8.质量为m的物体静止在光滑水平面上,在水平恒力F的作用下,经时间:走过位移1,动量变为p、动能变为E。若上述过程中F不变,物体的质量变为2m,下列说法正确的是A.经过时间1,物体动量变为pB.经过时间1,物体动能变为EC.经过位移1,物体动量变为2印D.经过位移1,物体动能变为2E高二物理第1页(共8页)高二物理第2页(共8页)2023~2024学年第二学期高二年级期中学业诊断物理参考答案及评分建议一、单项选择题:本题包含 10小题,每小题 3分,共 30分。题号 1 2 3 4 5 6 7 8 9 10选项 B A A A D D C A D C二、多项选择题:本题包含 5小题,每小题 3分,共 15分。题号 11 12 13 14 15选项 AC AC BC AB ABC三、实验题:共 14分。16.(6分)(1) C (1分)(2)19.80(2分)805.0(1分)(3) (2分)17.(8分)(1)天平(2分)(2)D (2分)(3) 1 1 0 = 1 2 0 + 2 3 0(2分) 1( 1 0) = 1( 2 0) + 2( 3 0) 其他正确答案均可得分(2分)四、计算题:共 41分。18.(8分)(1)甲车向右做减速运动,由动能定理得 f L 1 m v 2 11 m v20 ····························································(1分)2 2f 0 . 1m g ············································································· (1分) 1 = 2 m/s················································································ (1分)(2)甲、乙车相碰时,组成系统动量守恒,以右为正方向m v 1 2 m v 2 ·······································································(2分)两车碰撞后嵌套在一起,向右做减速运动2 f x 1 0 2 m v 22 ·····························································(2分)2 = 0.25m···················································································(1分)19.(9分)(1)设磁感应强度为 B,导线框边长为 l,导线框电阻为 R,质量为 m,导线框进入磁场时 2 2 = ···················································································(2分) 2 2 = ······················································································(1分) 安培力方向与运动方向相反,v减小,加速度 a减小····························(1分)(2)以右为正,导线框全部穿进磁场时速度为 v由动量定理,导线框穿入磁场,以右为正方向 = 1································································(1分) 导线框穿出磁场 = 2 ································································(1分)