2026年浙教版适配高一物理学业水平模拟卷声光热综合应用标准试卷第092套（含答案解析与可打印作答区）

高一物理学业水平模拟卷第092套2026年浙教版适配高一物理学业水平模拟卷声光热综合应用标准试卷第092套（含答案解析与可打印作答区）学校：______________班级：______________姓名：______________考号：______________考试时间：120分钟满分：120分答题说明：本卷共四大题，28小题。第1—12题为单项选择题，每题只有一个最佳答案；其余各题按题目要求在作答区内作答。计算题应写出必要公式、代入过程、单位和结论。诚信提示：独立完成，书写规范，保持卷面整洁；不得在答题区外书写与作答无关的内容。选择题答题栏题号123456789101112答案一、单项选择题（本大题共12小题，每小题3分，共36分）1.（3分）把正在发声的电铃放入抽气罩内，随着罩内空气逐渐被抽出，听到的铃声越来越小。下列判断正确的是A.声音在真空中传播更快B.声音只能由金属传播C.声音传播需要介质D.抽气后电铃振动频率变低2.（3分）某同学站在山崖前发出短促喊声，0.60s后听到回声。取空气中声速340m/s，则他到山崖的距离约为A.51mB.68mC.102mD.204m3.（3分）调紧吉他琴弦后，拨动同一根弦，声音的音调明显升高。音调升高主要说明声源振动的A.振幅变大B.频率变大C.传播速度变大D.传播距离变大4.（3分）高速公路两侧设置隔音屏，主要是为了A.在声源处减弱噪声B.使声音不能发生反射C.在人耳处减弱噪声D.在传播过程中减弱噪声5.（3分）一束光射到平面镜上，入射光线与镜面夹角为50°。则反射角以及入射光线与反射光线的夹角分别为A.50°，100°B.40°，80°C.50°，80°D.40°，100°6.（3分）光从空气斜射入水中时，下列关于折射光线的说法正确的是A.向法线偏折，光速变小B.远离法线偏折，光速变大C.传播方向不变，光速不变D.一定沿原路返回7.（3分）焦距为10cm的凸透镜前放一支蜡烛，物距为30cm。屏上所成的像应为A.倒立、缩小的实像B.倒立、放大的实像C.正立、放大的虚像D.正立、等大的虚像8.（3分）小明以0.50m/s的速度沿垂直镜面的方向走近平面镜。关于他在镜中的像，下列说法正确的是A.像逐渐变大B.像逐渐变小C.像相对镜面的速度为1.00m/sD.人与像的距离每秒减小1.00m9.（3分）阳光通过三棱镜后在白屏上形成彩色光带。该现象表明A.白光不是光B.白光由多种色光组成C.红光的传播速度一定最大D.彩色光带是屏幕发光产生的10.（3分）冬天用手同时触摸室内木桌和金属门把手，常觉得金属更冷，主要原因是A.金属温度一定更低B.木桌不能传热C.金属导热能力较强D.金属会主动吸收冷气11.（3分）质量相同的水和干沙在相同阳光下吸收相同热量，水的温度升高较小。最合理的解释是A.水没有吸热B.沙子的质量变小C.水的密度较大D.水的比热容较大12.（3分）冰水混合物在标准大气压下继续吸热，直到冰未完全熔化前，其温度A.不断升高B.保持0℃C.不断降低D.先升高后降低二、填空与基础应用题（本大题共6小题，每小题4分，共24分）13.（4分）某声源在空气中发出频率为680Hz的声音，取声速340m/s。该声音的周期约为__________s，波长为__________m。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________14.（4分）用噪声计测得同一闹钟在门外声强级为52dB，关门后室内测得42dB，则门对该声音的降噪量为__________dB；关门减弱噪声属于在__________减弱。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________15.（4分）一束光照到平面镜上，入射光线与镜面夹角为25°，则入射角为__________°，反射角为__________°。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________16.（4分）光从空气斜射入玻璃砖时，传播速度__________，折射光线向__________偏折。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________17.（4分）质量为0.25kg的水温度升高12℃，水吸收热量为__________J。若电加热器功率为60W且有70%的电能被水吸收，则加热时间为__________s。取水的比热容4.2×10³J/(kg·℃)。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18.（4分）常用液体温度计是利用液体__________的性质制成的。室温从18℃升到26℃时，空气分子平均动能__________。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________三、实验与材料分析题（本大题共4小题，每小题7分，共28分）19.（7分）某小组在空旷操场上利用教学楼墙面测量空气中的声速。同学甲站在距墙51.0m处击掌，并用手机记录从击掌到听到回声的时间，三次数据分别为0.31s、0.30s、0.29s。

（1）求本实验测得的声速；

（2）说明本实验为什么要取较大的距离并多次测量；

（3）写出一个会使测量结果产生误差的因素。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.（7分）在“探究平面镜成像特点”的实验中，小组使用薄玻璃板、两支完全相同的蜡烛、白纸、刻度尺进行实验。点燃蜡烛A，在玻璃板后移动未点燃的蜡烛B，使其与蜡烛A的像重合，并记录物距与像距。

（1）实验中选用薄玻璃板而不是普通平面镜的主要目的是什么；

（2）蜡烛B不点燃的原因是什么；

（3）若记录多组数据均满足物距与像距相等，可得到什么结论；

（4）使用较厚玻璃板可能带来什么问题。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21.（7分）某小组探究凸透镜成像规律，得到如下数据。

物距u/cm40302015像距v/cm13.315.020.030.0像的性质倒立缩小倒立缩小倒立等大倒立放大（1）由表格估算该凸透镜的焦距；（2）当物距为25cm时，所成像的性质是什么；（3）照相机工作时通常利用表中哪一类成像情况；（4）若用白纸遮住凸透镜上半部分，屏上像会发生什么变化。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（7分）某小组用两个完全相同的烧杯分别装入质量均为0.20kg的水和食用油，用相同加热器同时加热3min，记录温度变化如下。不计烧杯吸热，取水的比热容4.2×10³J/(kg·℃)。液体初温/℃末温/℃升高温度/℃水22286食用油223614（1）实验中应控制哪些量相同；（2）由数据比较水和食用油的吸热能力；（3）估算食用油的比热容；（4）说明本实验中热量损失会怎样影响估算结果。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________四、综合应用题（本大题共6小题，共32分）23.（5分）学校礼堂调试扩声系统。后排座位到扬声器的距离约68m，现场空气中的声速取340m/s，舞台灯光到达观众的时间可不计。

（1）求后排观众听到声音相对看到演员动作的延迟；

（2）若要求声画延迟不超过0.10s，请提出一种物理上可行的改进方法；

（3）夏季室温升高时，声速通常有所增大，延迟将怎样变化。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________24.（5分）潜望镜内有两块互相平行且均与水平面成45°角的平面镜。水平射入的光线经第一块镜反射后竖直向下，再经第二块镜反射后水平射出。

（1）分别说明两次反射都遵循的规律；

（2）若入射光线与第一块镜面的夹角为45°，求第一次反射的入射角和反射角；

（3）简述潜望镜能改变观察方向的原因。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________25.（5分）某保温杯中有0.50kg、90℃的热水，经过一段时间后降到50℃。取水的比热容4.2×10³J/(kg·℃)。

（1）求热水放出的热量；

（2）若杯壁主要通过传导、对流和辐射向外散热，请各写出一种减小散热的设计思路；

（3）若同样质量的油代替水且比热容更小，在相同放热条件下降温会更快还是更慢。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________26.（5分）某温室大棚白天吸收阳光，夜间用水桶储热。棚内有100kg水，白天水温由18℃升到26℃。取水的比热容4.2×10³J/(kg·℃)。

（1）求这些水白天储存的热量；

（2）说明透明薄膜有利于白天升温的光学原因；

（3）说明夜间水桶能缓和降温的热学原因。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________27.（6分）汽车倒车雷达向障碍物发出超声波，控制器从发射到接收回声共用0.020s，其中电子处理时间约0.002s。空气中声速取340m/s。车辆后视镜采用凸面镜。

（1）求障碍物到车尾的距离；

（2）说明倒车雷达不能用次声波替代超声波的一个原因；

（3）从成像特点说明后视镜采用凸面镜的优点与不足。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________28.（6分）科技馆“声光热协同演示台”中，扬声器、平面镜和加热灯安装在同一侧，观众站在距扬声器85m处。演示时，激光水平射向平面镜，入射角为35°；加热灯功率为120W，连续工作5min，其中40%的电能最终转化为展台附近空气的内能。取空气中声速340m/s。

（1）求观众听到声音相对看见激光亮斑变化的延迟；

（2）求反射角和入射光线与反射光线的夹角；

（3）求加热灯传给空气的能量，并说明为什么不能把这部分能量全部看成展品温度升高所需的热量。【作答区】________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析一、单项选择题答案题号123456789101112答案CCBDBAADBCDB1.C。电铃仍在振动，但抽气后声音明显减弱，说明声音从声源传到人耳需要空气等介质。真空不能传播声音，频率并不会因抽气必然降低。2.C。回声经历“人到山崖再返回人”的往返路程，距离s=vt/2=340×0.60/2=102m。3.B。音调由声源振动频率决定。琴弦调紧后振动频率增大，音调升高；响度主要与振幅有关。4.D。隔音屏位于声源和人耳之间，阻挡、吸收和反射部分声音，属于在传播过程中减弱噪声。5.B。入射角是入射光线与法线的夹角，