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2018-2019学年上海市浦东新区建平中学高三下学期月考试卷

2018-2019学年上海市浦东新区建平中学高三下学期月考试卷
2018-2019学年上海市浦东新区建平中学高三下学期月考试卷

2018学年上海市浦东新区建平中学高三第二学期月考试卷

一、选择题

1、下列物理概念的建立不属于用到“等效替代”方法的是( )

(A)质点(B)重心

(C)平均速度(D)合力

【答案】A

2、对电磁感应现象进行深入研究并获得巨大成就的科学家是( )

(A)奥斯特(B)法拉第

(C)库仑(D)欧姆

【答案】B

3、关于磁感线,下列说法正确的是( )

(A)磁感线是磁场中客观存在的曲线

(B)在磁场中由小铁屑排列成的曲线就是磁感线

(C)磁感线上的每一点的切线方向就是该处的磁场方向

(D)磁感线总是从磁体的N极出发到S极终止

【答案】C

4、伽利略为了研究自由落体的规律,将落体实验转化为著名的“斜面实验”,从而创造了一种科学研究的方法.利用斜面实验主要是考虑到,实验时便于测量小球运动的()

(A)速度(B)时间(C)路程(D)加速度

【答案】B

(A)(B)(C)(D)

5、关于布朗运动,下列说法中正确的是()

(A)液体分子的无规则运动就是布朗运动

(B)布朗运动的激烈程度跟温度无关

(C)布朗运动是悬浮在液体中的固体颗粒不断地受到液体分子的撞击而引起的

(D)悬浮在液体中固体颗粒越大,在某一瞬时撞击它的分子数越多,布朗运动越明显【答案】C

6、两列振幅和波长都相同而传播方向相反的波(如图甲所示),在相遇的某一时刻,两列波“消失”(如图乙所示).此时介质在x、y两质点的运动方向是()

(A)x向下,y向上(B)x向上,y向下

(C)x、y都向上(D)x、y都向下

【答案】A

7、如图,是一个质点做匀变速直线运动s-t图中的一段。从图中所给的数据可以确定质点在运动过程中经过图线上P点所对应位置时的速度大小一定( )

(A)大于2m/s

(B)等于2m/s

(C)小于2m/s

(D)无法确定

【解析】物体做的是匀变速直线运动,p点为位移的中间位置,而这段时间的平均速度的大小为2m/s,根据匀变速直线运动的规律可以知道,此段位移的中间时刻的瞬时速度即为2m/s,因为中间时刻的瞬时速度要小于中间位置的瞬时速度,所以P点速度要大于2m/s,所以A 正确。

【答案】B

8、如图所示电路,电源电压U恒定,由于某元件出现故障,使灯L1变亮,灯L2不亮,其原因可能是()

(A)R1断路(B)R2断路(C)R2短路(D)R1短路

【答案】D

9、如图所示,某人用托里拆利管做测定大气压强的实验时,由于管内漏进了空气,测得管内汞柱的高度仅为70cm,但当时的实际大气压强为一个标准大气压(相当于76厘米高的汞柱产生的压强)。今采用下述哪种方法,可使管内、外汞面的高度差大于70cm( )

(A)把托里拆利管逐渐倾斜(管子露出部分长度不变)

(B)把托里拆利管慢慢向上提,但下端不离开汞槽

(C)保持装置不动,往汞槽内加汞,以增大压强

(D)整个装置竖直向上做加速运动

【解析】:

A、压强只与高度差有关,把托里拆利管逐渐倾斜,

高度差不变,A错误

B、把托里拆利管慢慢向上提,假设气体体积不变,则压强增大,

使水银柱下移,体积增大,温度不变,所以实际气体压强应该减小,液面差减小,B错误C、保持装置不动,往汞槽内加汞,假设气体体积不变,则压强减小,水银柱上移,温度不变,压强增大,故液面差增大,C正确

D、整个装置竖直向上做加速运动,水银柱超重,液面差减小,D错误

所以C选项是正确的

【答案】:C.

10、一带点粒子射入一固定的正电荷Q的电场中,沿如图所示的虚线由a点经b运动到c,b点离Q最近。若不计重力,则()

(A)带电粒子带负电

(B)带电粒子到达b点时动能最大

(C)带电粒子从a到b电场力做正功

(D)带电粒子从b到c电势能减小

【解析】:

A、轨迹弯曲的方向大致指向合力(电场力)的方向,知电场力向左,该粒子带正电.故A错误.

B、电场力从a点经b运动到c,先做负功再做正功,根据动能定理,动能先减小,后增大.b点动能最小.故B错误.

C、带电粒子从a到b,电场力做负功.故C错误.

D、带电粒子从b到c,电场力做正功,电势能减小.所以D选项是正确的.

【答案】:D

11、有一个物体以初速度v0沿倾角为θ的足够长的粗糙斜面上滑,已知物体与该斜面间的动摩擦因数μ

【解析】:物体所受摩擦力为f=mgcosθμ<mgsinθ,则物体上滑时所受合外力沿斜面向下,大小为mgsinθ+f,物体下滑时所受合外力沿斜面向下,大小为mgsinθ-f,故物体先沿斜面向上做匀减速直线运动,后沿斜面向下做匀加速直线运动,根据牛顿第二定律可知:物体上滑的加速度大于下滑的加速度,则上滑的时间短于下滑的时间,返回出发点时速度比初速度v0小,则A图符合物体的运动情况。

【答案】:A

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