青浦区2017学年高三年级第二次学业质量调研测试
青浦区2017学年高三年级第二次学业质量调研测试
物理试卷
（时间60分钟，满分100分）2018.04全卷包括三大题，第一大题为单项选择题，第二大题为填空题，第三大题综合题。
一、单项选择题（共40分，1到8题每小题3分，9到12题每小题4分）
1．下列物理量中，属于标量的是（）
A．电流强度B．电场强度C．磁感应强度D．引力场强度（即重力加速度）
2．下列四种电磁波中，最容易表现衍射现象的是（）
A．红外线B．紫外线C．无线电波D．γ射线
3．下列论述中，能够揭示原子核有复杂结构的是（）
A．卢瑟福通过α粒子散射实验建立原子核式结构模型
B．汤姆孙发现了电子
C．贝克勒尔发现天然放射性现象
D．伦琴发现X射线
4．某单色光照射某金属时不能产生光电效应，下列措施中可能使该金属产生光电效应的是（）A．延长光照时间B．增大光的强度
C．换用频率较低的光照射D．换用波长较短的光照射
5．一个物体静止在斜面上时，正确表示斜面对物体作用力F方向的是图中的（）
6．如图所示，O是水面上一波源，实线和虚线分别表示某时刻的波峰和波谷，A是挡板，B是小孔。若不考虑波的反射因素，则经过足够长的时间后，水面上的波将分布于（）
A．整个区域B．阴影Ⅰ以外区域
C．阴影Ⅱ以外区域D．阴影II和III以外的区域
7．如图所示，一定质量的理想气体由状态A变化到状态B，则A、B 两个状态的密度相比（）
A．气体密度减小B．气体密度不变
C．气体密度增大D．气体密度先增大后减小
Ⅰ
ⅡⅢ
A
B
O
P
T 0
A
B
8．如图甲所示直线是一条电场线，A 、B 是该线上的两点。一正电荷由A 点以一定初速度沿电场线从A 运动到B ，运动过程中的v -t 图线如图乙所示，则下列说法中正确的是（ ）
A ．该电场是非匀强电场
B ．A 、B 两点的电势相比一定是φA ＜φB
C ．A 、B 两点的场强大小相比一定是E A ＜E B
D ．该电荷在两点的电势能大小相比一定是
E pA ＞E pB 9．雨滴从高空静止下落，受到的空气阻力随雨滴速度的增大而增大，则下落过程中雨滴的速度随时间变化的规律是（ ）
10．如图所示的电路，电池内阻不计，当R 2的滑动触头P 由a 端向b 端滑动过程中，电流表示数A 和电压表示数V 将（ ）
A ． V 不变， A 变大
B ． V 不变， A 变小
C ． V 变小， A 变大
D ． V 变小， A 变小
11．在同一平面上有a 、b 、c 三根等间距平行放置的长直导线，依次载有电流强度大小为1A 、2A 和3A 的电流，各电流的方向如图所示，则导线b 所受的合力方向是（ ）
A ．水平向左
B ．水平向右
C ．垂直纸面向外
D ．垂直纸面向里
12．如图所示，A 是一均匀小球，B 是个1/4圆弧形滑块，最初A 、B 相切于小球
的最低点，一切摩擦均不计。一水平推力F 作用在滑块B 上，使球与滑块均静止，现将滑块B 向右缓慢推过一较小的距离，在此过程中（ ）
A ．墙壁对球的弹力不变
B ．滑块对球的弹力增大
C ．地面对滑块的弹力增大
D ．推力F 减小
二、填空题 (共20分,每空2分)
13．如图所示，实线表示两个相干波源S 1、S 2发出的波的波峰位置，则图中的_______点为振动加强的位置，图中的_______点为振动减弱的位置。 14．若两颗人造卫星A 和B 绕地球做匀速圆周运动，角速度之比为8∶1，则A 和B 两颗卫星的周期之比为________，轨道半径之比为_________。
15．用细绳拴一个0.1kg 的小球，从地面以2m/s 2的加速度竖直向上提升了25m ，空气阻力不计，g =10m/s 2，则拉力对小球所做的功为______J ，若小球到达25m 处时撤去细绳的拉力，则小球落地时的速度为_______ m/s 。
S 1
S 2
P
Q
1A 2A 3a
b c
A V
R 1
R 2 S
P
E a b 0 v
B
A
B
A 甲
乙
t
F
B
A
图（乙）
U /V
I /A
0.5
1.0
1.5
2.0 4.0
6.0 16．如图（甲）所示的电路，R 1为定值电阻，R 2为滑动变阻器。闭合开关S 后，调节滑动变阻器，将滑片从a 端向b 端滑动的过程中，两只电压表随安培表示数变化的完整图像（实线）如图（乙）所示。则电源的内阻为_______Ω，R 1的最大功率为_______W 。
17．如图所示的单摆，摆长为L ，小球体积忽略不计，且Ob =L ，把小球拉到悬线与竖直方向成θ角后，静止释放使其做简谐振动，则球从a 运动到b 的时间为_______s ，当其运动到O 点正下方b 点时，摆线在p 处被烧断，不计小球在b 处的能量损失，小球继续沿光滑水平轨道运动，此轨道与光滑竖直的圆轨道的最低点相切，小球沿圆轨道运动时恰能通过最高点，则圆轨道的半径为_______m 。 三、综合题（第18题10分，第19题14分，第20题16分，共40分）
18．（10分）简易温度计构造如图所示。两内径均匀的竖直玻璃管下端与软管连接，在管中灌入水银后，将左管的上端通过橡皮塞插入玻璃泡。在标准大气压下，调节右管的高度，使左右两管的液面相平，在左管液面位置标上相应的温度刻度。多次改变温度，重复上述操作。 （1）玻璃泡内封闭气体发生的状态变化可近似的认为是_______变化。初始温度为t 1＝0o C ，当温度为t 2＝25 o C ，则封闭气体的压强为_______cmHg 。 （2）（单选题）此温度计的特点是（ ）
A ．刻度均匀，刻度值上小下大
B ．刻度均匀，刻度值上大下小
C ．刻度不均匀，刻度值上小下大
D ．刻度不均匀，刻度值上大下小
（3）当大气压减小时，左右两管的液面相比________（选填“左管高”“一样高”或“右管高”），用该温度计测得的温度值_______（选填“偏大”或“偏小”）。
19．（14分）如图所示，质量m =4kg 的小物块从倾角θ=37°的光滑斜面上的A 点由静止开始下滑，经过B 点后进入粗糙水平面，已知AB 长度为3m ，斜面末端B 处与粗糙水平面平滑连接。（g =10m/s 2，sin37°=0.6，cos37°=0.8 ）试求： （1）小物块滑到B 点时的速度大小。
（2）若小物块从A 点开始运动到C 点停下，一共经历时间t =2.5s ，求BC 的距离。
（3）上问中，小物块与水平面的动摩擦因数μ多大？
a
b
c
O
θ
L
v
L
P P S
R 1
R 2
V 2 A 图（甲）
V 1
a b
E r
20．（16分）如图所示，足够长光滑的两平行金属导轨，间距为L ，导轨平面与水平面成θ角，定值电阻大小为R 。导轨上停放着一根质量为m 、电阻为r 的金属杆CD ，导轨电阻不计，整个装置处于磁感应强度为B 的匀强磁场中，磁场方向垂直导轨平面向上。现用一垂直于金属杆CD 的外力F ，沿导轨斜面向上拉金属杆，使之由静止开始沿导轨向上做加速度为a 的匀加速直线运动，并开始计时。 试求：
（1）推导外力F 随时间t 的变化关系； （2）在时间t 0内通过电阻的电量；
（3）若时间t 0末，撤销拉力F ，试分析并说明导体棒在撤力以后的速度、加速度的变化情况。
参考答案
一、选择题
1、A
2、C
3、C
4、D
5、B
6、B
7、A
8、B
9、D 10、A 11、A 12、B 二、填空题
13、P ； Q 14、1:8 ； 1:4 15、30 ； 610 16、0.5 ；7.875 17、
g
L
2π
; )cos 1(52θ-L 三、综合题
18、（10分）每空2分（1）等容（等体积）；82.96 （2）A （3）右管大； 偏大 19、（14分） （1）根据机械能守恒得：2
2
137sin B AB mv mgs =
? （2分） A
B
C
D
37°
B
C D
R
r
θ F
s m s m gs v AB B /6/6.0310237sin 2=???=?=（2分）
（2）物块在斜面上，受力如图所示， （1分）
由牛顿第二定律，F 合=ma ， F 合= mgin θ （1分） 物块在斜面上的加速度为：a 1=gsin θ=6m/s 2， （1分） 在斜面上有：2
12
1t a s AB =
， 代入数据解得：t 1=1s （1分） 物块在BC 段的运动时间为：t 2=t -t 1=1.5s BC 段的位移为：m m t v s B BC 5.45.162
1
212=??==
（2分） （3）在水平面上，受力如图所示， （1分）
有：2222m/s 4m/s 5
.16===
t v a B （1分） 根据牛顿第二定律有：F 合=ma ，F 合=f =μmg （1分）
a 2=μg 代入数据解得：μ=0.4 （1分）
20.（16分）
（1）如图所示，当CD 棒向上运动时，对其进行受力分析： （1分）
由 闭合电路欧姆定律 r
R BLv
r R +=+=
E I （1分）
r
R v
L B BIL F A +==22 （1分）
由物体匀加速直线运动 at v t =
（1分）
由牛顿运动定律： ma mg =+θsin -r R v
L B -F 22 （1分） 所以：ma mgsin t r
R a
L B 22+++=
θF （1分） （2）由电流的定义式：t
q
I =
可知：It q = （1分） 由t r
R BLa
I +=+=
r R BLv 可知：电流I 随时间t 线性变化，如图所示 （2分）
G N G
N f
G
N F F
A
)
(22q 2
0r R BLat t I I t I t
+=
+=
= （1分）
（3）撤去拉力F 后。
①在上升阶段，如图受力分析：因为运动方向和合外力方向相反，所以CD 减速运动。
（1分）
由牛顿第二定律：ma r R v
L B mg =++
22sin θ )
(sin 22r R m v
L B g a ++=θ （1分）
因为速度减小，所以CD 棒做加速度减小的减速运动。 （1分）
②在下降阶段阶段，如图受力分析： 开始时，速度较小，F A 较小，合外力方向向下，与速度方向相同， CD 棒做加速运动。 （1分）
由牛顿第二定律：ma r R v
L B mg =+22—
sin θ )
(—sin 22r R m v
L B g a +=θ （1分）
因为速度增大，所以CD 棒先做加速度减小的加速运动，最后做匀速直线运动。（1分）
以上答案仅供参考，如有错误，请自行改之！
I
t
N
G
F
N
F G
2017届上海市徐汇区高三英语二模卷(含听力文本和答案)
2016学年第二学期徐汇区高三模拟考英语试卷2017.4 I. Listening Comprehension Section A Short Conversations Directions: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard. 1. A. He knows who is knocking. B. He is eager to know who it is. C. He doesn’t want to open the door. D. He is ready to open the door. 2. A. By plane. B. By bus. C. By taxi. D. By train. 3. A. $100. B. $200. C. $300. D. $400. 4. A. She went to cinema. B. She went to an exhibition. C. She stayed at home. D. She stayed with her classmates. 5. A. In a doctor’s office. B. In a professor’s office. C. In an operating room. D. In an emergency ward. 6. A. The man paid the tuition for learning physics. B. The man got a lot of money for his hard work. C. His hard work was not rewarding at all. D. His work before the test led to a good result. 7. A. A furnished house. B. A recent book. C. A further study. D. A new record. 8. A. They will go swimming. B. They will climb mountains. C. They will buy some clothes. D. They will forecast the weather conditions. 9. A. He has another lecture to attend. B. He has no interest in the lecture. C. He’s attended the same lecture given by Professor Wilson before. D. He might miss the lecture, if the woman didn’t remind him. 10.A. She fully agrees with the man. B. They are uncertain about the weather. C. She disagrees with the man. D. She thought the man was always late. Section B Directions: In Section B, you will hear several longer conversation(s) and short passage(s), and you will be asked several questions on each of the conversation(s) and the passage(s). The conversation(s) and the passage(s) will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard. Questions 11 through 13 are based on the following passage. 11. A. People are encouraged to be a craftsman. B. Learning woodworking is not as hard as you think. C. Learning woodworking will help you know more people. D. Taking a class in woodworking will be very helpful. 12. A. Because I am a talent in this art and want to share it with others. B. Because I am interested in it and want to show it to others. C. Because I wonder how to pick materials and how to do it well. 1 / 16
2017年上海市青浦区高考数学一模试卷(解析版)
2017年上海市青浦区高考数学一模试卷 一.填空题（本大题满分54分）本大题共有12题，1-6每题4分，7-12每题5分考生应在答题纸相应编号的空格内直接填写结果，每个空格填对得分，否则一律得零分. 1．已知复数z=2+i（i为虚数单位），则． 2．已知集合，则A∩B=． 3．在二项式（x+）6的展开式中，常数项是． 4．等轴双曲线C：x2﹣y2=a2与抛物线y2=16x的准线交于A、B两点，|AB|=4，则双曲线C的实轴长等于． 5．如果由矩阵=表示x，y的二元一次方程组无解，则实数a=．6．执行如图所示的程序框图，若输入n=1的，则输出S=． 7．若圆锥的侧面积为20π，且母线与底面所成的角为，则该圆锥的体积为． 8．设数列{a n}的通项公式为a n=n2+bn，若数列{a n}是单调递增数列，则实数b 的取值范围为． 9．将边长为10的正三角形ABC，按“斜二测”画法在水平放置的平面上画出为△A′B′C′，则△A′B′C′中最短边的边长为．（精确到0.01） 10．已知点A是圆O：x2+y2=4上的一个定点，点B是圆O上的一个动点，若满
足|+|=|﹣|，则?=． 11．若定义域均为D的三个函数f（x），g（x），h（x）满足条件：对任意x∈D，点（x，g（x）与点（x，h（x）都关于点（x，f（x）对称，则称h（x）是g（x） 关于f（x）的“对称函数”．已知g（x）=，f（x）=2x+b，h（x）是g（x）关于f（x）的“对称函数”，且h（x）≥g（x）恒成立，则实数b的取值范围是． =ka n+3k﹣3，其中k为不等于0 12．已知数列{a n}满足：对任意的n∈N*均有a n +1 与1的常数，若a i∈{﹣678，﹣78，﹣3，22，222，2222}，i=2，3，4，5，则满足条件的a1所有可能值的和为． 二.选择题（本大题满分20分）本大题共有4题，每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分. 13．已知f（x）=sin x，A={1，2，3，4，5，6，7，8}现从集合A中任取两个不同元素s、t，则使得f（s）?f（t）=0的可能情况为（） A．12种B．13种C．14种D．15种 14．已知空间两条直线m，n两个平面α，β，给出下面四个命题： ①m∥n，m⊥α?n⊥α； ②α∥β，m?α，n?β?n⊥α； ③m∥n；m∥α?n∥α ④α∥β，m∥n，m⊥α?n⊥β． 其中正确的序号是（） A．①④B．②③C．①②④D．①③④ 15．如图，有一直角墙角，两边的长度足够长，若P处有一棵树与两墙的距离分别是4m和am（0＜a＜12），不考虑树的粗细．现用16m长的篱笆，借助墙角围成一个矩形花圃ABCD．设此矩形花圃的最大面积为u，若将这棵树围在矩形花圃内，则函数u=f（a）（单位m2）的图象大致是（）
2017届浦东区高三英语二模试卷(含答案)
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