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北京市东城区2010-2011学年第二学期初三英语二模

北京市东城区2010-2011学年第二学期初三英语二模
北京市东城区2010-2011学年第二学期初三英语二模

北京市东城区2010-2011学年第二学期初三综合练习(二)

物理试卷

考生须知

1.本试卷共8页,共五道大题,38道小题,满分100分。考试时间120分钟。

2.在试卷、答题纸上准确填写学校名称、姓名和考号。

3.试卷答案一律填涂或书写在答题纸上,在试卷上作答无效。

4.考试结束,将试卷和答题纸一并交回。

一、单项选择题(下列各小题中只有一个选项符合题意。共24分,每小题2分)

1.在国际单位制中,电功率的单位是

A.焦耳(J)B.伏特(V)C.欧姆(W)D.瓦特(W)

2.如图1所示的实例中,没有应用连通器原理的是

3.下列措施中,能使蒸发加快的是

A.拧紧香水瓶的盖子B.用电吹风机吹潮湿的头发

C.用地膜覆盖农田D.把新鲜的蔬菜装入塑料袋

4.如图2所示的现象中,属于光的反射现象的是

5.对下列物体的质量和长度的估计值最接近实际的是

A.一角硬币的直径大约是2 .5cm B.一支签字笔的质量约为500g

C.初中物理课本的长度约为0.26m D.一个鸡蛋的质量约为0.5kg

6.如图3所示的电路中,开关闭合后,三个灯泡并联的电路是

7.在内燃机的一个工作循环中,既有机械能转化为内能的过程,也有内能转化为机械能的过程,这两个能量转化的过程分别存在于

A.吸气冲程和压缩冲程B.压缩冲程和做功冲程

C.做功冲程和排气冲程D.吸气冲程和排气冲程

8.如图4所示的图象中能,正确描述各物理量之间关系的是

9.关于家庭电路,下列说法正确的是

A.空气开关自动切断,一定是出现了短路

B.同时工作的用电器越多,总电阻越大

C.使用试电笔时,人的手不能接触笔尾金属体

D.控制灯泡的开关应串联在火线和灯泡之间

10.如图5所示,甲、乙两个实心正方体对水平面的压强相同,边长a甲=2a乙。如果沿竖直方向将甲、乙分别切去厚度为各自边长的部分,然后将甲切去部分叠放在乙的剩余部分上,将乙切去部分叠放在甲的剩余部分上,此时甲对水平面的压强

p甲,乙对水平面的压强p乙,甲对水平面的压力F甲,乙对水平面的压力F乙,下列判断正确的是

A.p甲> p乙B.p甲< p乙

C.F甲< F乙D.F甲= F乙

11.如图6所示,A、B重G A:G B=1:3,用甲、乙两滑轮组分别匀速

提升A、B两物体,在相同时间内两物体被提升高度分别为hA和hB,

2hA=5hB。已知两滑轮组动滑轮重分别为G甲和G乙,且G甲? G乙=1?3。

两滑轮组的机械效率分别为η1和η2,功率分别为P1和P2。

若不计绳重和滑轮轴处摩擦,则下列判断正确的是

A.η1 :η2 = 1 : 1 B.F1:F2= 3 : 8

C.P1 :P 2 = 5 : 3 D.vA:vB =2 : 5

12 .如图7所示电路,开关S闭合后,电流表A1的示数大于A2的示数。下列说法正确的是

A.若只将R1与R2的位置对调,电流表A1的示数一定不变,

A2的示数一定变小

B.若只将R1与R3的位置对调,电流表A1的示数一定变大,

A2的示数一定变小

C.若用电压表V1 、V2分别替换电流表A1 、A2,电压表

V1的示数比V2的大

D.若只用电压表V1替换电流表A1,电压表V1无示数,

电流表A2的示数变大

二、多项选择题(下列各小题中符合题意的选项均多于一个。本大题共12分,每小题3分。每小题全部选对的得3分,选对但不全的得2分,有错选的不得分)

13.下列说法正确的是

A.在空中匀速下落的物体,重力势能减小,动能增大

B.物体的运动状态改变时,物体所受的合力一定不等于零

C.晶体熔化需要吸热,非晶体熔化不需要吸热

D. 发电机是利用电磁感应现象的原理制成的

14.下列说法正确的是

表1:几种晶体的熔点(在标准大气压下)表2:几种物质的比热容

A.质量相同的水和沙石吸收相同的热量,沙石升高的温度大于水升高的温度

B.因为水的凝固点比较低,机械的散热系统一般用水做制冷剂

C.水结冰时放出热量,所以冰的温度一定比水的温度低

D.在-20℃的寒冷环境中,水银是液态的

15.在A、B、C三个相同的烧杯内装有深度相同的液体,三种液体的密度关系是

ρ1=ρ2>ρ3。将甲、乙、丙三个重力为G甲、G乙、G丙的实心小球分别放在A、B、C的液体中,其中甲球在A中沉入液体底部,乙球在B中恰好悬浮,丙球在C中漂浮在液面上。三个球的密度分别为ρ甲、ρ乙、ρ丙,体积关系是V甲>V乙=V丙,三个球所受浮力分别为F1、F2、F3,三个烧杯里的液体对杯底的压强分别为p1、p2、p3。以下关系中正确的是

A.ρ甲>ρ乙=ρ丙

B.G甲>G乙>G丙

C.F1 > F2 > F3

D.p1 = p2 > p3

16.如图8所示电路图,把端点A、B接在电压U不变的电源两端,只闭合开关S1,电压表的示数为U1,电流表A1的示数为I1。把端点C、D 接同一电源两端,只闭合开关S2,电压表的示数为U2,电流表A2的示数为I2。把端点A、D接同一电源两端,只闭合开关

S1,电压表的示数为U3,电流表A1的示数为I3。把端点B、C接同一电源两端,只闭合开关S2,电压表的示数为U4,电流表A2的示数为I4 。已知:R1=R3=12Ω,U1=6V, U1∶U3 =4∶3,I3∶I4=3∶5。则

A.定值电阻R2=3Ω

B.电源两端电压U =30V

C.把端点B、C接同一电源两端,只闭合开关S1,总功率P =300W

D.把端点B、C接同一电源两端,闭合开关S1、S2,总电阻R =2.5Ω

三、填空题(共14分,每空2分)

17.一束激光射到平面镜上,当入射角是65°时,反射角是°。

18.拉萨与北京相比,海拨高度要高出数千米,所以通常拉萨地区的大气压强

(选填“高于”、“等于”或“低于”)北京地区的大气压强。

19.天然气是一种清洁能源。1m3的天然气完全燃烧能放出J的热量。(天然气的热值取7.1×107J/m3)

20.0.1kW×h的电能可以使标有“220V 40W”的灯泡正常工作分钟。

21.如图9所示,A、B两段同种材料制成的实心导体电阻值相等,A导体比B导体的长度和横截面积都小。将它们串联在电路中,通电一段时间后,A导体升高的温度

(选填“大于”、“等于”或“小于”)B导体升高的温度。

22.如图10甲所示电路中,R1是滑动变阻器,R2是定值电阻,电源两端电压保持6V 不变,当滑动变阻器滑片P从最右端滑到最左端,电压表示数随电流变化的图象如图10乙所示,则滑动变阻器的最大阻值为Ω。

23.如图11甲所示;一个底面积为50 cm2的烧杯装有某种液体,把小石块放在木块上,

静止时液体深h1=16cm;如图11乙所示;若将小石块放入液体中,液体深h2=12 cm,石块对杯底的压力F=1.6N;如图11丙所示,取出小石块后, 液体深h3=10cm。则小石块的密度ρ石为kg/m3。

(g取10N/kg)

四、实验与探究题(共34分,24-27、29、30、34题各2分;28、33题各3分;31、32、题各4分;35题6分)

24.如图12所示,电流表的读数是A。

25.在图13中,画出入射光线AO的反射光线。

26.如图14所示,温度计的读数是℃。

27.小东同学探究重力大小与质量的关系时,把不同质量的钩码挂在弹簧测力计下,分别测出它们所受的重力,并记录在下表中。根据数据归纳出物体所受重力大小与其质量的关系式为___________。

钩码的质量m/kg0.050.10.150.20.250.3

钩码受到的重力G/N0.490.981.471.962.452.94

28.在下列各组实验中:

(1)如图15(a)所示, 从密封罩向外抽气的过程中,铃声越来越小,最后几乎听不到了。这个实验说明: 。

(2)如图15(b)所示,将压强计的探头先后放入两种不同的液体中,这个实验主要是探究:液体的压强与__________的关系。

(3)如图15(c)所示,闭合开关S,根据电磁铁甲、乙吸引小铁钉的情况,可以得出结论:。

29.如图16是“探究平面镜成像特点”的情景:竖立的透明玻璃板下方放一把直尺,直尺与玻璃板垂直,两支相同的蜡烛A、B竖立于玻璃板两侧的直尺上,以A蜡烛为成像物体。

(1)采用透明玻璃板代替平面镜,能在观察到A蜡烛像的同时,也观察到,

解决了确定像的位置和大小的问题。

(2)点燃A蜡烛,移动B蜡烛,直到与A蜡烛的像重合为止。这时发现像与物的大小相等,进一步观察A、B两支蜡烛在直尺上的位置发现,像与物到玻璃板的距离。

30.如图17甲所示,小文在探究凸透镜成像规律时,将凸透镜A固定在光具座上保持位置不变,使烛焰在光屏上成清晰的像。把烛焰与光屏的位置对换后,在光屏上将观察到倒立、(选填“放大”、“缩小”或“等大”)的实像。小文将光屏先后放在如图17乙所示位置,在光屏上能观察到大小相等的光斑,凸透镜B的焦距为cm。

31.做测定“小灯泡的电功率”实验时,所用器材有:电压为3V的电源,额定电压为2.5V 的小灯泡,以及符合实验要求的滑动变阻器、已调零的电流表和电压表、开关、导线若干。

(1)用笔画线代替导线,将图18甲的实物电路连接完整。

(2)连好电路闭合开关S,移动滑动变阻器滑片P,发现小灯泡始终不亮,电压表有示数,电流表无示数,则故障的原因可能是出现断路。(选填“小灯泡”或“滑动变阻器”)

(3)排除故障后闭合开关S,移动滑片P到某处,电压表的示数如图18乙所示。要测量小灯泡的额定功率,应将滑片P向(选填“左”或“右”)滑动。

(4)通过移动滑片P,分别记录多组对应的电压表和电流表的读数,绘制成了图18 丙所示的U-I图象,小灯泡的额定功率是W。

32.为探究“电流通过导体产生的热量跟哪些因素有关”,小成设计了图19所示的实验电路。烧瓶中盛有质量相同的煤油,闭合开关S1、S2,用电阻丝(其中R甲=R丙<R乙)给煤油加热相同时间,观察并记录烧瓶中温度计示数的变化情况。

(1)比较烧瓶甲和烧瓶乙中温度计示数的变化,研究电热与的关系。

(2)比较烧瓶甲和烧瓶丙中温度计示数的变化,研究电热与的关系。

(3)小成对此实验装置稍做改装,用改装后的装置测量未知液体比热容。测量时,分别在两烧瓶中装入水和待测液体,闭合开关,一段时间后分别用温度计测出水和待测液体升高的温度Δt水和Δt,在忽略热损失的情况下,则待测液体的比热容c=Δt水c水/Δt。

小成对此实验装置做出的改装是_________。

为了得出待测液体比热容的表达式c=Δt水c水/Δt,实验中还需要增加一种测量仪器,说出该仪器名称和作用_________。

33.小东猜想动能的大小可能与物体质量和运动速度有关,于是设计了如图20甲、乙所示的实验,探究动能的大小与哪些因素有关。

(1)如图20甲所示,让不同质量的两个小球沿同一光滑斜面从B处分别开始运动,然后与放在水平面上的纸盒相碰,纸盒在水平面上移动一段距离后静止。小球从同一高度开始运动的目的是使两球到达水平面时具有相同的__________。得出的结论是:__________。

(2)如图20乙所示,让质量相同的两个小球沿同一光滑斜面分别从A处和B处开始运动,然后与放在水平面上的纸盒相碰,纸盒在水平面上移动一段距离后静止,得出的结论是:__________。

34.小亮认为:滑轮组的机械效率由滑轮组本身决定,与被提升的重物无关。请利用如图21所示的实验装置和若干个质量已知的钩码设计实验,证明小亮的看法是不正确的。

35.如图22所示,在一个已经调平衡、有刻度的杠杆右侧挂重力为G0 的钩码,钩码的位置B固定不动;左侧挂重力未知的金属圆柱体,悬挂点A的位置可以在杠杆上移动,细线长短可以调整,此时杠杆在水平位置平衡;量筒中盛有适量的水。要求利用上述实验装置和器材,探究“浸在水里的物体所受浮力大小与物体浸入水中的体积成正比”。

写出主要实验步骤,设计实验数据记录表。

五、计算题(共16分,36题3分,37题7分,38题6分)

36.热水器内盛有25℃的水20kg,吸收了2.52×106J的热量后,这些水的温度能升高到多少摄氏度?[c水=4.2×103J/(kg×℃) ]

37.如图23所示,电源两端电压保持不变,当开关S1断开、S2闭合时,电流表的示数为I1,电阻R2的电功率为P2,总功率为P总。当开关S1闭合、S2断开,滑动变阻器的滑片P在B点时,变阻器连入电路的电阻为RB,此时电流表的示数为I2,电阻R2的电功率为P2,,总功率为P总,。当开关S1、S2都断开,滑动变阻器的滑片P在C点时,变阻

器接入电路的电阻为Rc,电压表V1的示数为U1,电压表V2的示数为U2,电流表的示数为I3,。已知P2∶P2,=9∶4,

U1∶U2=3∶1,I2∶I3=7∶9。求:

(1)P总∶P总,;

(2)I1 ∶I3;

(3)RB ∶Rc。

38.如图24是利用电子秤显示水库水位装置的示意图。该装置主要由滑轮C、D,物体A、B以及轻质杠杆MN组成。物体A通过细绳与滑轮C相连,物体B通过细绳与杠杆相连。杠杆可以绕支点O在竖直平面内转动,杠杆始终在水平位置平衡,且MO:MN=1:3。物体B受到的重力为100N,A的底面积为0.04m2,高1 m。当物体A恰好浸没在水中时,物体B对电子秤的压力为F1;若水位下降至物体A恰好完全露出水面时,物体B对电子秤的压力为F2,已知:每个滑轮的重力为20N,F1﹕F2=27﹕7。滑轮与转轴的摩擦、杠杆与轴的摩擦均忽略不计,g取10N/kg。求:

(1)物体A的底部距水面深为0.75 m时,

A受到的浮力F浮。

(2)物体A的密度ρA。

(3)如果把细绳由N端向左移动到N,处,电子

秤的示数恰好为零,NN …﹕MN =17﹕60,此

时物体A露出水面的体积V露。

理试卷参考答案

一、单项选择题(每题2分,共24分。错选、多选不得分)

题号123456789101112

答案DDBACDBBDBAC

二、多项选择题(每题3分,共12分。选对但不全者得2分,错选,不选者不得分)

题号13141516

答案B DA DB CA B C

三、填空题(每空2分,共14分)

题号答案得分

17652

18低于2

197.1×1072

201502

21大于2

22502

232.4×1032

四、实验与探究题(共34分)

24.(2分)0.18

25.(2分)略

26.(2分)24

27.(2分)G =(9.8N/kg)m 或G =9.8m (N)

28.(3分) (1) 真空不能传声(2) 液体密度

(3) 电流相同时,线圈的匝数越多,磁性越强

29.(2分)B蜡烛相等

30.(2分)缩小30

31.(4分)(1) 图略(2) 小灯泡(3)右⑷ 0.5

32 .(4分)(1)电阻(2)电流(3)去掉烧瓶乙,闭合开关S1、S2;(或去掉烧瓶乙,烧瓶甲与烧瓶丙串联,闭合开关S2)天平用天平称量等质量的水和待测液体

33.(3分)(1)速度速度相同时,物体质量越大,动能越大

(2)质量相同时,物体速度越大,动能越大

34.(2分)在动滑轮下挂1个钩码,用已调零的弹簧测力计竖直向上匀速拉绳端,记录绳端移动距离s和钩码移动距离h,所挂钩码的质量m,弹簧测力计的示数F,利用公式η=W 有/W总=mgh/Fs求滑轮组的机械效率η1;增加钩码,仿照上述方法测机械效率η2。η1不等于η2,说明小亮的看法不正确。

35.(6分)实验步骤:

(1)记录杠杆平衡时OB的距离L0,AO的距离L,钩码的重力G0;水面对应的量筒的刻度V0在表格中;(1分)

(2)把量筒移到圆柱体下方,调整细线长度,使部分圆柱体浸入水中,此时水面对应的量筒的刻度V1,向左调整杠杆左侧悬点位置至A1,使杠杆在水平位置平衡,A 1O的距离L?1,并把L?1、V1记录在表格中;(1分)

(3)依次改变圆柱体浸入水中的体积,仿照步骤2,再做5次实验,并将各次V2、V3、V4、V5、V6和L?2、L?3、L?4、L?5、L?6记录在表格中;(1分)

(4)根据,根据杠杆平衡条件G圆柱体= ;(1分),解得,计算出圆柱体所受F 浮1、F浮2、F浮3、F浮4、F浮5、F浮6,并记录在表格中。(1分)

(其他答案正确的,均可相应得分)

实验数据记录表(1分)

G0/N

V0/m3

V/m3

V排/m3

L0/m

L/m

L…/m

F浮/N

五、计算题(共16分)

36.Q吸=c水m(t-t0) ……………………………(1分)

=4.2×103J/(kg?℃)×20kg×(t-25℃) ……………………………(1分)

=2.52×106J

t =55℃……………………………(1分)

37.解:当闭合S2、断开S1时,等效电路如图1;当P在B,闭合S1、断开S2时,等效电路如图2;当P在C,断开S1、S2时,等效电路如图3;

解③④⑤可得:……(1分)

38.解:(1)由阿基米德原理可知:…(1分)

(2)当A恰好完全浸没在水中时:受力分析图………………(1分)

对A和动滑轮C受力分析如图1(a):……………⑴

对动滑轮D受力分析如图1(b):……………⑵

对杠杆MN受力分析如图1(c)所示:……………⑶

对B受力分析如图1(d):……………⑷

当A恰好完全露出水面时:

对A和动滑轮C受力分析如图2(a):……………⑸

对动滑轮D受力分析如图2(b):……………⑹

对杠杆MN受力分析如图2(c)所示:……………⑺

对B受力分析如图2(d):……………⑻

又因为:……………⑼

解⑴⑵⑶⑷⑸⑹⑺⑻⑼得:GA=600N……………⑽

(3)当把N移到N′处时:

对A和动滑轮C受力分析如图3(a):……………⑾

对动滑轮D受力分析如图3(b):……………⑿

对杠杆MN受力分析如图3(c)所示:……………⒀

对B受力分析如图3(d):……………⒁

解⑽⑾⑿⒀⒁得:F浮3=200N …………………(1分)......

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