麓山国际2014-2015-1初三第二次限时训练物理

麓山国际2014-2015-1初三第二次限时训练物理
麓山国际2014-2015-1初三第二次限时训练物理

一、选择题(每小题只有一个正确答案,每题3分,共36分)

1.(3分)扩散现象的发生是由于()

A.分子在永不停息地做无规则运动

B.分子之间有作用力

C.分子间的斥力大于引力

D.物质是由分子组成的

2.(3分)在一个电源两端接有两个不同的灯泡,用电流表测得通过这两个灯泡的电流不相等,则这两个灯泡的连接方式()

A.一定是串联B.一定是并联

C.串联、并联都可以D.条件不足,无法判断

3.(3分)实验装置如图所示,在一个厚壁玻璃筒里放一块浸有少量乙醚(乙醚极易挥发)的棉花,用力把活塞迅速下压,棉花就会立即燃烧.根据该实验现象得出的下列结论正确的是()

A.气体比液体更容易被压缩

B.浸有少量乙醚可以降低棉花的着火点

C.活塞迅速下压,乙醚蒸气液化放出热量,使棉花燃烧

D.外界对物体做功时,物体的内能会增加

4.(3分)如图所示的四个电路图中,电源电压均为U,每个电阻的阻值都为R,开关闭合后,哪个电路中电流表的示数最大?()

A.B.C.D.

5.(3分)如图电路中,当开关S闭合后,电压表测出的电压是

()

A.L1两端的电压

B.L2两端的电压

C.电源和L2串联后的总电压

D.电源电压

6.(3分)将质量相同的甲、乙、丙三块金属加热到相同的温度后,放到上表面平整的冰块上.经过一定时间后,冰块形状基本不再变化时的情形如图所示.则三块金属的比热容c甲、c乙、c丙大小相比()

A.c甲最大B.c乙最大C.c丙最大D.c甲=c乙=c丙

7.(3分)如图所示的各种电路,同种元件的参数均相等,能利用滑动变阻器调节电灯从亮到熄灭的电路是()

A.B.C.D.

8.(3分)教室里投影仪的光源是强光灯泡,发光时必须用风扇给予降温.在使用投影仪时,要求先启动带动风扇的电动机,再使灯泡发光;如果风扇不启动,灯泡就不能发光.如图所示的电路图中能符合要求的是()

A.B.C.D.

9.(3分)冰在熔化过程中,下列判断正确的是()

A.吸收热量,温度不变

B.内能不变,比热容不变

C.比热容、内能、温度都不变

D.比热容变大、内能增加,温度升高

10.(3分)如图所示,电源电压为4.5V且恒定不变,当开关s闭合后,滑动变阻器的滑片P从a端滑向b端的过程中,三只理想电压表的示数变化的绝对值分别为△U1、△U2、△U3,下列可能出现的情况是()

A.△U1=OV、△U2=2V、△U3=1V

B.△U1=OV、△U2=2V、△U3=2V

C.△U1=0.5V、△U2=1V、△U3=1.5V

D.△U1=0.2V、△U2=1V、△U3=0.8V

11.(3分)如图所示电路中,电源电压保持不变,闭合开关S后,将滑动变阻器R2的滑片P向右移动,在此过程中()

A.电流表A示数变大,电压表V2示数变小

B.电流表A示数变大,电压表V1示数变小

C.电压表V1示数变小,电压表V2示数变大

D.电压表V1示数不变,电压表V2示数变大

12.(3分)标有“4V 1W”的小灯泡和“20Ω 1A”的滑动变阻器连接在图所示的电路中,电源电压为6V,电流表量程为“O~O.6A”,电压表量程为“0~3V”.为确保电路安全,闭合开关时,滑动变阻器接入电路的阻值变化范围应控制在(不考虑温度对电阻的影响)()A.O~8Ω

B.8~16Ω

C.16~20Ω

D.以上答案都不对

二、填空题(每空2分,共22分)

13.(4分)有一个导体,两端的电压为10V时,通过的电流为1A,则它的电阻是

Ω;当导体两端的电压为零时,该导体的电阻是Ω.

14.(4分)一只小灯泡正常工作时的电压为6V,正常发光时的电流为0.5A,现手边只有一个电压为9V的电源,为了使小灯泡能正常发光,应联一只阻值为的电阻.

15.(6分)用煤气灶把1kg、初温为20℃的水烧到70℃,消耗了1×10﹣2kg煤气.已知水的比热容是4.2×103J/(kg?℃),煤气的热值为4.2×107J/kg,则水吸收的热量为J,燃烧煤气放出的热量为J煤气灶烧水的效率为%.

16.(4分)如图,用铁锤连续敲打铁块,铁块变热,内能;铁块的内能是通过

的方式改变的.

17.(2分)利用如图的电路,测量Rx的阻值.电源电压保持不变,S是单刀双掷开关、R′是电阻箱、R0是已知阻值的电阻、Rx是待测电阻.将开关S拨到a,电流表的示数为

I;再将开关S拨到b,调节电阻箱R′的阻值,当电流表的示数为时,读出电阻箱的阻值为R,则被测电阻Rx的阻值为.

18.(2分)如图的电路中,三个电阻R1、R2、R 3的阻值分别为1Ω、3Ω、6Ω,电流表

A1、

A2和A3的内阻均可忽略,它们的示数分别为I1、I2和I3,则I1:I2:I3=.

三、实验探究题(每空2分,共24分)

19.(8分)实验测得1kg某物质温度随时间变化的图象如图.已知该物质在固态下的比热容c1=2.1×103J/(kg?℃),设物质从热源吸热的功率恒定不变,根据图象解答下列问题:(1)在最初2min内,物质温度升高了40℃,则吸收的热量J,吸热功率为

W.

(2)在t2=6min时,物质处于固液共存状态,且仍在(填“吸热”、或“放热”).(3)该物质在液态下的比热容c2为J/(kg?℃).

20.(8分)在探究“电压一定时,电流跟电阻的关系”的实验中,设计电路图如图甲.

(1)根据电路图将实物连接成完整.连接电路时,开关闭合前、滑动变阻器的滑片应该处于(填“最左端”或“最右端”).

(2)连接好电路,闭合开关,发现电流表没有示数,移动滑动变阻器的滑片,电压表示数始终接近电源电压.造成这一现象的原因可能是

A.电流表坏了B.滑动变阻器短路

C.电阻处接触不良D.电阻短路

(3)排除电路故障进行实验,多次改变R的阻值,调节滑动变阻器,使电压示数保持不变,实验数据记录如表.其中第5次实验电流表示数如图丙其读数为A 实验次数12345

电阻R/Ω510152025

电流V/A0.60.30.20.15

)实验结论是:.

21.(8分)在“探究串联电路中各点的电流有什么关系”时,小明设计实验如右图.把两个灯泡L1、L2串联起来接到右图所示电路中,分别把图中A、B、C各点断开,把接入,测量流过的电流,看看它们之间有什么关系.

换上另外两个小灯泡,再次测量三点的,看看是否还有同样的关系.

A点的电流I A B点的电流I B C点的电流I C

第一次测量0.3A0.3A0.3A

第二次测量0.2A0.2A0.2A

(1)在拆接电路时,开关S必须处于状态.

(2)结论:串联电路中各处相等.

四、综合题(4+6+8共18分)

22.(4分)阅读下面短文,并回答问题:

日本福岛第一核电站引发核危机

北京时间2011年3月11日13时46分,日本本州岛仙台港附近发生了9.0级地震,地震引发的海啸造成福岛第一核电站冷却系统失灵,第1、2、3、4号反应堆相继发生爆炸,导致放射性物质泄漏,从而引发核危机.

核电站的核心设备是核反应堆,核反应堆是通过重核裂变释放核能的装置,图是核电站的原理图.核反应堆重核裂变,释放出核能传递给水,产生高温高压蒸汽,推动蒸汽轮机带动发电机发电.

3月11日回答问题:

(1)根据图,请在方框内完成核电站发电过程中能量的转化流程图:

(2)仔细观察核电站的原理图.在发电过程中,水发生的物态变化有汽化和液化.核电站在运行过程中要产生巨大热量,常用水来冷却,这是因为水具有的特性,所以,核电站一般建在海边.

23.(6分)如图所示电路中,R1=10Ω.当开关S闭合时,电流表示数为0.2A,电压表为4V.

求:

(1)R2的阻值有多大?

(2)R1两端的电压是多少?

(3)电源电压是多少?

24.(8分)如图电路中,电源电压不变,已知R l=6Ω,R2=4Ω,当S1和S2都闭合时,电流表A1的示数为1A,电流表A的示数为2.5A,求:

(1)电源电压.

(2)电阻R3的阻值.

(3)当S1和S2都断开时,电流表A l和电压表V的示数.

中考物理培优专题训练三

中学中考物理培优专题训练 一、例题 例1.下文是摘自某刊物的文章,请仔细阅读后,按要求回答问题。 《光污染,来自靓丽的玻璃幕墙》 最近张小姐十分苦恼,因为她的房子正对着一座新大厦的玻璃幕墙,有时站在她的窗前,看到对面玻璃幕墙就像平面镜一样,将同楼居民家的一举一动看得请请楚楚,玻璃幕墙的反光也使她苦不堪言,只要是晴天,她的房间就被强烈的反射光线照得通亮,无法正常休息。尤其是那种凹形建筑物,其玻璃幕墙在客观上形成一种巨型聚光镜,一个几十甚至几百平方米的凹透镜,其聚光功能是相当可观的,能使局部温度升高,造成火灾隐患…… ⑴从文中找出一个光学方面的物理知识填入横线:___________________________________; ⑵文中有一处出现了科学性的错误,请在错误句子下面画上横线; ⑶从上文找出一个玻璃幕墙给居民生活带来的不便或危害的实例:_____________________ _______________________________________________________________________________ ⑷如何预防光污染?请你提出一条合理化建议:_____________________________________ ______________________________________________________________________________。 [分析与解答]: ⑴平面镜成像是光的反射。 ⑵一个几十甚至几百平方米的凹透镜,其聚光功能是相当可观的,划线部分是错误的,因为凹透镜对光束是发散的,不可能会聚,作者将凹透镜和凸面镜混淆,显然为一误。 ⑶危害居民的身体健康,暴露居民的生活隐私,聚光甚至要引起火灾。 ⑷要限制开发商在居民区造商务大楼。对玻璃墙的设计、制作及使用范围要有统一的技术标准,防止光污染进入居民家或公共设施内。 例2.夏天,我们使用的吊扇配有一个调速器,当我们将吊扇的速度调慢时,过一会儿,调速器摸上去有点热。当我们将吊扇的速度调到最慢时,调速器摸上去温度最高。而当吊扇转速最快时,调速器几乎不发热。你是否发现这个问题,请试一下,并解释这个现象。 [分析与解答]:吊扇的调速器实际上相当于一个变阻器,它与吊扇串联连接,通过的电流相等。当它的 电阻变大时,吊扇的转速变慢,根据公式P =I 2 R 调速器的电阻也将消耗电功率,而这部分电能将转化为内能,所以调速器发热,当吊扇转速最小,调速器的电阻最大,消耗电功率最大,所以调速器温度也最高,反之亦然。 例3.家庭使用的可调的台灯是有一个小灯、一个滑动变阻器和开关连接成的,若小灯上标有“220伏,0.44安”,滑动变阻器的阻值变化范围是0-1000欧,请你分析图1 所示两个电路的台灯在使用时的优点和缺点。 [分析与解答]:由题意得R 灯=U 灯/I 灯=220伏/0.44安=500欧,如图a 所示,在变阻器电阻为零时,小灯两端电压为220伏。此时小 灯最大电流I 1=U/R 灯=220伏/500欧=0.44安小灯发光最亮;当变阻器电阻最大时,小灯 电流最小I 1’=U/R 总=U/(R 灯+R 滑)=220 伏/(500欧+1000欧)=0.15安小灯不会熄灭,灯的亮度调节范围较小。 对于图b 所示电路,当滑片位于最右端时,电路为并联电路,小灯两端为电源电压。此时小灯电流最大I 2=U/R 灯=220伏/500欧=0.44安小灯发光最亮;当滑片位于最左端,小灯短路,此时灯两端电压为零,则I 2’=0,小灯完全熄灭,灯的亮度调节范围较大。 在小灯同样亮度下,图a 所示电源输出电流小于图b 电路。如:(1)在小灯最亮时,图a 电源输出电流I 1=U/R 灯=220伏/500欧=0.44安,图b 电源输出电流I 2=I 灯+I 滑 =U/R 灯+U/R 滑=220伏/500欧+220伏/1000欧=0.44安+0.22安=0.66 安。⑵在通过小灯电流为零时,图a 中电源输出电流I 1’=0图b 中电源输出电流I 2’=U/R 滑=220伏/1000欧=0.22安,因此图b 电路较消耗电 能,不经济。 例3.饮水机的电路如图2 所示,水沸腾前红灯亮、绿灯灭,加热管正常 1 温控开关2 加 热 管 图2 (a ) 图1 (b)

高一英语阅读限时训练

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