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2016静安中考物理二模Word解析版

2016静安中考物理二模Word解析版
2016静安中考物理二模Word解析版

6 1 2 3 4 5 0 0.61.2 1.8 2.4

t /秒

甲车

12 2 4 6 8 10 0 0.9

1.8

2.7

3.6 t /秒 乙车

s /米

s /米

(a ) 图1 (b ) 静安、青浦区2015-2016学年初三年级第二次质量调研测试 九年级理化试卷

(满分:150分,时间:100分钟) 2016.4

物 理 部 分

考生注意:

1.本试卷的物理部分含五个大题。

2.考生务必按要求在答题纸规定的位置上作答,在草稿纸、本试卷上答题一律无效。 一、单项选择题(共16分)

下列各题均只有一个正确选项。请用2B 铅笔在答题纸的相应位置上填涂所选的选号;更改答案时,用橡皮擦去,重新填涂。

1.成语“震耳欲聋”形容的是声音的 A .响度大

B .音调高

C .音色好

D .频率高

2.依据卢瑟福的原子行星模型理论,在原子中绕核高速旋转的是 A .质子

B .中子

C .电子

D .核子

3.若光线从空气垂直射入水中,则反射角为 A .0°

B .30°

C .45°

D .90°

4.一本九年级物理教科书的质量约为

A .0.02千克

B .0.2千克

C .2千克

D .20千克

5.在掌心中滴一滴水珠,水珠下会有手纹的“清晰”像,这个“清晰”像是手纹的 A .倒立放大的像 B .倒立缩小的像

C .正立等大的像

D .正立放大的像

6.甲、乙两小车同时同地沿同一直线做匀速直线运动,它们的s -t 图像分别如图1(a )和(b )所示。两小车的速度关系及运动6秒时两小车之间的距离s 为

A .v 甲>v 乙;s 可能为1.2米

B .v 甲>v 乙;s 可能为0.6米

C .v 甲<v 乙;s 可能为1.2米

D .v 甲<v 乙;s 可能为4.2米

7.如图2所示,底面积不同的圆柱形容器分别盛有甲、乙两种液体,液体对各自容器底部的压

强相等。若在两容器中分别抽出相同高度的液体,则剩余液体对各自容器底部的压强p 、压力F 的关系是

A .p 甲>

p 乙;F 甲> F 乙 B .p 甲

F 乙 D .p 甲=p 乙;F 甲< F 乙

8.在图3所示的电路中,电源电压保持不变。当电键S 闭合后,电压表V 1的示数小于电压表V 2的示数。电路中的灯L 、电阻R 可能存在故障。下列关于电路的判断,正确的是

① 若灯L 发光,电路一定完好。 ② 若灯L 不发光,一定是灯L 短路。 ③ 若灯L 不发光,可能是电阻R 短路。

④ 若灯L 不发光,可能是灯L 短路且电阻R 断路。

A. ①和②

B. ②和③

C. ①和④

D. ③和④ 二、填空题(共23分)

请将结果填入答题纸的相应位置。

9.上海地区家庭中,电热水器正常工作的电压为 (1) 伏,它与空调器是 (2) 连接的,家庭每月的耗电量用 (3) 表计量。

10.如图4所示,跳水运动员起跳时,跳板弯曲说明力可以使物体 (4) ,运动员向上跃起,可以说明物体间力的作用是 (5) 的。运动员跃起,在上升过程中,其重力势能 (6) (选填“变大”、“变小”或“不变”)。

11.小红拉着拉杆式书包在水平地面上前进,如图5所示,书包底部的小轮与地面间的摩擦为 (7) 摩擦;前进中小轮温度升高,这是通过 (8) 方式增加了小轮的内能。当拉杆式书包水平放置地面时,拉杆可看做为杠杆,且动力臂是阻力臂的5倍,若书包重为60牛,则小红在把手上用力向上提起时的拉力为 (9) 牛。

12.若10秒内通过某导体横截面的电量为3库,导体两端的电压为3伏,

则通过该导体的电流为 (10) 安,电流做功为 (11) 焦。如果通过该导体的电流为0.6安,则该导体的电阻为 (12) 欧。

13.把一个重为3牛的苹果竖直向上抛出,苹果在空中受到的阻力大小始终为0.6牛,苹果上升过程中,所受合力大小为 (13) 牛。若苹果在0.5秒内从最高点

图3

V 1

S V 2

L

R

R 1

S A 1

图4

图5

图7

密度

概念

定义、定义式 单位 物质特性之一 测量

原理: (18)

方法

应用

鉴别物质

(19)

下落了1米,此过程中,重力做的功为 (14) 焦,功率为 (15) 瓦。

14.在图6所示的电路中,电源电压保持不变。闭合电键S 后,当滑动变阻器R 2的滑片P 由中点向右端移动时,

① 三个电表中示数不变的电表有 (16) ;

② 电压表V 示数与电流表A 2示数的比值 (17) (选填 “变大”、“变小”或“不变”)。

15.如图7所示为某同学所建立的“密度概念图”。 ① 请你将“概念图”的相关内容填写完整 (18) ; (19) 。 ② 请你另举一个反映“物质特性”的物理量, 并简要说明其反映的是物质哪方面的特性:

(20) 。

三.作图题(共7分)

请将图直接画在答题纸的相应位置,作图必须使用2B 铅笔。

16.重为8牛的物体A 静止在水平面上。请在图8中用力的图示法画出A 所受的重力G 。 17.根据平面镜成像的特点,在图9中画出像A′B′所对应的物体

AB 。

18.在图10电路中缺少两根导线,请按要求用笔线代替导线完成电路连接。要求:闭合电键S 后,向左移动变阻器的滑片时,电压表的示数变小。

四.计算题(共26分)

请将计算过程和答案写入答题纸的相应位置。

19.将质量为5千克的铜加热,铜的温度升高了20℃,求:铜吸收的热量Q 吸。

[c 铜=0.39×103焦/(千克·℃)]

20.体积为3×10-4

米3的金属块浸没在水中。求:该金属块所受到的浮力F 浮。

图8

A

图10

R

- 3 15

V

图9

A ′

B ′

测力计 单位:N

012345

012345

21.如图11所示,圆柱体甲和薄壁圆柱形容器乙置于水平地面。甲的质量为8千克、底面积为4×10-2米2。乙的质量为4千克、底面积为5×10-2米2。乙容器中装有质量为8千克的水。 ① 求乙内水的体积V 水。

② 求乙内水面下0.1米深处的压强p 水。 ③ 将甲浸没在乙容器的水中后(无水溢出), 求乙容器对水平地面的压强p 乙。

22.在图12(a )所示的电路中,电流表A 1、A 2的表盘均如图12(b )所示,电源电压为6伏且保持不变,电阻R 1的阻值为10欧。

① 求电键S 闭合后电流表A 2的示数。

② 现有上表所列的A 、B 、C 三种不同规格的变阻器,请在不损坏电路元件的情况下,按以下要求各选择一个变阻器取代电路中的变阻器R 2。

(a )能使电路的电功率获得最大值,并求电功率最大时通过变阻器R 2的电流I 2。 (b )使电路中电流表A 1示数的变化量最大,并求出该变化量ΔI 。 五.实验题(共18分)

请根据要求在答题纸的相应位置作答。

23.使用如图13(a )所示的电流表时,电流表应 (1) 联接入待测电路,若其所测电流约为1.0安,则应选的量程为 (2) 。图13(b )所示为某托盘天平称量物体时分度盘的情况,此时应将游码向 (3) 移动,使天平在 (4) 位置平衡。

24.在用弹簧测力计测铁块重力的实验中,测力计应按 (5) 方向进行调零,如图14所示,测力计的示数为 (6) 牛。在“探究二力平衡的条件”实验中,物体应在 (7) 的作用下,且分别处于匀速直线运动状态或 (8) 状态。

(a ) (b )

R 1

R 2

S

A 2

A 1

图12

序号 变阻器的规格 A

5Ω 3A B 20Ω 2A C

50Ω 1.5A 乙

图11

0.6 3

A

图13

(a ) (b )

25.小华同学利用焦距f 为15厘米的凸透镜、一个高度为3厘米的发光体、光屏和光具座等做“验证凸透镜成实像的规律”实验。实验中,所测得的物距u 、像距v 以及所成像的像高h 分别记录在表一中。在验证得到凸透镜成缩小实像、放大实像的初步规律后,又做了进一步研究。

① 分析比较实验序号1~6数据中物距u 、像距v 的变化情况及成像变化的情况,可得出的初步结论是: (9) 。

② 分析比较实验序号1~6数据中像距v 与物距u 的比值及像高与物高的比值,可得出的初步结论是:凸透镜成实像时, (10) 。

③ 小华同学在进一步分析比较表中的数据及成像情况后,提出了一个猜想:“

当物体处于凸透镜的两倍焦距处,可能会成等大的实像”。

(a )在分析比较表中的数据及成像情况后,你是否同意小华的猜想,并简述理由 (11) 。

(b )为了验证该猜想,小华准备再添加些器材继续进行实验。在下列所提供的器材中,你认为最需要添加的实验器材是 (12) 。(选填序号)

A .若干个高度不同的发光体

B .若干个焦距不同的凸透镜

C .若干个大小不同的光屏

D .若干个长度不同的光具座

26. 小红同学做“测定小灯泡的电功率”实验,电源由串联节数可变的干电池组成(其电压为1.5伏的整数倍),待用的滑动变阻器有两种规格(分别标有“5Ω 3A”、“20Ω 2A”的字样),所用小灯是额定电压为 “4.5V” 和“6.3V”中的一个。 她选定某一电源电压和滑动变阻器进行实验,实验电路连接及操作正确,在移动滑动变阻器滑片的过程中,发现小灯始终无法正常发光,且所接电压表最大的示数如图15(a )所示。她经过思考得出原因所在,于是在不改变其他实验器材情况下重新选定电源电压并正确实验。当小灯接近正常发光时,她再将电压表并联在电路中,接着移动滑片使小灯正常发光,此时变阻器的滑片在中点附近某位置,且观察到电流表示数如图15(b )所示。最后,她使小灯两端的电压略高于其额定电压,并计算得到相应的功率为2.1瓦。

① 图15(b )中电流表的示数为 (13) 安。 ② 求小灯正常发光时电源电压U 。 (14)

图15 (a ) (b )

实验 序号 物距u

(厘米) 像距v

(厘米) 像高h

(厘米) 1 60.00 20.00 1.00 2 45.00 22.50 1.50 3 33.75 27.00 2.40 4 27.50 33.00 3.60 5 25.00 37.50 4.50 6

22.50

45.00

6.00

表一

③ 说明判定小灯正常发光的方法,并求小灯的额定电功率P 额。 (15) (求解电源电压及小灯的额定电功率需要写出解答过程)

2016年静安区中考物理二模卷

一、选择题

1.A 【解析】本题考查的是声音的三个特征。“震耳欲聋”是声音大的意思,即指声音的响度大。故选A 。

2.C 【解析】本题考查的是原子的结构。原子分为原子核和核外电子,而核外电子是绕核高速旋转的。故选C 。

3.A 【解析】本题考查的是光的反射。由光的反射定律可知:光线发生反射时,反射角等于入射角。当光线垂直入射水面时,入射角为0°,则反射角也为0°。故选A 。

4.B 【解析】本题考查的是物理常识。一本物理教科书的质量约为0.2 千克,故选B 。

5.D 【解析】本题考查的是凸透镜成像的特点。掌心中的水珠相当于凸透镜,掌纹与水珠的距离就是物距,在一倍焦距之内,相当于放大镜的作用,成一个正立放大的虚像。故选D 。

6.B 【解析】本题考查的是对s -t 图象的理解与分析。由图像可知甲车的速度

2.40=

=0460v --甲米/秒.米/秒,乙车的速度为 3.60

==03120

v --乙米/秒.米/秒,所以v 甲>v 乙;当甲车与乙车同向行驶时,6 秒时两车之间的距离s =2.4 米-1.8 米=0.6 米,当甲车与乙车背向行驶时,6 秒时两车之间的距离s =2.4 米+1.8 米=4.2 米。故选B 。

7.A 【解析】本题考查的是液体压强。由题可知,在抽出之前甲液体对容器底部的压强 p 甲=ρ甲gh 甲,乙液体对容器底部的压强p 乙=gh ρ乙乙,由于p 甲=p 乙,h 甲>h 乙,所以ρρ<甲乙;

在两容器中分别抽出相同高度?h 的液体,则两容器中液体对各自容器底部的压强减小量为:?p 甲=ρ甲g ?h ,?p 乙=ρ乙g ?h ,所以,?p 甲<?p 乙;由于剩余液体对各自容器底部的压强p =p 0-?p ,所以,p 甲>p 乙。由压强公式F

p S

=

可知:剩余液体对各自容器底部的压力F =pS ,所以,F 甲>F 乙。故选A 。

8.C 【解析】本题考查的是对电路故障的分析。若灯L 发光,即正常工作时,电键S 闭合后,电压表V 1测得是灯L 两端的电压,电压表V 2测得是电源两端的电压,则电压表V 1的示数小于电压表V 2的示数,电路一定完好,故①正确;若灯L 不发光,则灯L 可能短路或断路,或者灯L 短路且电阻R 断路,此时电压表V 1的示数为0,电压表V 2测得是电源两端的电压,满足条件,故④正确。故选C 。 二、填空题

9.(3分) (1)220 (2)并联 (3)电能 【解析】 本题考查了家庭电路中的电压、家庭电路中电器的连接以及电能的计量。我国家用电器的额定电压都是220V ,电热水器是家用电器,额定电压是220V ,也就是正常工作的电压是220V ;电热水器和空调都是家用电器,家庭电路中为了使各用电器正常工作,并且互不影响,各用电器之间应该是并联的;电能表是测量消耗电能的仪表,家庭电路中,测量消耗电能的多少需要用到电能表。

10.(3分) (4)形变 (5)相互 (6)变大 【解析】 本题考查了力的作用效果、物体间力的作用特点以及重力势能的变化。力的作用效果有两个:①力可以改变物体的形状即使物体发生形变;②力可以改变物体的运动状态,包括物体的运动速度大小发生变化、运动方向发生变化。跳水运动

员起跳时,运动员对跳板施加向下的力把跳板压弯,即跳板发生形变,所以力可以使物体发生形变;运动员对跳板施加向下的力的同时,跳板也对运动员有一个向上的力,使得运动员向上跃起,说明力的作用是相互的;运动员跃起,在上升过程中,运动员质量不变,重力做负功,重力势能变大。

11.(3分) (7)滚动 (8)做功 (9)12 【解析】 本题考查了摩擦力的区分、做功与能量转化以及杠杆平衡条件。在前进过程中,底部的小轮在地面上滚动,所以小轮与地面间的摩擦是滚动摩擦,可以减小摩擦力;凡是由于摩擦使物体温度升高内能增大的,都是机械能转化为内能的过程,都属于以做功的方式改变物体的内能;设阻力臂L ,则动力臂为5L ,书包的重力为阻力,则由杠杆的平衡条件可得拉力60

555

L G G F L =

== 牛=12牛。 12.(3分) (10)0.3 (11)9 (12)10 【解析】 本题考查了电流的物理意义、电功的计算以及欧姆定律。通过该导体的电流3

10

Q I t =

=安=0.3安;电流做功W =UIt =3×0.3×10安=9焦;由欧姆定律U I R =

可得,3

0.3

U R I ==欧=10欧,电阻是导体的性质,不会随着导体通过的电流的变化而

变化,所以通过改导体的电流变为0.6安,导体的电阻依然是10欧。

13.(3分) (13)3.6 (14)3 (15)6 【解析】 本题考查了物体的受力分析、做功的计算以及功率的计算。苹果在上升过程中,受到竖直向下的重力作用以及与运动方向相反的阻力的作用,所以在上升过程中的苹果所受合力大小=F 合3牛+0.6牛=3.6牛;根据功的计算公式W Fs =可得,重力做功G W Gh ==3牛×1米=3焦;根据功率的计算公式W

P t

=

可得,重力做功的功率3

=

0.5

G G W P t =瓦=6瓦。 14.(4分) (16)A 1、V (17)变大 【解析】 本题考查欧姆定律以及电路的动态分析。观察分析图示电路可得,电阻R 1与滑动变阻器R 2并联,电流表A 1、A 2分别测的是通过电阻R 1、R 2的电流,电压表测的是电阻R 1、R 2两端的电压,也即电源电压,所以电压表V 示数始终保持不变。当滑动变阻器的滑片有中点向右移动时,滑动变阻器接入电路中的阻值变大,两端的电压不变,根据欧姆定律可知,通过滑动变阻器的电流变小,即电流表A 2的示数变小;电阻R 1与滑动变阻器R 2并联,互不影响,两端电压不变,阻值不变,通过的电流不变,即电流表A 1的示数不变;电压表的示数是滑动变阻器两端的电压,电流表A 2的示数是通过滑动变阻器的电流的大小,所以两者的比值是滑动变阻器的阻值,所以两者的比值是变大的。 15.(4分) (18)m

V

ρ=

(19)计算不可测的质量或体积 (20)比热容,反映的是物质吸放热的本领 【解析】本题考查了密度的相关概念。测量密度的实验包括测固体的密度和测液体的密度,它们的原理都是相同的,都要依据密度公式m

V

ρ=

;根据密度公式,若某一物体的质量或体积不可

测,就可以利用物质的密度,再测量质量或体积其中的一个物理量,通过计算就可以得出该物质的体积或质量;物质的特性指的是这种物质特有的,可以用于区别其它物质的性质。密度是物质的基本特性之一,每种物质都有自己的密度,各种物质的密度是一定的,不同物质的密度一般不同;不同的物质,在质量相等、温度升高的度数相同时,吸收的热量是不同的,单位质量的某种物质,温度升高1℃所吸收的热量叫做这种物质的比热容,比热容是物质的一种属性,每种物质都有自己的比热容,所以密度、比热容是表示物质的特性的物理量。

三、作图题

16.(3分)如图所示

JAF01

【解析】本题考查力的图示法。作用点在物体A 的几何中心,标度为2牛,重力的的长度为4格,方向竖直向下。(力的大小、方向、作用点正确3分) 17.(2分)如图所示

JAF02

【解析】本题考查平面镜成像特点。物体在平面镜中所成的像是虚像,像和物体的大小相等,位于镜面两侧,它们到镜面的距离相等且连线垂直于镜面,故首先作关于平面镜对称的A B ''、的对称点A 、B ,连接AB 。(图像对称1分、虚像1分,共2分) 18.(2分)如图所示

JAF03

【解析】本题考查的是电路的正确连接。电压表测的是滑动变阻器上的电压,向左移动变阻器的滑片时,电阻减小,电流增大,电阻R 上的电压增大,则电压表示数减小,正确连接如图。(串联正确1分,电压表正确1分) 四、计算题

19.(4分)3.9×104焦 【解析】Q 吸=cm Δt

=0.39×103焦/(千克·℃)×5千克×20℃

=3.9×104焦 4分

20.(4分)2.94牛 【解析】F 浮=ρ液g V 排 =1.0×103千克/米3×9.8牛/千克×3×10-4米3

=2.94牛4分

21.(9分)①8×10-3米3②980帕③3920帕

【解析】①V水=m水/ρ水=

8千克/(1.0×103千克/米3) =8×10-3米33分

②p水=ρ水gh

=1×103千克/米3× 9.8牛/千克×0.1米=980帕3分

③ p乙=F乙/S=G/S=mg/S

=(8+4+8)千克× 9.8牛/千克/5×10-2米2

=3920帕3分

22.(9分)①0.6安②(a)2.4安(b)1.7安

【解析】① I1=U/R1=6伏/10欧=0.6安3分

② (a)选择变阻器A

I2=I-I1=3安-0.6安=2.4安3分

(b) 选择变阻器B

I2ma x=2安

I2mi n=U/ R2ma x=6伏/20欧=0.3安

ΔI=ΔI2=2安-0.3安=1.7安3分

五、实验题

23.(4分)(1)串(2)0-3安(3)左(4)水平【解析】本题考查电流表和天平的使用。电流表在电路中可以看作通路,是串联在待测电路中,因为待测电流约为1.0安,所以电流表的量程应选用0-3安;根据托盘天平的分度盘显示,天平的右边偏重,所以为了使天平在水平位置平衡,应将游码向左移动。

24.(4分)(5)竖直(6)3.2 (7)两个力(8)静止【解析】本题考查弹簧测力计的使用及其“探究二力平衡的条件”的实验。本实验是测铁块重力的实验,所以测力计应在竖直方向上进行调零。根据图示,测力计的示数为3.2牛。根据“探究二力平衡条件”的实验原理,物体应在两个力的作用下,并且分别处于匀速直线状态或静止状态。

25.(4分)(9)凸透镜成实像时,物距u减小,像距v变大,像变大(10)像距v与物距u的比值等于像高与物高的比值(11)同意,根据实验数据可知:大于2倍焦距成缩小的像,小于2倍焦距(大于焦距)成放大的像,那么就有可能在2倍焦距处成等大的像

(12)B 【解析】本题考查“验证凸透镜成实像的规律”实验。①分析实验1~6的数据,物距不断减小,像距不断增大,像的高度不断增大,由此我们可以得出:凸透镜成实像时,物距u减小,像距v变大,像变大②根据表中的数据,虽然像距不断增大、物距不断减小、像高不断增大,但是像距与物距的比值都是等于像高与物高的比值。由此我们可以得出:凸透镜成实像时,像距v与物距u的比值等于像高与物高的比值。③(a)同意,根据实验数据可知:大于2倍焦距成缩小的像,小于2倍焦距(大于焦距)成放大的像,那么就有可能在2倍焦距处成等大的像(b)为了验证实验结论的有效性,需要若干个焦距不同的凸透镜来进行实验,而高度不同的发光体、大小不同的光屏、长度不同的光具座都不能验证实验结论的有效性。故选B。

26.(4分)(13)0.28安(14)9.0伏(15)电压表并联在滑动变阻器两端,当电压表数为2.7伏时小灯正常发光 1.764瓦【解析】本题考查“测定小灯泡的电功率”的实验。①根据电流表的示数得,电流为0.28安或1.4安。若电流为1.4安,根据P=I2R,U=IR,得到此时小灯泡的电压为1.5

伏,远小于灯泡的额定电压,与题目中“小灯泡两端的电源略高于其额定电压”不符,所以电流表的示数为0.28安。②因为电流I=0.28安,根据①的方法,计算得出当滑片移到中点附近时,灯泡的电压为7.5伏,所以小灯泡的额定电压为6.3伏。U=U额+U变

若变阻器为“5 Ω 3 A” ,U=6.3伏+0.28安×2.5欧=7.0伏(舍去)

若变阻器为“20 Ω 2 A” ,U=6.3伏+0.28安×10欧=9.1伏

电源电压为9.0伏。

③电压表并联在滑动变阻器两端,当电压表示数为2.7伏时,小灯泡正常发光。

P额=U额I额=6.3伏×0.28安=1.764瓦。

上海市静安区中考英语二模(含答案).医生.doc

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