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2014年浦东二模高三物理

2014年浦东二模高三物理
2014年浦东二模高三物理

浦东新区2014年高考预测物理试卷

一.单项选择题.(共16分,每小題2分,每小题只有一个正确选项,答案涂写在答题卡上.) 1.下列叙述中符合物理学史实的是

A .惠更斯发现了单摆的等时性

B .奥斯特发现了电磁感应现象

C .托马斯·杨通过双缝干涉实验证实了光具有波动性

D .汤姆孙通过α粒子散射实验提出了原子的核式结构模型 2.如图所示,在白光照射下,能在竖直放置的肥皂膜上看到

A .上疏下密的黑白条纹

B .上疏下密的彩色条纹

C .上密下疏的黑白条纹

D .上密下疏的彩色条纹

3.关于固体和液体的微观结构,下列说法正确的是

A .固体分子静止在平衡位置

B .固体分子没有固定的平衡位置

C .液体分子没有平衡位置

D .液体分子没有固定的平衡位置

4.下列核反应方程中X 表示α粒子,且属于α衰变的是

A .X Th U +→234

9023892 B .X Pa Th +→234

9123490 C .n X H H 103121+→+

D .n C X Be 1012694+→+

5.一定质量的乙醚液体全部蒸发,变为同温度的乙醚气体,在这一过程中

A .分子引力增大,分子斥力减小

B .分子势能减小

C .乙醚的内能不变

D .分子平均动能不变

6.物体做平抛运动,下列说法正确的是

A .加速度的方向时刻改变

B .速度的变化率不断增大

C .任意一段时间内速度变化量的方向均竖直向下

D .第1秒内、第2秒内、第3秒内的位移之比为1:3:5

7.如图所示为用“与”门构成的报警器电路。下列哪种情况下,报警器一定会报警

A .S 断开,P 在a 端

B .S 闭合,P 在a 端

C .S 断开,P 在b 端

D .S 闭合,P 在b 端

8.某质点作简谐运动的振动图像如图所示,则

A .t =0.2s 时,质点速度方向沿x 正方向

B .t =0.2s 时,质点加速度方向沿x 负方向

C .0.2s -0.4s 内质点速度先减小后增大

D .0.2s -0.4s 内质点加速度先增大后减小

二.单项选择题.(共24分,每小题3分,每小题只有一个正确选项,答案涂写在答题卡上.) 9.物体正在沿x 轴做直线运动,可能出现的情况是

A .加速度沿+x 方向逐渐增大,速度沿+x 方向逐渐减小

B .加速度沿+x 方向逐渐减小,速度沿+x 方向逐渐减小

C .加速度沿-x 方向逐渐增大,速度沿-x 方向逐渐减小

D .加速度沿-x 方向逐渐减小,速度沿-x 方向逐渐增大

10.如图所示,当长木板从水平位置开始缓慢转动至竖直位置 过程中,原来静止在长木板上的物块

A .所受合外力始终为零

B .受到的支持力一直减小

C .受到的摩擦力一直增大

D .对长木板作用力先不变后增大

11.在光滑水平面上,一物体在水平力F 的作用下,由静止开始做直线运动,F 随位移s 变化的图线如图所示,下列描述其位移-时间(s -t )和速度-时间(v -t )关系图像正确的是

12.如图所示,在两个等量异种点电荷的电场中,C 、D 两点位于两电荷连线的中垂线上,A 、B 两点位于两电荷的连线上,且关于上述中垂线左右对称,则以下判断正确的是

A .将检验电荷+q 从A 点移动到

B 点,电场力不做功 B .将检验电荷-q 从A 点移动到B 点,电场力做正功

C .将检验电荷+q 从

D 点移动到C 点,电场力不做功 D .将检验电荷-q 从D 点移动到C 点,电场力做负功

13.美国物理学家蔡特曼(Zarman )和我国物理学家葛正权于1930—1934年对施

特恩测定分子速率的实验作了改进,设计了如图所示的装置。半径为R 的圆筒B 可绕O 轴以角速度ω匀速转动,aOcd 在一直线上,银原子以一定速率从d 点沿虚线方向射出,穿过筒上狭缝c 打在圆筒内壁b 点,∠aOb =θ,ab 弧长为s ,其间圆筒转过角度小于90°,则下列判断正确的是

A .圆筒逆时针方向转动

B .银原子在筒内运动时间ω

s

t =

C .银原子速率为θ

ωR

D .银原子速率为s

R 2

14.如图所示,一形状不规则、质量分布不均匀的棒,可绕固定转轴O 在竖直面内自由转动,当棒自由平衡时下端A 恰位于转轴O 点正下方。现用水平恒力F 作用于A 端,在转动角度小于90°的过程中

A .棒的重力力矩先减小后增大

B .恒力F 的力矩先减小后增大

C .棒转动的角速度先增大后减小

D .棒上A 、B 两点线速度方向始终相同

15.如图所示,竖直放置的弯曲玻璃管a 端封闭,b 端开口,水银将两段空气封闭在管内,管内各液面间高度差为h 1、h 2、h 3且h 1=h 2=h 3;K 1、K 2为两个阀门,K 2位置与b 管水银面等高,打开阀门后可

与外界大气相通。打开K 1或K 2,下列判断正确的是

A .打开K 1,h 1、h 2和h 3均变为零

B .打开K 1,h 1增大,h 2和h 3均变为零

C .打开K 2,h 1、h 2和h 3均变为零

D .打开K 2,h 1、h 2、h 3的长度保持不变

16.在平直公路上行驶的a 车和b 车,其位移-时间图像分别为图中直线a 和曲线b ,由图可知

A .b 车运动方向始终不变

B .在t 1时刻a 车的位移大于b 车

C .t 1到t 2时间内a 车的平均速度小于b 车

D .t 1到t 2时间内某时刻两车的速度可能相同

三.多项选择题.(共16分,每小题4分,每小题有两个或三个正确选项,全选对的,得4分,选对但不全的,得2分,有选错或不答的,得0分,答案涂写在答题卡上.)

17.将小球从地面以初速度v 0竖直向上抛出,运动过程中小球受到的空气阻力大小不变,最终小球又落回到地面,以地面为零势能面,则小球

A .上升时间大于下落时间

B .上升的最大高度小于g

v 22

C .上升过程中到达最大高度一半时其动能大于重力势能

D .下降过程中到达最大高度一半时其动能等于重力势能

18.如图所示,在竖直平面内有一半径为R 的3/4圆弧轨道,半径OA 水平、OB 竖直,一个质量为m 的小球自A 的正上方P 点由静止开始自由下落,不计空气阻力,小球沿轨道到达最高点B 时恰好对轨道没有压力。已知PA=2.5R ,重力加速度为g ,则小球从P 到B 的运动过程中

A .重力做功1.5mgR

B .动能增加mgR

C .克服摩擦力做功mgR

D .离开B 点后小球能落在圆轨道内

19.如图所示,两个完全相同的光滑球的质量均为m ,放在竖直挡板和倾角为α的固定斜面间。若缓慢转动挡板到与斜面垂直,在此过程中

A .P 、Q 两球间的弹力不变

B .P 球对斜面的压力逐渐增大

C .Q 球对斜面的压力逐渐增大

D .Q 球对挡板的压力逐渐减小

20.如图所示的电路中,定值电阻R 1、R 2、R 3、R 4的阻值均为R 0,理想电压表读数U ,变化量的绝对值ΔU ,理想电流表读数I ,变化量的绝对值ΔI ,在滑动变阻器的滑动端自右向左滑动的过程中,下列判断正确的是

A .U 增大,I 减小

B .

I U

增大 C .

I

U

??增大 D .

I

U

??

21.如图所示,边长为L 的正方形导线框abcd 放在纸面内,在ad 边左侧有足够大的匀强磁场,磁感应强度大小为B ,方向垂直纸面向里,现使导线框绕a 点在纸面内顺时针转动,经时间t 转到图中虚线位置,则在时间t 内导线框abcd 中感应电流方向为__________方向(选填“顺时针”或“逆时针”),平均感应电动势大小等于__________。

22A .在光滑的水平面上有A 、B 两辆质量均为2m 的小车,保持静止状态,A 车上站着一个质量为m 的人,当人从A 车跳到B 车上,并与B 车保持相对静止,则A 车与B 车速度大小之比等于______,A 车与B 车动量大小之比等于______。

22B .已知地球半径R 、自转周期T 、地球表面重力加速度g ,则其第一宇宙速度为_________。地球同步卫星离地面的高度为___________。

23.如图所示,为一列沿x 轴正方向传播的简谐横波在 t =0时刻的波的图像。经Δt =0.1s ,质点M 第一次回到平衡位置,则这列波的波速为v =_____m/s ,周期T =______s 。

24.长为L 的轻杆上端连着一质量为m 的小球,杆的下端用铰链固接于水

平地面上的O 点,斜靠在质量为M 的正方体上,在外力作用下保持静止,如图所示。忽略一切摩擦,现撤去外力,使杆向右倾倒,当正方体和小球刚脱离瞬间,杆与水平面的夹角为θ,小球速度大小为v ,此时正方体M 的速度大小为________,小球m 落地时的速度大小为_______。

25.如图所示,AB、BC、CA三根等长带电棒绝缘相连成正三角形,每根棒上

电荷均匀分布,AB、BC带电均为+q,CA带电为-q,P点为正三角形中心,P、

Q两点关于AC边对称。现测得P、Q两点的电场强度大小分别为E1和E2,若

撤去CA棒,则P、Q两点的电场强度大小分别为E P=_______,E Q=________。

五.实验题.(本大题4小题,共24分.)

26.(5分)在用力矩盘做“研究有固定转动轴物体平衡条件”的实验中

(1)(多选题)力矩盘静止时,盘上有一污点A正好在转轴正下方,如图所示。轻轻转动力矩盘,每次都能转动较长时间,最终污点A都停在转轴正下方,这说明()

A.力矩盘质量太大

B.力矩盘不在竖直平面内

C.力矩盘与转轴间的摩擦足够小

D.力矩盘的重心不在盘的中心

(2)某次实验力矩盘静止后,污点A位置如图所示,则该次实验测

量结果,力矩M顺_____M逆。(选填“大于”、“等于”或“小于”)

27.(5分)某研究性学习小组在“利用DIS研究一定质量的气体在体积不

变时,其压强与温度的关系”实验中,先后用了大、小不同的两个试管甲、

乙,封闭了质量不同但温度和压强都相同的同种气体做实验,实验装置如

图所示。

(1)该实验用到了______传感器和______传感器。

(2)(单选题)若将测得的实验数据在同一p-T坐标系作图,得到的图像应是()

28.(6分)在“用DIS研究感应电动势大小与磁通量变化快慢关系”的实验中

(1)(多选题)有关实验原理、过程和操作,下列说法正确的是()

A.实验中控制磁通量的变化ΔΦ不变

B.实验中保持螺线管、光电门位置不变

C.实验中必须保持轨道倾角一定

D.小车每次必须由静止释放

A

E/V

t

1

/s-1(2)(多选题)一名同学某次实验得到的实验数据通过计算机拟合得到如图所示图线,其中有一数据

点A明显偏离直线,那么造成其偏离的可能原因是该次操作中()

A.小车释放的位置移动了

B.光电门与螺线管间的距离变化了

C.给了小车一个初速度

D.小车上固定的磁铁相对小车移动了

29.(8分)(1)用DIS测电源电动势和内电阻电路如图(a)所示,R0为定值电阻。调节电阻箱R,

记录电阻箱的阻值R和相应的电流值I,通过变换坐标,经计算机拟合得到如图(b)所示图线,则该

图线选取了_______为纵坐标,由图线可得该电源电动势为________V。

(2)现有三个标有“2.5V,0.6A”相同规格的小灯泡,其I-U特性曲线如图(c)所示,将它们与图(a)

中电源按图(d)所示电路相连,A灯恰好正常发光,则电源内阻r=____Ω,图(a)中定值电阻R0=____Ω。(3)若将图(a)中定值电阻R0换成图(d)中小灯泡A,调节电阻箱R的阻值,使电阻箱R消耗的

电功率是小灯泡A的两倍,则此时电阻箱阻值应调到_______Ω。调节电阻箱R的阻值,能否使小灯

泡A、电阻箱R和电源内阻r上消耗的电功率相同?答____________(选填“能”、“不能”或“不

能确定”)

六.计算题.(本大题4小题,共50分.)

30.(10分)如图所示,面积S=100cm2的轻活塞A将一定质量的气体封闭在导热性能良好的汽缸B

内,汽缸开口向上竖直放置,高度足够大。在活塞上放一重物,质量为m=20kg,静止时活塞到缸底的

距离为L1=20cm,摩擦不计,大气压强为P0=1.0×105Pa,温度为27℃,g取10m/s2。

(1)若保持温度不变,将重物去掉,求活塞A移动的距离;

(2)若加热汽缸B,使封闭气体温度升高到177℃,求活塞A移动的距离。

31.(12分)如图所示,截面是直角梯形的物块静置于光滑水平地面上,其两个侧面恰好与两个固定在地面上的挡板X 和Y 相接触。图中AB 高H =0.3m 、AD 长L =0.5m ,斜面倾角θ=37°。可视为质点的小物块P (图中未画出)质量m =1kg ,它与斜面的动摩擦因数μ可以通过更换斜面表面的材料进行调节,调节范围是0≤μ<1。(sin37°=0.6,cos37°=0.8,重力加速度g 取10m/s 2) (1)令μ=0,将P 由D 点静止释放,求小物块P 在斜面上的运动时间;

(2)令μ=0.5,在D 点给P 一个沿斜面向下的初速度v 0=2m/s ,求小物块P 落地时的动能;

(3)对于不同的μ,每次都在D 点给小物块P 一个沿斜面向下足够大的初速度以保证它能滑离斜面,

求μ的取值在哪个范围内,挡板X 始终受到压力的作用。

32.(14分)如图所示,在光滑绝缘水平面放置一带正电的长直细棒,其周围产生垂直于带电细棒的辐射状电场,场强大小E 与距细棒的垂直距离r 成反比,即r

k

E

。在带电长直细棒右侧,有一长为l 的绝缘细线连接了两个质量均为m 的带电小球A 和B ,小球A 、B 所带电荷量分别为+q 和+4q ,A 球距直棒的距离也为l ,两个球在外力F=2mg 的作用下处于静止状态。不计两小球之间的静电力作用。 (1)求k 的值;

(2)若撤去外力F ,求在撤去外力瞬时A 、B 小球的加速度和A 、B 小球间绝缘细线的拉力; (3)若剪断A 、B 间的绝缘细线,保持外力F =2mg 不变,A 球向左运动的最大速度为v m ,求从剪断绝缘细线到A 球向左运动达到最大速度,A 球所处位置电势怎样变化?变化了多少?

33.(14分)如图(a),平行长直导轨MN、PQ水平放置,两导轨间距L=0.5m,导轨左端MP间接有一阻值为R=0.2Ω的定值电阻,导体棒ab质量m=0.1kg,与导轨间的动摩擦因数μ=0.1,导体棒垂直于导轨放在距离左端d=1.0m处,导轨和导体棒电阻均忽略不计。整个装置处在范围足够大的匀强磁场中,t=0时刻,磁场方向竖直向下,此后,磁感应强度B随时间t的变化如图(b)所示,不计感应电流磁场的影响。当t=3s时,突然使ab棒获得向右的速度v0=8 m/s,同时在棒上施加一方向水平、大小可变化的外力F,保持ab棒具有大小为恒为a=4m/s2、方向向左的加速度,取g=10m/s2。

(1)求前3s内电路中感应电流的大小和方向;

(2)求前3s内ab棒所受的摩擦力随时间变化的关系式;

(3)求ab棒的位移s1=3.5m时外力F;

(4)从t=0时刻开始,当通过电阻R的电量q=2.25C时,ab棒正在向右运动,此时撤去外力F,此后ab棒又运动了s2=6.05m后静止。求撤去外力F后电阻R上产生的热量Q。

浦东新区2014年高考预测物理参考答案及评分标准

一、单项选择题(每小题2分,共16分) 1.C

2.B

3.D

4.A

5.D

6.C

7.D

8.B

二、单项选择题(每小题3分,共24分) 9.D

10.B

11.C

12.C

13.D

14.C

15.D

16.D

三、多项选择题(每小题4分,共16分) 17.BC

18.AC

19.AD

20.ABD

四、填空题(每空2分,共20分)

21.顺时针,t

BL 22

22A .3∶2,3∶2

22B .gR

,R R gT -3

2

2

24π 23.1,1.2

24.θsin v ,θsin 22

gL v +(或υ3) 25.

21E ,

212E E -(写成2

12E

E +也给分) 五、实验题(共24分)

26.(5分)(1)CD (3分)(漏选得1分) (2)小于(2分) 27.(5分)(1)压强,温度(每空1分)

(2)C (3分)

28.(6分)(1)AB (3分)(漏选得1分) (2)BD (3分)(漏选得1分) 29.(8分)(1)(2分)1/I ,4.5(每空1分) (2)(3分)2.5(2分),2(1分)

(3)(3分)4.60-4.90(2分),不能(1分)

六、计算题 30.(10分)

解:(1)(5分)初态压强55104200

(1.010)Pa 1.210Pa 10010mg p p S -=+

=?+=?? 气体做等温变化 p 1V 1=p 2V 2

0002()mg

p L S p L S S

+

?= 5521.21020 1.010L ??=? 解得 L 2=24cm 2124204cm d L L =-=-=

(2)(5分)加热气缸,气体做等压变化

3

3

11T V T V = 3

311273273t S L t S L +=+

450

30020

3L = 解得L 3=30cm

31(3020)cm 10cm d L L '=-=-=

31.(12分)

解:(1)(3分)当μ=0时,P 沿斜面下滑的加速度为a ,由牛顿第二定律

mg sin θ=ma 得 a =g sin θ=6m/s 2

由运动学公式:2

2

1at L =

得 220.56s s 0.41s 6

6

L t a

?=

=

=

(2)(4分)设落地时P 的动能为E k ,P 沿斜面下滑到落地,由动能定理得:

2

1sin )cos 2

k mg H L mgL E mv θμθ+-=-(

代入数据,解得

2[10(0.30.50.6)0.5100.80.50.512]J 6J k E =?+?-???+??=

(3)(5分)P 在斜面上下滑过程中梯形物块受力如图:

由平衡条件可得 F +N sin θ=f cos θ

所以

F =μmg cos θco sθ-mg cos θsin θ 代入数据,得

F =6.4μ-4.8

挡板X 受压力,则F >0,所以 6.4μ-4.8>0,得μ>0.75

依题意,0≤μ<1,所以μ的取值范围为 0.75<μ<1

方法Ⅱ:

欲使挡板X 始终受到压力作用,须使小物块P 对AD 面的摩擦力f 与P 对AD 面的正压力N 的合力(全反力)方向偏向左下方,即tan tan 370.75f N

θμ==

>=

依题意,0≤μ<1,所以μ的取值范围为 0.75<μ<1 32.(14分)

解:(1)(3分)对小球A 、B 及细线构成的整体,受力平衡,有

mg l

k

q l k q

224=+ 得 q

m gl

k 32=

(2)(5分)若撤去外力瞬时, A 、B 间细线拉力突然变为零,则

对A 球:A k

q ma l

?

= 得 ml

kq

a A =,方向向右 对B 球:42B k

q ma l

?

= 得 ml

kq

a B 2=

,方向向右 因为a A

因此,对A 、B 整体,由牛顿第二定律,有422k k

q q ma l l

?

+?=

即 2mg =2ma 得a=g

对A : k

q T ma l ?

+= 解得 mg T 3

1=

Mg

F N '

N f

θ θ

故撤去外力瞬时,A 、B 的加速度a=g ;A 、B 小球间绝缘细线的拉力mg T 3

1=

(3)(6分)当A 向左运动达到速度最大时 2k

q mg r

?

= 代入k ,得

l r 3

1=

设从剪断绝缘细线到A 球向左运动达到最大速度,电场力做功为W ,由动能定理

21

2()02

m mg l r W mv -+=

- 解得 241

(

)32

m mgl W mv =--,负号表示电场力做负功。 又由电场力做功可知W=qU 因此 2

4()32m W mgl mv U q q q

==--

在A 球向左运动达到最大速度的过程中,所在处电势升高了。 变化量为 2

432m mgl mv U q q

?=-

33.(14分)

解:(1)(3分)由图知

0.2

0.1T/s 2

B t ?==? 0.10.51V 0.05V BLd

E t t

???=

=

??=??= 0.05A 0.25A 0.2

E I R === 根据楞次定律,电流方向为abPMa 。

(2)(2分)ab 棒在水平方向受安培力和摩擦力,二力平衡

f =BIL =(B 0-kt )IL =(0.2-0.1t )×0.25×0.5N=0.0125(2-t )N (t <3s )

(3)(5分)ab 棒做匀变速运动,位移s 1=3.5m 时,速度大小设为v 1,则

22

102v v as -= 代入数据,得 a =-4m/s 2 解得

v 1=6 m/s

对应安培力大小 2222

10.10.56N 0.075N 0.2

A B L v F B I L

R ??==== 向右运动时,F +μmg +F A =ma F +0.1+0.075=0.1×4 F =0.225N ;方向向左 向左运动时,F -μmg -F A =ma

F -0.1-0.075=0.1×4

F =0.575N ;方向向左

(4)(4分)前3s 内通过电阻R 的电量q 1=I ×Δt =0.25×3 C =0.75C

设3s 后到撤去外力F 时又运动了s 则

21BLs

q I t t q q t R R R

????=??=

??===-??=1.5C 解得 s =6m

此时ab 棒的速度设为v 2则 22

202v v as -= 解得 v 2=4 m/s

此后到停止,由动能定理 W A +W f =ΔE k W A =-Q

2221(0)(0.6050.8)J 0.195J 2

f k Q W E mgs mv μ=-?=---=-+=

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闵行区 2017 学年度第二学期初三年级学业质量调研 英语学科试卷2018 年4 月 (满分150 分,考试时间:100 分钟) 考生注意:本卷有7 大题,共94 小题。试题均采用连续编号,所有答案务必按照规定在答 纸上完成,做在试卷上不给分) Part 1 Listening(第一部分听力) Part 2 Phonetics , Grammar and Vocabulary (第二部分语音、语法和词汇) II.Choose the best answer(选择最恰当的答案):(共20 分) 26. Which of the following underlined parts is different in pronunciation from the others? A.I need double milk for my coffee. B.We are proud to be Chinese. C.The women shouted loudly for help D.They are going to buy a new house. 答案:A; A 发音[?] ,BCD 发音[a?] 27. Take a look at booklet on the shelf.It tells you how to use the new cooker. A.a B.an C.the D./ 答案:C;特指加定冠词the. 28. Linda came back home half past ten last night.That's really too late. A.at B.on C.in D.for 答案:A; 具体时刻前用at 29.It's important for to know that honesty is the best solution when in trouble. A.we https://www.wendangku.net/doc/7f4906601.html, C.our D.ours 答案:B; 介词后做宾语用宾格 30. The government says that they are going to build the third road in the town. A.wide B.wider C.widest D.the widest 答案:C; 有序数词修饰的最高级前省略the 31. The audience was surprised the result of the singing contest. A.from B.with C.at D.about 答案:C; be surprised at sth. 对某事感到惊讶 32. Though it was very dangerous, hesitated when entering the burning lab. A.everybody B.nobody C.anybody D.somebody 答案:B; 出现though, 让步状从,尽管很危险,没有人犹豫 33. He's put on so much weight that his clothes aren't any more. A.wonderfully B.beautifully C.expensive D.fit 答案:D;put on so much weight 表示增重很多,所以对应衣服不再适合。 34.Alice would rather in an air-conditioned room in such hot weather. A.stay B.to stay C.stays D.staying 答案:A,would rather do 35.Tracey suggested basketball in Sports Club after school.

2016年上海市闵行区中考英语二模试卷及解析

2016年上海市闵行区中考英语二模试卷 一、I Listening Comprehension(听力理解)(共30分)A.Listen and choose the right picture(根据你听到的内容,选出相应的图片)(6分) 1.(6分) 123456. 一、B.Listen to the dialogue and choose the best answer to the question you hear (根据你听到的对话和问题,选出最恰当的答案)(8分) 7.(1分)A.A watch B.A book C.A case D.A stamp. 8.(1分)A.By car. B.By train. C.By plane D.By bus. 9.(1分)A)Sunny. B)Rainy. C)Windy. D)Snowy. 10.(1分)A.She doesn't like parties

B.Her doctor will visit her. C.She is busy working. D.She is not feeling well. 11.(1分)A.At an airport. B.At a supermarket. C.At a hotel D.At a bank. 12.(1分)A.At 7:00. B.At 8:00. C.At 9:00. D.At 10:00. 13.(1分)A.Manager and secretary. B.Shop assistant and customer. C.Teacher and student. D.Wife and husband. 14.(1分)A.They can buy vegetables together. B.They can go out for dinner that day. C.He has never eaten in that restaurant. D.He disliked the food the woman cooked. 一、 C. Listen to the passage and tell whether the following statements are true or false ((判断下列是否符合你听到的内容,符合的用“T“表示,不符合的用“F“表示. (6分. 15.(6分)15.An old couple went to their doctor because they were seriously ill.16.The doctor told them to write things down in order not to forget. 17.The old lady told her husband to get her a bowl of rice after they got home.18.The old lady told her husband to write something down.but he refused.19.The husband could remember clearly what to do before he went to the kitchen.20.From the story we can learn that the old lady was less forgetful than her husband.

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