第一节　 听力理解(5段共15小题；每小题2分，满分30分)

每段播放两遍。各段后有几个小题，各段播放前每小题有5秒钟的阅题时间。请根据各段播放内容及其相关小题，在5秒钟内从题中所给的A、B、C项中，选出最佳选项，并在答题卡上将该项涂黑。

听第一段对话，回答第1至3题。

1.What is main topic of the dialogue?

　 A. Anxiety　　　 B. Sympathy　　 C. Examination

2. Why does Jack turn to Grace for help?

　 A. She is older than he is.

　 B. She has had a similar experience.

　 C. She is a medical student

3. What does the school offer to students in Jack’s condition?

　 A. A course about how to handle stress.

　 B. A book on how to deal with anxiety.

　 C. A lecture on how to cope with stress.　

听第二段对话，回答第4至6题。

4.What did the man do after he read the newspapers that morning?

　 A. Went to his office.　　 B. Turned on the computer.　　 C. Checked his e-mail.

5. How did the man know there was a big fire in Hong Kong yesterday?

A. His father sent him the message.

B. He got the news from his brother.

　 C. He read it from the newspaper.

6.Where does the conversation take place?

A. At the man’s home.　

B. At the airline reservation agency.

C. At the man’s office.

听第三段独白，回答第7至9题。

7. What does the speaker mainly focus on?

A. Haka, an ancient Maori dance.

B. Maori people, native New Zealanders.

C. A match between Italy and New Zealand.

8. Why do the All Blacks players perform before a rugby match really starts?

A. They want to get their fans excited.

B. They want to tell the enemy they are violent.

C. They want to make the enemy frightened.

9. What can you conclude from the speaker?

A. In order to be successful, now all opponents look down upon the war dance.

B. Italians didn’t think much of the war dance, so the Italians won the match.

C. The war dance still played a role in the match between Italy and New Zealand.

听第四段对话，回答第10至12题。

10. What does the man want to do?

A. Play basketball with friends from work.

B. Try out for the company basketball team.

C. Get in shape and compete in a cycling race.

11. What’s the woman’s main concern?

A. She is worried her husband will spend too much time away from home.

B. She is afraid her husband will become a fitness fan.

C. She is concerned about her husband’s health.

12. What does the woman advise about the man’s diet?

A. He should eat less salt.

B. He should eat less fatty foods.

C. He should add more protein products to his diet.

听第五段独白，回答第13至15题。

13. How is the melting of Arctic sea ice going on?

A. It’s slowing down.　　 B. It has stopped.　　 C. It’s accelerating.

14. What can we learn from the speaker?

A. The colder weather in the northern hemisphere has caused less melting of Arctic sea ice.

B. The latest data shows that Arctic sea ice is not melting as fast as before.

C. In the past few months the Arctic sea ice has been melting faster than last summer.

15. What disaster has the melting of Arctic sea ice caused?

　 A. Drought in some areas.　　 B. Flood in some areas.　　 C. Colder weather.　

36．(1)带电小球在正交的匀强电场和匀强磁场中因受到的重力G＝mg＝0.1N (1分)

电场力F₁＝qE₁＝0.1N　　　　　(1分)

即G＝F₁，故小球在正交的电磁场中由A到C做匀速圆周运动．(2分)

根据牛顿第二定律得：　　　 (2分)

解得：　　R＝　(1分)

由几何关系得：(2分)

(2)带电小球在C点的速度大小仍为v₀=4m/s，方向与水平方向成45°．由于电场力F₂＝qE₂＝0.1N，与重力大小相等，方向相互垂直，则合力的大小为F＝，方向与初速度垂直，故小球在第二个电场中作平抛运动．(2分)

建立如图所示的x、y坐标系．

沿y方向上，

小球的加速度a=F/m＝(1分)，

位移y=(1分)

沿x方向上，小球的位移x=v₀t(1分)

由几何关系有：(1分)

即：＝v₀t，解得：t＝0.4s(1分)

Q点到P点的距离s₂＝．(2分)

35．⑴A所受B的滑动摩擦力f₁=μmg=0.2×2×10kg=4N　　　　　　　　　 (1分)

则A反作用于B的滑动摩擦力f₁′=4N　　　　　　　　　　 　(1分)

B所受地面的滑动摩擦力f₂=μ(M₁+m)g=0.2×(3+2)×10kg=10N　　　　 (2分)

因B获得动力f₁′小于受到的阻力f₂，故在A滑动的过程中，B始终没有滑动(1分)

对A，根据牛顿第二定律得其加速度 (2分)

1s内运动的位移　 　　　　　　　　　　　(1分)

因s=L₁，说明在1s末A刚好滑上C的左端　　　　　　(1分)

1s末A获得的速度v₁＝at＝2×1m/s=2m/s　　　　　　　　　　　　　 (1分)

⑵A滑上C之后，它对C的滑动摩擦力使C向右做匀加速运动，而C对A的滑动摩擦力使A做匀减速运动，要保证A不会从C上掉下，则长木板C的最小长度L₂必须符合在A滑至C的右端时双方获得共同速度v₂．　　　　　　　　　　　　　 (1分)

在双方相对运动的过程中所受合外力等于零，系统动量守恒，有：

　mv₁=(M₂+m)v₂　　　　　　　　　　　　　　　　　　　　　　 (2分)

又根据系统的能量守恒定律可得：

　f₁L₂＝mv₁²－(M₂+m)v₂²　　　　　　　　 　 (2分)

由以上两式解得：L₂＝　　　　　　　　　　　　　　　 (2分)

代入数据解得：L₂＝0.6m　　　　　　　　　　　　　　　　　　　　 (1分)

34．⑴甲(3分)；先释放纸带再接通电源(3分)．

⑵(I)

③作出－R图象(4分)

④2.0，1.0(共2分)

(Ⅱ)如右图所示(2分)②A₁；(1分)

④(3分)

33.(16分)(1)醛基、酯基(2分)　　 (2) (2分)

(3)加成(2分)　 取代(2分)

(4)CH₃OCH₃(2分)

(5)②2CH₃CH₂OH+O₂ 2CH₃CHO+2H₂O(2分)

④2CH₃COOH+HOCH₂-CH₂OHCH₃COOCH₂-CH₂OOCCH₃+2H₂O(2分)

(6)CH₃CHO+2Cu(OH)₂CH₃COOH+Cu₂O↓+2H₂O(2分)

32．(16分,每空2分)(1)弱酸；

(2) 大于；c(A^－)＝c(Na⁺)

(3) 大于；c(A^－) ＞c(Na⁺) ＞c(H⁺)＞ c(OH^－)

(4) 10 ^-5； (5)10 ^-5－10 ^-9；　 10 ^-9．

31.(16分)(1)强氧化性(漂白性)(1分)．

(2)a、b、c(各1分，全对给3分，选有d不给分)．

(3)盐析(1分)；使2 Na₂CO₃·3H₂O₂在水中的溶解降低(1分)，析出更多晶体(1分)．

(4)催化双氧水的分解(1分)．

(5)温度高时双氧水易分解(1分)，温度低时反应缓慢(1分)．

(6)晶体的洗涤(2分) 向置于过滤器上的沉淀加蒸馏水至刚好淹没沉淀，静置，待水自然流出后，再重复操作两到三次。(2分)

(7)因为通过气泵(或其它方法)使过滤器下方的压强降低，滤液上下形成压强差，从而加快过滤的速度。(2分)

30.(16分)1)假设1　 碳酸钠、氯化钠；(2分)

　　　　　　　 假设2　 碳酸钠、硫酸钠.(2分)

(1)错的(1分)。溶液中含有CO₃^2-，可与BaCl₂溶液产生BaCO₃沉淀，因此无法确定该沉淀是BaSO₄(3分)。

　(2)：

29.(16分)(1) ①BbRR(1分)　　 BbRr(1分).

　②子叶深绿抗病∶子叶深绿不抗病∶子叶浅绿抗病∶子叶浅绿不抗病(2分).

　3∶1∶6∶2 (2分).

③80%(2分).

④BR与BR、BR与Br(2分).

⑤用组合一的父本植株自交，在子代中选出子叶深绿类型即为纯合的子叶深绿抗病大豆材料(2分).

(2)①限制性内切酶和DNA连接酶(2分).

②定向改变生物的遗传性状；克服杂交的不亲和性(2分).

28．(14分)

(1)A(2分)。病毒无细胞结构，需要寄生在活细胞才能生存和繁殖(其它合理答案也可给分)(2分)

(2)蛋白质与双缩脲试剂显紫色(2分)

(3)①　 RNA蛋白质(2分)

② 基因突变(2分)

(4)效应T细胞与被侵染的靶细胞密切接触，使靶细胞裂解死亡，病毒因失去藏身之处而被相应抗体或吞噬细胞消灭(2分)。

(5)产生针对甲型H1N1流感病毒的抗体，对该病毒所致流感起到免疫预防作用(2分)。