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永安一中第一次月考高一英语试卷和答案五校联考 2008-2009学年下

永安一中第一次月考高一英语试卷和答案五校联考

2008-2009学年下学期第一次月考

高一物理试题

(考试时间:90分钟总分:100分)

一、单项选择题(本题共11小题,每小题3分,共33分。在每小题给出的四个选项中,只有一个选项正确,选

对的得3分,选错或不答的得0分。)

1.水平地面上放一个重4N的足球,一个小孩用10N的力踢球,使球向前运动2m.下列关于小孩对球做功的

说法中正确的是

A. 小孩对球做功20J

B. 小孩对球做功8J

C. 小孩对球做功28J

D. 题中条件不足,无法算出小孩对球做功的多少

2.如图所示,两个互相垂直的力F1与F2作用在同一物体上,使物体通过一段位移,此过程中,力F1对物体做功4J,力F2对物体做功3J,则力F1与F2的合力做功为

A.7J Array B.1J

C.5J

D.3.5J

3. 下列说法错误的是

A.-10J的功大于+5J的功

B.功是标量,正、负表示外力对物体做功,还是物体克服外力做功

C.一个力对物体做了负功,则说明这个力一定阻碍物体的运动

D.功是矢量,正、负表示方向

4. 如图所示,用F=40N的恒力通过距水平地面h=4m高处的光滑定滑轮把静止在水平面上的质量m=5kg的小

物体从A点拉到B点,A、B两点到定滑轮正下方的C点的距离分别为s1=9.6m、s2=3m.则在此过程中绳

子的拉力对物体所做的功为Array

A. 264J

B. 216J

C. 108J

D. 无法求解

5. 如图所示,F 1=F 2,α1=α2,m 1>m 2,μ1<μ2,在F 1和F 2的作用下,物体m 1、m 2移动了相同的距离,若F 1

做的功为W 1,F 2做的功为W 2,则 A.W 1>W 2 B.W 1=W 2 C.W 1

6. 下列说法中正确的是

A .功就是能,能就是功

B .做功越多,物体的能就越大

C .外力对物体不做功,这个物体就没有能量

D .能量转化的多少可用功来量度 7. 关于功率的说法,正确的是

A .由P=

t

W

可知,力做功越多,功率就越大 B .由P=Fv 可知,物体运动越快,功率越大 C .功率越大,机械效率一定越高 D .物体做功越快,其功率越大

8. 起重机的钢索将重物由地面沿竖直方向吊起到空中某个高度,其速度一时间图象如图甲所示,则钢索对

重物的拉力的功率随时间变化的图像可能是图乙中的

9. 质量为m 的小球,从离桌面H 高处由静止落下,桌面离地面的高度为h ,如图所示,若以桌面作为重力

势能0势能面,那么当小球落地时的机械能为

A .mgH

B .mgh

C .mg (H +h )

D .mg (H -h )

10.速度为v 的子弹,恰可穿透一固定着的木板,如果子弹速度为2v ,子弹

穿透木板

的阻力视为不变,则可穿透同样的木块 A .2块 B .3块 C .4块 D .1块

11.如图所示,物体沿倾角30°的固定斜面以

g 2

1

(g 为本地的重力加速度大小)的加速度匀减速上升,则在此过程中,物体的机械能是 A. 不变的 B. 减小的 C. 增加的 D. 不能判断的

二、不定项选择题 (本题共5小题,每小题4分,共20分。每小题至少有一个选项符合题意,全部选对的得4分,选对但不全的得2分,错选或不答的得0分。) 12.关于机械能是否守恒的的叙述,下面说法中正确的是

A .做匀速运动的物体机械能一定守恒

B .做变速运动的物体机械能可能守恒

C .合力对物体做的功的等于零时,机械能一定守恒

D .不论物体受力情况如何,若只有重力做功,则物体的机械能守恒 13.下列说法正确的是

A .内能能够从高温物体传到低温物体,但不能从低温物体传到高温物体

B .根据能量守恒定律,经过不断的技术改造,热机的效率可以达到100%

C .自然界中的能量是守恒的,但有的能量便于利用,有的不便于利用,因此要节约能源

D .因为能量守恒,所以“能源危机”是不可能真正出现的

14.质量为m 1、m 2的两部小车,静止在光滑的水平面上,质量为m 的人站在m 1上用恒力F 拉绳子,经过一

段时间后,两小车的速度大小分别为v 1和v 2,位移分别为s 1和s 2,如图所示. 则这段时间内此人所做的功的大小等于 A .Fs 1

B .F (s 1+s 2)

C .

22221121)(21v m v m m ++ D .211)(2

1v m m +

15. 构建和谐、节约型社会深得民心,遍布于生活的方方面面。自动充电式电动车就是很好的一例,电动车

的前轮装有发电机,发电机与电动机的蓄电池连接。当骑车者用力蹬车或电动自行车自动滑行时,自行车就可以连通发电机向蓄电池充电,将其它形式的能转化成电能存储起来。现有某人骑车以500J 的初动能在粗糙的水平路面上滑行,第一次关闭自充电装置,让车自由滑行,其动能随位移变化关系如图中线①所示;第二次启动自充电装置,其动能随位移变化关系如图中线②所示,则第二次向蓄电池所充的电能是 A. 200J B. 250 J C. 300 J

D. 500 J

16.如下图所示,物体由静止开始分别沿不同斜面由顶端A

滑至底端B ,两次下滑的路径分别为图中的Ⅰ和Ⅱ,两次物体与斜面间动摩擦因数相同,且不计路径Ⅱ中转折处的能量损失,到达B 点时的动能分别为EK1和EK2,重力的瞬时功率分别为P1和P2 ,则

A. EK1>EK2

B. EK1=EK2

C. P1>P2

D. P1=P2

三、填空题(每空3分,共15分。把答案填在相应的横线上) 17.在探究“恒力做功与动能改变的关系”的实验时,某实验小组组装了一套如图所示的装置。另外,他们还找到了打点计时器所用的学生电源、导线、复写纸、纸带、小木块、细沙、天平。当滑块连接上纸带,用细线跨过滑轮挂上空的小沙桶时,释放小桶,滑块处于静止状态。

若你是该实验小组中的一位成员,要完成该实验,你认为:

①还需要的实验器材有

②实验时为了保证小车受到的合力与沙和沙桶的总重力大小基本相等,沙和沙桶的总质量应满足的实验条件是沙和沙桶的总质量远小于小车的质量,实验时首先要做的步骤是

18.在探究“动能与重力势能的转化和守恒”的实验中,已知打点计时器所用电源的频率为50 Hz ,查得当地

的重力加速度g=9.80 m/s 2,测得所用的重物的质量为1.00 kg ,按实验要求正确地选出一条纸带进行测量,如图所示,把第一个点记作O ,另选连续的4个点A 、B 、C 、D 作为测量的点,经测量知道A 、B 、C 、D 各点到O 点的距离分别为62.99 cm 、70.18 cm 、77.76 cm 、85.73 cm

,根据以上数据,可知重物

小车

由O 点运动到C 点,重力势能的减少量等于__ ___J ,动能的增加量等于___ __J (取三位有效数字).实验的结论是 .

四、计算题(本题共3小题,共32分,其中19、20小题各10分,21小题12分。答题时请写出必要的文字说明、方程式和重要的演算步骤,只写出最后答案的不能得分。)

19.一辆汽车的额定功率为P 0=8kW ,运行中所受阻力恒为4×102

N ,汽车质量为400kg 。

(1)在额功率不变的条件下,汽车可达到的最大速度v m 为多大?

(2)如果保持牵引力F=8×102

N 不变,汽车从静止开始匀加速运行的最长时间t.

20.滑板运动是一项惊险刺激的运动,深受青少年的喜爱.如图所示是滑板运动的轨道,AB 和CD 是一段圆弧

形轨道,BC 是一段长4 m 的水平轨道.一运动员从AB 轨道上P 点以6 m/s 的速度下滑经BC 轨道后冲上CD 轨道,到Q 点时速度减为零.已知运动员的质量为50 kg,h=2 m ,H=3 m ,不计圆弧轨道上的摩擦,g=10m/s 2.求:

(1)运动员与BC 轨道的动摩擦因数; (2)运动员最后停在BC 轨道上何处?

21.如图所示是一个横截面为半圆,半径为R 的光滑柱面,一根不可伸长的细线两端分别系着可视为质点的

物体A 、B,且m A =2m B =2m ,由图示位置从静止开始释放A 物体,当物体B 达到半圆顶点时,求此过程中绳的张力对物体B 所做的功.

永安一中2008-2009学年下学期第一次月考

高一物理试题答题卷

(考试时间:90分钟 总分:100分)

一、单项选择题(每小题3分,共33分)

二、不定项选择题(每小题4分,共20分,每题至少有一个以上选项是正确的,全部选对的得4分,选

三、填空题(每空3分,共15分)

17、___________ ;_______ .

18、____________ ;_________ ___ ; ________________

______ .

四、计算题(3小题,共32分,解答应写出必要的文字说明、方程式和重要演算步骤)

学校_______________高一 _____ 班 姓名___________ 座号_______

-------------------------装

------------------------------------------ ----订

------------------------------------------

线

2008-2009学年下学期第一次月考

高一物理参考答案

二、不定项选择题(每小题4分,共20分,每题至少有一个以上选项是正确的,全部选对的得4分,选不

全得2三、填空题(每空3分,共15分)

17、刻度尺;平衡摩擦力. 18、7.62 ;7.57 ;在实验误差允许范围内,机械能守恒 四、计算题(3小题,共32分,解答应写出必要的文字说明、方程式和重要演算步骤) 19 (10分) 、(1)当汽车加速度为零时,速度达到最大,有P 0=fv m (2分)

代入数据解得v m =20m/s (2分)

(2)汽车匀加速运动的最后时刻,功率达到额定功率,有 P 0=Fv t ① (2分)

该过程对汽车由牛顿第二定律得 F-F=ma ② (1分)

匀加速的最长时间 a

v t t

=

③ (1分) 代入数据解得 t =10s (2分)

20 (10分) 、(1)物体由P 到Q ,据动能定理 22

10)(p BC mv mgs h H mg -

=---μ (2分) 代入数据解得 μ=0.2 (2分)

(2)物体最终停在水平面上,BC 间的路程为s ,则 22

10p mv mgs mgh -

=-μ (2分) 代入数据解得 s=19m (2分)

最终停在BC 间,距B 点3m 处 (2分)

21 (12分) 、系统势能的减少量gR m R

g

m E B A P -=?2

π (2分)

系统动能的增加量2)(2

1

v m m E B A K +=

? (2分)

系统机械能守恒 K P E E ?=? (2分)

得 gR v )1(3

2

2

-=

π (2分) 绳的张力对B 球做的功 gR m v m W B B +=2

2

1 (2分)

得 mgR W 3

2

+=π (2分) 龙海二中估计难度系数为:0.55

高三第二次月考英语试卷Word版含答案

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