2007年全国理科综合卷1答案解析
14:B 解析:由题意可以得到g ’=1.6g;由黄金代换GM=gR 2可以得到22'''M R g MR g
=解得R ’=2R
15:A 解析:由振动图像得到原点处的质点在y 正半轴向下运动,由于向负x 轴传播,所以只有A 选项正确.
16:AC 解析: 由于两种状态下压强相等,所以在单位时间单位面积里气体分子对活塞的总冲量肯定相等;由于b 状态的温度比a 状态的温度要高,所以分子的平均动量增大,因为总冲量保持不变,所以b 状态单位时间内冲到活塞的分子数肯定比
a 状态要少.
17:C 解析:如图所示,光线射到A 或B 时,入射角大于临界
角,发生全反射,而后由几何关系得到第二次到达界面的时候垂直
打出.O 点为?ABC 的重心,设EC=x,则由几何关系得到: 23
x x r =+.解得光斑半径x=2r
18:C 解析 ,选向下为正方向,由动量定理分别得到对于A 图:
11.520.51mg mg mv ?+?=
对于B 图: 20.511 1.51mg mg mg mv ?+?+?=
对于C 图: 31 1.52mg mg mv ?+?=
对于D 图: 41.52mg mv ?=
综合四个选项得到3v 最大
19:AD 解析: 存在两种可能,第一种n=2到n=4,由于是电子轰击,所以电子的能量必须满足13.6-0.85 20:B 解析:运用一个结论:在匀强电场中,任意 一族平行线上等距离的两点的电势差相等,所以 Uab=Ucd,所以c 点电势为8v; :D 解析:由初始位置可得,切割的有效长度在逐 渐变大,且为逆时针,所以BD 中选一个,由于BD 两项 中第2秒是一样的,没有区别.在第3秒内,线框已经 有部分出上面磁场,切割的有效长度在减少,且为顺时针方向,所以只有D 选项是正确的. 22.(1)①竖直位移或↑↓ 衰减或衰减调节 y 增益 ② 扫描范围 1k 挡位 扫描微调 (2)①P 点是在实验的第一步中小球1落点的平均位置 M 点是小球1与小球2碰后小球1落点的平均位置 N 点是小球2落点的平均位置 ②小球从槽口C 飞出后作平抛运动的时间相同,假设为 t,则有 10op v t = 1OM v t = 2ON v t = 小球2碰撞前静止,即200v = 2110200v v ON OM ON OM e v v OP OP ---===-- ③OP 与小球的质量无关,OM 和ON 与小的质量有关 222121111222 M mv mv MV =+ ⑥ 解得:21m M v v m M -=+ ⑦ 整理得:220( )m M v v m M -=-+ ⑧ 所以:0()n n m M v v m M -=+ ⑨ 而偏离方向为450的临界速度满足:021(1cos 45)2 mgl mv -=临界 ⑩ 联立① ⑨ ⑩代入数据解得,当n=2时,2v v >临界 当n=3时,3v v <临界 所以,最多碰撞3次 25 解:对于y 轴上的光屏亮线范围的临界条件如图1所示:带电粒 子的轨迹和x=a 相切,此时r=a ,y 轴上的最高点为y=2r=2a ; 对于 x 轴上光屏亮线范围的临界条件如图2所示:左边界的极限情况 还是和x=a 相切,此刻,带电粒子在右边的轨迹是个圆,由几何知识得到 在x 轴上的坐标为x=2a;速度最大的粒子是如图2中的实线,又两段圆弧 组成,圆心分别是c 和c ’ 由对称性得到 c ’在 x 轴上,设在左右两部分磁 场中运动时间分别为t1和t2,满足1225 t t = 12712 t t T += 解得116t T = 2512 t T = 由数学关系得到:32R a = OP=2a+R 代入数据得到:3OP=2(1+ )3a 所以在x 轴上的范围是2a x 2(1+3 a ≤≤ 2008年普通高等学校招生全国统一考试答案 14、D 解析:竖直速度与水平速度之比为:tanφ = gt v 0 ,竖直位移与水平位移之比为:tanθ = 0.5gt 2 v 0t ,故tanφ =2 tanθ ,D 正确。 15、AD 解析:对小球水平方向受到向右的弹簧弹力N ,由牛顿第二定律可知,小球必定具有向右的加速度,小球与小车相对静止,故小车可能向右加速运动或向左减速运动。 16、C 解析:由图可看出波长为4m ,t =0时刻x =3m 处的质点向上振动,可得该波向左传播。将整个波形图向左平移1.5m 时,a 质点到达波峰,此时b 质点正好在平衡位置,与t =0时刻平衡位置在7m 处的质点振动状态一样,故a 质点到达波峰时,b 质点正在平衡位置并向上振动,A 错;将图像整体向左平移1m ,即波传播T /4时,a 的振动状态与与t =0时刻平衡位置在3.5m 处的质点振动状态一样,即处在平衡位置上方并向y 轴正方向运动,B 错;将图像整体向左平移3m ,即波传播3T /4时,a 的振动状态与与t =0时刻平衡位置在 9.5m 处和1.5m 的质点振动状态一样,即处在平衡位置下方并向y 轴负方向运动,C 对;a 、b 质点相隔3m ,即相差3T /4,速度相同的质点应该在半周期内才会出现,故D 错。 17、B 解析:设太阳质量M ,地球质量m ,月球质量m 0,日地间距离为R ,月地间距离为r ,日月之间距离近似等于R ,地球绕太阳的周期为T 约为360天,月球绕地球的周 期为t =27天。对地球绕着太阳转动,由万有引力定律:G Mm R 2=m 4π2R T 2,同理对月球绕着地球 转动:G mm 0r 2=m 04π2r t 2,则太阳质量与地球质量之比为M : m =R 3T 2 r 3t 2;太阳对月球的万有引力F = G Mm 0R 2,地球对月球的万有引力f = G mm 0r 2,故F : f = Mr 2 mR 2,带入太阳与地球质量比,计算出比值约为2,B 对。 18、CD 解析:设原子核X 的质量数为x ,电荷数为y ,依题意写出核反应方程,根据质量数守恒和电荷数守恒,可得原子核Y 的质量数为x ,电荷数为y -1,原子核Z 的质量数为x -3,电荷数为y -2。由此可得X 核的质子(y )比Z 核的质子(y -2)多2个,A 错;由此可得X 核的中子(x -y )比Z 核的中子(x -y -1)多1个,B 错;X 核的质量数(x )比Z 核的 质量数(x -3)多3个,C 对;X 核与Z 核的总电荷(2y -2)是Y 核电荷(y -1)的2倍,D 对。 19、B 解析:大气压是由大气重量产生的。大气压强p =mg S =mg 4πR 2,带入数据可得地球表面大气质量m =5.2×1018kg 。标准状态下1mol 气体的体积为v =22.4×10-3m 3,故地球表面大 气体积为V =m m 0 v =5.2×1018 29×10-3×22.4×10-3m 3=4×1018m 3,B 对。 20、D 解析:0-1s 内B 垂直纸面向里均匀增大,则由楞次定律及法拉第电磁感应定律可得线圈中产生恒定的感应电流,方向为逆时针方向,排除A 、C 选项;2s-3s 内,B 垂直纸面向外均匀增大,同理可得线圈中产生的感应电流方向为顺时针方向,排除B 选项,D 正确。 21、B 解析:设折射角为α,玻璃砖的厚度为h ,由折射定律n =sinθsinα ,且n =c v ,在 玻璃砖中的时间为t =h v cosα ,联立解得t 2∝n 4 n 2- sin 2θ,红光频率较小,θ为零时,t 1<t 2,θ 为90°时,趋近渐近线,初步判定该函数为单调函数,通过带入θ为其它特殊值,仍然有t 1<t 2,故B 对。 22、I (1)答案:①②或①③解析:通过连结 在一起的A 、B 两物体验证机械能守恒定律,即验证 系统的势能变化与动能变化是否相等,A 、B 连结在 一起,A 下降的距离一定等于B 上升的距离;A 、B 的速度大小总是相等的,故不需要测量绳子的长度 和B 上升的距离及时间。 (2)答案:①③。解析:如果绳子质量不能忽 略,则A 、B 组成的系统势能将有一部分转化为绳子的动能,从而为验证机械能守恒定律带来误差;若物块摇摆,则两物体的速度有差别,为计算系统的动能带来误差;绳子长度和两个物块质量差应适当。 (3)答案:对同一高度进行多次测量取平均值;或选取受力后相对伸长量尽量小的绳;(个人补充:尽量减小滑轮的质量、对滑轮转动轴进行润滑、选择质量相对较大的物块A 、 B ) 解析:多次取平均值可减少测量误差,绳子伸长量尽量小,可减少测量的高度的准确度。(个人补充解析:实验过程中,滑轮也会转动,其能量同样来源于A 、B 组成的系统,故应减小滑轮的质量。) II.答案:7,6.3(1)将待测电压表与标准电阻串联后与电源连接即可。设电源电动势为 E,则由闭合电路欧姆定律,当两开关都闭合时,R2被短路,有:U1=E当S1闭合,S2断开时, E=U2+U2 R0(R1+R2);解两式得:R1=6000Ω,E=6.3V;根据串联分压原理,可得电压表量程为 7V。 23、解析:设物体的加速度为a,到达A点的速度为v0,通过AB段和BC点所用的时间为t,则有 l1=v0t+1 2at2 ············································································································································① l1+l2=2v0t+2at2 ·····································································································································②联立①②式得 l2-l1=at2··················································································································································③3l1-l2=2v0t··············································································································································④设O与A的距离为l,则有 l=v02 2a ·························································································································································⑤ 联立③④⑤式得 l= (3l1-l2)2 8(l2-l1) 个人解析:设物体在OA段的距离为s,用时t,在AB、BC段用时均为t1,由运动学公式: 在OA段:s = 1 2at2·······························································································································① 在OB段:s +l1= 1 2a(t+t1)2······················································································································② 在OC段:s +l1+l2= 1 2a(t+2t1)2 ···············································································································③ 联立①②③解得s= (3l1-l2)2 8(l2-l1) 24、个人解析:(1)对系统,设小球在最低点时速度大小为v1,此时滑块的速度大小为v2,滑块与挡板接触前 由系统的机械能守恒定律:mgl = 1 2mv12 + 1 2mv22 ···············································································① 由系统的水平方向动量守恒定律:mv1 = mv2 ·····················································································②对滑块与挡板接触到速度刚好变为零的过程中,挡板阻力对滑块的冲量为: I = mv2······················································································································································③联立①②③解得I= m gl方向向左 ····································································································④ (2)小球释放到第一次到达最低点的过程中,设绳的拉力对小球做功的大小为W ,对小球由动能定理: mgl +W = 12mv 12 ····································································································································· ⑤ 联立①②⑤解得:W =-12mgl ,即绳的拉力对小球做负功,大小为12mgl 。 25、个人解析:(1)从A 点进入磁场后从O 点离开磁场的过程是匀速圆周运动,画出粒子运动的轨迹图,依题意由几何关系可得圆弧的圆心正好是两条虚线的交点。 故经过A 点的速度方向为x 轴正方向。 设圆周的半径为R ,有:∠OO 1A =30° ·············································· ① 根据向心力公式:Bqv = m v 2 R ····························································· ② A 点到x 轴的距离:x = R -R cos30° ·················································· ③ 联立①②③解得:x = 2-32 mv Bq (2)粒子能从O 点进入电场且能由O 点返回,对正电荷,说明电场的方向垂直于OC 向左,设电场强度大小为E ,电场中的时间为t 1,由动量定理: Eqt 1=2mv ····························································································· ④ 粒子从A 点射入到第二次离开磁场所用的时间恰好等于粒子在磁场中做圆周运动的周期T ,由: T = 2πR v ································································································ ⑤ 从O 点返回磁场后的轨迹如图,圆心角为120°,故: T =t 1+112T +13T ···················································································· ⑥ 联立②④⑤⑥解得:E = 12Bv 7π ·························································· ⑦ (3)第二次离开磁场后到再进入电场,如图轨迹。 则DF =OD =2R cos30° ··········································································· ⑧ 时间t 2= OD v = 3m Bq v v O O D F