6555福州三中第一学期高一期考

（1）写出B装置中反应的化学方程式____________________________________。实验结束后， 立即将B中溶液滴几滴在pH试纸上，可观察到的现象是______________________________。 （2）装置C的实验目的是验证氯气是否具有漂白性，为此C中 I、II、III处依次放入的物质正 确的是 ____________ 。①干燥的有色布条 ②湿润的有色布条 ③浓硫酸 ④碱石灰(固体 氢氧化钠和氧化钙组成的混合物)
A．浓硫酸
B．饱和食盐水
C．碘化钾溶液 D．氢氧化钠溶液
10．下列说法中正确的是
A．从碘水中提取碘单质可选用酒精作有机萃取剂
B．分液操作时上层液体经分液漏斗下口放出
C．浓硫酸不慎溅到手，立即用大量水冲洗，再用饱和碳酸氢钠溶液洗，最后用水冲洗
D．过滤时，为了加快过滤，可以用玻璃棒搅拌漏斗内待过滤的液体
中洗气
HCl 会与 Na2CO3 溶液反应 且产生更多 CO2
除去 NaCl 固体中混有的少 加水溶解配成热的浓溶
D
量 KNO3 杂质
液，冷却结晶，过滤
A．A
B．B
C．C
NaCl 和 KNO3 溶解度随温度 变化有较大差异
D．D
17．下图所示为海水综合利用部分流程，有关说法不正确的是
A．实验室进行①的操作需用到得玻璃仪器有漏斗、玻璃棒、烧杯 B．②步中除去粗盐中的 SO42-、Ca2＋、Mg2＋等杂质加入的药品顺序为：Na2CO3 溶液 →NaOH 溶液→BaCl2 溶液→过滤后加盐酸 C．③、④反应都涉及氧化还原反应
D．古剑“沈卢”“以剂钢为刃，柔铁为茎干，不尔则多断折”，剂钢属于混合物
2．下列状态的物质，既能导电又属于电解质的是

2024-2025学年福州市高三年级第一次质量检测数学答案一、单项选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的。

有多项符合题目要求。

全部选对的得6分，部分选对的得部分分，有选错的得0分。

15. （13分）已知数列{}n a 满足12a =，132n n a a +=+． （1）证明：数列{}1n a +是等比数列； （2）求{}n a 的前n 项和n S ．【解法一】（1）证明：因为132n n a a +=+，且12a =，所以10n a +≠， ··················································································· 1分 所以1132111n n n n a a a a ++++=++ ········································································ 3分 3(1)31n n a a +==+， ···································································· 5分 又113a +=，所以数列{}1n a +是以3为首项，3为公比的等比数列． ······························· 6分 （2）由（1）得13n n a +=，所以31n n a =−， ············································· 8分 所以()()()2313131n n S =−+−++−()233333n n =++++− ···························································· 10分13313n n +−=−− ············································································ 12分 133.2n n +−=− ············································································ 13分【解法二】（1）证明：因为132n n a a +=+，所以()113331n n n a a a ++=+=+， ····························································· 2分 因为12a =，所以1130a +=≠，所以10n a +≠， ········································ 4分 所以1131n n a a ++=+, 所以数列{}1n a +是以3为首项，3为公比的等比数列． ······························· 6分 （2）略，同解法一． 16. （15分）已知ABC △的内角,,A B C 的对边分别为,,a b c，且2cos cos cos a C C B =⋅． （1）求角C ；（2）若4a =，b =D 为AB 中点，求CD 的长． 【解法一】（1）因为2cos cos cos a C C B =+⋅， 由正弦定理，得2sin cos cos sin A C B C B C =·············································· 2分()B C =+ ·································································· 4分()πA −A =，······································································ 6分 因为0πA <<，则sin 0A ≠，所以cos C =， ·········································· 7分 由于0πC <<，则π6C =； ···································································· 8分 （2）因为D 为AB 中点，故()12CD CA CB =+， ······································ 10分22111πcos 4426CA CB CA CB =++ ············································ 13分 1113164442=⨯+⨯+ 314=，················································································· 14分 所以CD ． ······································································ 15分 【解法二】（1）因为2coscos cos a CC B =⋅，由余弦定理，得2222222cos 22a b c a c b a C ab ac+−+−=··························· 2分 ， ···························································· 4分所以cos C =， ················································································ 6分 由于0πC <<，则π6C =； ···································································· 8分 （2）由（1）知，π6ACB ∠=， 在ABC △中，由余弦定理，得2222cos c a b ab ACB =+−∠··································································· 10分 22424=+−⨯ 7=， ··························································································· 11分 故c =， ······················································································· 12分 因为D 为AB 中点，所以cos cos 0ADC BDC ∠+∠=，故222222022AD CD AC BD CD BC AD CD BD CD +−+−+=⨯⨯⨯⨯， ·········································· 13分22222240CD CD +−+−=，故CD ． ··········································································· 15分 【解法三】（1）略，同解法一或解法二； （2）由（1）知，π6ACB ∠=， 在ABC △中，由余弦定理，得2222cos c a b ab ACB =+−∠··································································· 10分22424=+−⨯ 7=， ··························································································· 11分故c =， ······················································································· 12分 所以222cos 2b c a A bc+−=2224+−==， ············································································· 13分 在ACD △中，由余弦定理， 得2222cos CD AC AD AC AD A =+−⋅222⎛=+− ⎝⎭⎝314=， ······················································································· 14分故CD ． ··········································································· 15分 17. （15分）如图，在四棱锥S ABCD −中，BC ⊥平面SAB ，∥AD BC ，1SA BC ==，SB =，o 45SBA ∠=．（1）求证：SA ⊥平面ABCD ；（2）若12AD =，求平面SCD 与平面SAB 的夹角的余弦值． 【解法一】（1）在△SAB 中， 因为1SA =，o 45SBA ∠=，SB =， 由正弦定理，得sin sin SA SBSBA SAB=∠∠， ········································································· 1分所以1sin 45︒， ······································································ 2分 所以sin 1SAB ∠=，因为0180SAB ︒<∠<︒，所以90SAB ∠=︒，所以SA AB ⊥． ··················································································· 4分 因为BC ⊥平面SAB ，SA ⊂平面SAB ，所以BC SA ⊥， ··················································································· 5分 又BCAB B =，所以SA ⊥平面ABCD ； ········································································· 6分 （2）解：由（1）知SA ⊥平面ABCD ，又,⊂AB AD 平面ABCD ，所以SA AB ⊥，SA AD ⊥，因为BC ⊥平面SAB ， ··········································································· 7分 ⊂AB 平面SAB ，所以BC AB ⊥，因为∥AD BC ，所以AD AB ⊥，所以,,SA AD AB 两两垂直． ··································································· 8分 以点A 为原点，分别以AD ，AB ，AS 所在直线为x 轴，y 轴，z 轴建立如图所示的空间直角坐标系， ················································································ 9分 则1(1,1,0),,0,0,2(0,0,1),D S C ⎛⎫ ⎪⎝⎭所以()1,1,1SC =−，1,0,12SD ⎛⎫=− ⎪⎝⎭，设平面SCD 的法向量为1(,,)x y z =n ，则11,,SC SD ⎧⊥⎪⎨⊥⎪⎩n n 即110,10,2SC x y z SD x z ⎧⋅=+−=⎪⎨⋅=−=⎪⎩n n 取2x =，则()12,1,1=−n ， ·································································· 11分显然平面SAB 的一个法向量()21,0,0=n ， ················································ 12分 所以cos ⋅=⋅121212n n n ,n n n ····································································· 13分==········································································· 14分 所以平面SCD 与平面SAB ． ··································· 15分 【解法二】（1）证明：设AB x =，在△SAB 中， 因为1SA =，o 45SBA ∠=，SB =， 由余弦定理，得2222cos SA SB AB SB S AB BA =∠+−⋅， · (1)分 所以212co 5s 4x =+−︒， (2)分所以221x +−=， 所以2210x x −+=，解得1x =． ························································································ 3分 所以2222SA AB SB +==，所以SA AB ⊥． ················································ 4分 因为BC ⊥平面SAB ，SA ⊂平面SAB ，所以BC SA ⊥， ··················································································· 5分 又BCAB B =，所以SA ⊥平面ABCD ； ········································································· 6分 （2）略，同解法一．【解法三】（1）设AB x =，在△SAB 中， 因为1SA =，o 45SBA ∠=，SB =， 由余弦定理，得2222cos SA SB AB SB S AB BA =∠+−⋅， (1)分所以212co 5s 4x =+−︒， ································································ 2分所以221x+−=，所以2210x x−+=，解得1x=． ························································································3分所以2222SA AB SB+==，所以SA AB⊥．················································4分因为BC⊥平面SAB，BC⊂平面ABCD，所以平面ABCD⊥平面SAB；·································································5分又平面ABCD平面SAB AB=，SA AB⊥，SA⊂平面SAB，所以SA⊥平面ABCD；·········································································6分（2）由（1）知SA⊥平面ABCD，过B作BM SA，则BM⊥平面ABCD，又,AB BC⊂平面ABCD，所以BM AB⊥，BM BC⊥，因为BC⊥平面SAB，···········································································7分又⊂AB平面SAB，所以BC AB⊥，所以,,BM BA BC两两垂直．··································································8分以点B为原点，分别以BA，BC，BM所在直线为x轴，y轴，z轴建立如图所示的空间直角坐标系， ················································································9分则1(0,1,0),1,(,0,21,0,1),CS D⎛⎫⎪⎝⎭所以()1,1,1SC=−−，11,,02CD⎛⎫=−⎪⎝⎭，设平面SCD的法向量为1(,,)x y z=n，则11,,SCCD⎧⊥⎪⎨⊥⎪⎩nn即110,10,2SC x y zCD x y⎧⋅=−+−=⎪⎨⋅=−=⎪⎩nn取2y=，则()11,2,1=n， ···································································· 11分显然平面SAB的一个法向量()20,1,0=n， ··············································· 12分所以cos⋅=⋅121212n nn,nn n····································································· 13分=。

2019-2020学年福建省福州三中高一（上）期中物理试卷一、单选题（本大题共10小题，共20.0分）1. 第31届夏季奥林匹克运动会于2016年8月在巴西里约成功举行．在考察下列运动员的比赛成绩时，运动员可视为质点的是( )A.B.C.D.2. 关于加速度，下列说法正确的是( )A. 加速度表示速度“增加”B. 加速度表示速度“变化”C. 加速度表示速度变化的快慢D. 加速度表示速度变化的大小3. 一个朝某方向做直线运动的物体，在时间t 内的平均速度是υ，紧接t2内的平均速度是v2，则物体在这段时间内的平均速度是( )A. υB. 2v3C. 3v4 D. 5v64. 一物体从空中某一点自由下落，关于下落过程，下列说法错误 的是( )A. 第1秒、第2秒、第3秒末的速度之比为1：2：3B. .第1秒、第2秒、第3秒内的位移之比为1：3：5C. 第1秒、第2秒、第3秒内的路程之比为1：4：9D. 第1秒、第2秒、第3秒的时间之比为1：1：15. (v 2−v 02=2ax)关于公式x =v 2−v 022a，下列说法正确的是 ( )A. 此公式只适用于匀加速直线运动B. 此公式适用于匀减速直线运动C. 此公式只适用于位移为正的情况D. 此公式不可能出现a、x同时为负值的情况6.一质点做初速度为4m/s的匀变速直线运动，经过1s速度大小变为10m/s，现以初速度方向为正方向，关于加速度说法正确的是()A. 一定为B. 可能为C. 可能小于，且为D. 以上说法均错误7.下列说法正确的是()A. 木块放在桌面上受到一个向上的弹力，这是由于木块发生微小形变而产生的B. 重力就是地球的吸引力C. 有受力物体就必有施力物体D. 由磁铁间有相互作用可知：力可以离开物体而单独存在8.如图所示，位于水平地面上的质量为M的小木块，在大小为F、方向与水平方向成α角的拉力作用下沿地面做加速运动．若木块与地面之间的动摩擦因数为μ，则木块的加速度为()A. FM B. FcosθMC. Fcosθ−μMgM D. Fcosα−μ(Mg−Fsinα)M9.如图所示，质量2kg的木块放在粗糙的水平桌面上，木块与地面的滑动摩擦因数μ=0.5，外力F1、F2沿水平方向作用在木块上，木块处于静止状态，其中F1=9N，F2=2N.若撤去力F2，则木块受到的摩擦力是()．A. 10N，方向向右B. 10N，方向向左C. 9N，方向向右D. 9N，方向向左10.如图所示，劲度系数为k1的弹簧两端分别与质量为m1、m2的物块相连，劲度系数为k2的弹簧上端与质量为m2的物块相连，整个装置放在水平地面上，开始时m1、m2都处于静止状态，现缓慢用力提上面的木块，直到下面的弹簧离开刚离开地面时，上面木块移动的距离为(重力加速度为g)()A. (m1+m2)gk1B. (m1+m2)gk1+k2C. (m1+m2)g(1k1+1k2) D. (m1+m2)gk2二、多选题（本大题共6小题，共24.0分）11.如图所示，质量为m的物块静止在倾角为θ的斜面上。

福建省福州第一中学2024-2025学年高一上学期期中考试物理试卷一、单选题1．“每天锻炼一小时，健康生活一辈子”，如图为福州一中某老师绕学校四百米跑道跑步的轨迹记录图及相关数据，以下说法正确的是（）A．本次跑步位移大小为6.67km B．本次跑步平均速度大小约为10km/hC．本次跑步过程该老师可视为质点D．5分54秒表示时刻2．图甲所示的医用智能机器人在某医院大厅巡视，图乙是该机器人在某段时间内做直线运动的位移—时间图像，20~30s的图线为曲线，其余为直线。

则机器人在（）A．0~10s内做匀加速直线运动B．0~20s内平均速度大小为零C．0~30s内的位移大小为5m D．5s末的速度与15s末的速度相同3．中国自主研发的“暗剑”无人机，时速可超过2马赫．在某次试飞测试中，起飞前沿地面做匀加速直线运动，加速过程中连续经过两段均为120m的测试距离，用时分别为2s和l s，则无人机的加速度大小是A．20m/s2B．40m/s2C．60m/s2D ．80m/s 24．如图所示，打印机进纸槽里叠放有一叠白纸，进纸时滚轮以竖直向下的力F 压在第一张白纸上，并沿逆时针方向匀速转动，滚轮与第一张纸不打滑，但第一张纸与第二张纸间发生相对滑动。

设最大静摩擦力与滑动摩擦力相等。

滚轮与白纸之间的动摩擦因数为1μ，白纸之间、白纸与纸槽底座之间的动摩擦因数均为2μ，每张白纸的质量为m ，不考虑静电力的影响，重力加速度为g ，则下列说法正确的是（ ）A ．滚轮对第一张白纸的摩擦力大小为1F μB ．第二、三张白纸间的摩擦力大小为()22F mg μ+C ．第三、四张白纸间的摩擦力大小为()2F mg μ+D ．越靠近底座，白纸间的摩擦力越大二、多选题5．如图所示，用手握住竖直的瓶子，瓶子静止在手中，下面说法中正确的是（ ）A ．手对瓶子的压力恰好等于瓶子的重力B ．手对瓶子的摩擦力方向竖直向上C ．手握得越紧，手对瓶子的摩擦力越大D ．手对瓶子的摩擦力大小一定等于瓶子的重力大小6．如图所示是用频闪周期为t ∆的相机拍摄的一张真空中羽毛与苹果从同一高度自由下落的局部频闪照片。