B.F1=F2學(xué)科網(wǎng) 查看更多

 

題目列表(包括答案和解析)

已知雙曲線E的離心率為e,左、右兩焦點(diǎn)分別為F1F2,拋物線CF2為頂點(diǎn),F1為焦點(diǎn),點(diǎn)P為拋物線與雙曲線右支上的一個(gè)交點(diǎn),若a|PF2|+c|PF1|=8a 2(其中a、c分別為雙曲線的實(shí)半軸長(zhǎng)和半焦距),則e的值為  (    )學(xué)科網(wǎng)

A.               B. 3              C.             D. 學(xué)科網(wǎng)

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已知雙曲線E的離心率為e,左、右兩焦點(diǎn)分別為F1F2,拋物線CF2為頂點(diǎn),F1為焦點(diǎn),點(diǎn)P為拋物線與雙曲線右支上的一個(gè)交點(diǎn),若a|PF2|+c|PF1|=8a 2(其中a、c分別為雙曲線的實(shí)半軸長(zhǎng)和半焦距),則e的值為  (  A  )學(xué)科網(wǎng)

A.   B. 3    C.   D. 學(xué)科網(wǎng)

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△ABC中,a,b,c分別是角A,B,C的對(duì)邊,向量m=(2sinB,2-cos2B),學(xué)科網(wǎng)

 ,m⊥n,學(xué)科網(wǎng)

(I)求角B的大;學(xué)科網(wǎng)

  (Ⅱ)若,b=1,求c的值.學(xué)科網(wǎng)

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、是兩條不同的直線,是三個(gè)不同的平面,則下列命題中為真命題的是學(xué)科網(wǎng)

     A.若     B.若  學(xué)科網(wǎng)

     C.若      D.若學(xué)科網(wǎng)

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已知函數(shù),的零點(diǎn)分別為

,則的大小關(guān)系是(      )]

     B       

     D  [來(lái)源:學(xué)科網(wǎng)ZXXK]

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一、本題共10小題,每小題5分.每小題給出的四個(gè)選項(xiàng)中,有的小題只有一個(gè)選項(xiàng)正確,有的小題有多個(gè)選項(xiàng)正確,全部選對(duì)得5分,選不全得3分,有選錯(cuò)或不答得0分.

題號(hào)

1

2

3

4

5

6

7

8

9

10

答案

AB

D

B

D

AC

C

A

D

B

BCD

學(xué)科網(wǎng)(Zxxk.Com)二、本題共2小題,共18分

  11(4分)偏小

12.(8分)(1)電路圖(4分)

 

 

(2)9.0,10.0 (每空2分)

 

13.(4分)右  (2分)          t=  (2分)

14、(13分)解:(1)飛船在運(yùn)行過(guò)程中萬(wàn)有引力提供向心力

 

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2分)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

由(2) (3)得:飛船運(yùn)行周期:

2分)

 

 

 

 

 

 

 

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15.(13分)解(1)如圖所示,帶電質(zhì)點(diǎn)在重力mg(大小及方向均已知)、洛倫茲力qv0B(方向已知)、電場(chǎng)力qE(大小及方向均未知)的作用做勻速直線運(yùn)動(dòng)。根據(jù)力三角形知識(shí)分析可知:當(dāng)電場(chǎng)力方向與磁場(chǎng)方向相同時(shí),場(chǎng)強(qiáng)有最小值。

學(xué)科網(wǎng)(Zxxk.Com)根據(jù)物體的平衡規(guī)律有

                     (1分)

                     (1分)

解得                     (2分)

                          (2分)

(2)當(dāng)電場(chǎng)力和重力平衡時(shí),帶點(diǎn)質(zhì)點(diǎn)才能只受洛倫茲力作用做勻速圓周運(yùn)動(dòng)

則有:                 (1分)

得:                   (1分)

要使帶電質(zhì)點(diǎn)經(jīng)過(guò)x軸,圓周的直徑為   (2分)

根據(jù)(1分)

                           (2分)

16.(12分)解:⑴小球做勻速圓周運(yùn)動(dòng),突然放繩則小球以原有的速度做勻速直線運(yùn)動(dòng)到C,在C點(diǎn)處一瞬間8m繩突然拉直,沿繩方向的速度vyP突變?yōu)?,而小球?qū)⒁詖x做勻速圓周運(yùn)動(dòng),到達(dá)B點(diǎn),

學(xué)科網(wǎng)(Zxxk.Com)由幾何關(guān)系可知:S1=AC=   (1分)

∠AOC=60°=π/3

     (2分)                      

∠BOC=120°=2π/3     (1分)

在C點(diǎn),由矢量三角形可知:vx=vcos60°=v/2    (1分)

    (2分)

 (2分)

學(xué)科網(wǎng)(Zxxk.Com)(2)在B點(diǎn),則有:    (3分)

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        解:(1)小物塊沿斜面下滑的過(guò)程中受力情況如圖所示

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