D.無(wú)論場(chǎng)強(qiáng)大小如何.小球通過(guò)所有無(wú)電場(chǎng)區(qū)的時(shí)間均相同學(xué)科網(wǎng) 查看更多

 

題目列表(包括答案和解析)

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)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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    • 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(xiàn)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(xiàn)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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