17.如圖所示.光滑水平面的左端固定一斜面.斜面傾角θ=370.物塊與斜面的接觸面的動(dòng)摩擦因數(shù)μ=0.25(水平部分摩擦不計(jì).與斜面連接處無機(jī)械能損耗).緊靠斜面底端A處放置一質(zhì)量為m=1kg表面光滑的L型長(zhǎng)木板.木板左端上表面與斜面底部齊平.木板左端厚度d=10cm.一輕質(zhì)彈簧的右端固定在長(zhǎng)木板的右端上.彈簧勁度系數(shù)k=450N/m.一質(zhì)量也為m的小物塊從距斜面底端L=4.5m處無初速地滑下.在A處滑上長(zhǎng)木板的左端.此后小物塊壓縮彈簧.當(dāng)彈簧壓縮到長(zhǎng)度最短時(shí)將小物塊鎖定.與長(zhǎng)木板一起向右運(yùn)動(dòng).光滑水平面的右端B處有一固定物塊.長(zhǎng)木板在B處與固定物塊相碰且粘合在一起.此時(shí)解除鎖定.(sin370=0.6.cos370=0.8.g=10m/s2) 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)行過程中萬有引力提供向心力

 

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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)

        解:(1)小物塊沿斜面下滑的過程中受力情況如圖所示

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