题目内容
(1)某电动机的三组线圈①、②、③阻值相同,均为几欧姆,接法可能是图1中甲、乙两种之一,A、B和C是外接头。现有一组线圈断路,维修人员通过多用电表测量外接头之间的电阻来判断故障,若测A和B之间、B和C之间、A和C之间的电阻时,多用电表指针偏转分别如图2(a)、(b)、(c)所示,则测量中使用的欧姆档的倍率是 (填
、
、
或
),三组线圈的接法是 (填甲或乙),断路线圈是 (填①、②或③)。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241242165455094.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241242165929176.png)
图2
(2)某同学设计了如图3所示的装置,利用米尺、秒表、轻绳、轻滑轮、轨道、滑块、托盘和砝码等器材来制定滑块和轨道间的动摩擦因数μ。滑块和托盘上分别放有若干砝码,滑块质量为M,滑块上砝码总质量为m‘,托盘和盘中砝码的总质量为m,实验中,滑块在水平轨道上从A到B做初速为零的匀加速直线运动,重力加速度g取10m/s2。
①为测量滑块的加速度
,须测出它在A、B间运动的 与 ,计算
的运动学公式是 ;
②根据牛顿运动定律得到
与
的关系为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241242167011299.png)
他想通过多次改变
,测出相应的
值,并利用上式来计算
。若要求
是
的一次函数,必须使上式中的 保持不变,实验中应将从托盘中取出的砝码置于 ;
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412421682615809.png)
③实验得到
与
的关系如图4所示,由此可知μ= (取两位有效数字)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412421690444988.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216077274.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216093355.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216483378.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216514390.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241242165455094.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241242165929176.png)
图2
(2)某同学设计了如图3所示的装置,利用米尺、秒表、轻绳、轻滑轮、轨道、滑块、托盘和砝码等器材来制定滑块和轨道间的动摩擦因数μ。滑块和托盘上分别放有若干砝码,滑块质量为M,滑块上砝码总质量为m‘,托盘和盘中砝码的总质量为m,实验中,滑块在水平轨道上从A到B做初速为零的匀加速直线运动,重力加速度g取10m/s2。
①为测量滑块的加速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216623304.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216623304.png)
②根据牛顿运动定律得到
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216623304.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216685346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241242167011299.png)
他想通过多次改变
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216685346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216623304.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216763306.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216623304.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216685346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412421682615809.png)
③实验得到
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216623304.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216685346.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412421690444988.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412421695131989.png)
分析:殴姆表是测量电阻阻值的电表,它的读数是刻度盘的读数与倍率的乘积.通过殴姆表去探测故障也是故障分析的一种方法.
通过求出加速度的大小,从而列出加速度与质量的关系式,并作出这两量的图象,由图象斜率去算出摩擦因数.
解:(1)由于所测电阻均为几欧姆,串联后还是几欧姆,所以欧姆档的倍率选择×1;所测值分别是4Ω、4Ω和8Ω,说明接法是乙,断路的是③.因为采用甲接法时测量值应该有两个是无穷大,一个是8Ω.
(2)①滑块在水平轨道上从A到B做初速为零的匀加速直线运动,根据s=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216966338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124216997492.png)
所以需要测量的是位移和时间.
②根据整体法有a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241242170751066.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241242170911168.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124217107923.png)
若要求a是m的一次函数必须使
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124217107923.png)
所以实验中应将从托盘中取出的砝码置于滑块上.
(3)将
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124217107923.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124217200679.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824124217216709.png)
解得μ=0.23(在0.21到0.25之间是正确的)
故答案为:(1)×1,乙,③;
(2)m+m′滑块上;
(3)0.23(0.21~0.25).
![](http://thumb2018.1010pic.com/images/loading.gif)
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