题目内容
如图甲是利用激光测转速的原理示意图,图中圆盘可绕固定轴转动,盘边缘侧面上有一小段涂有很薄的反光材料,当盘转到某一位置时,接收器可以接收到反光涂层所反射的激光束,并将所收到的光信号立即转变成电信号,在示波器显示屏上显示出来。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082423182074812411.png)
(1)若图乙中示波器显示屏上横向的每小格对应的时间为1.00×10-2s(即乙图中t轴每大格表示5.00×10-2s),则圆盘的转速为________r/s。(保留3位有效数字)
(2)若测得圆盘直径为11cm,则可求得圆盘侧面反光涂层的长度为________cm。(保留3位有效数字)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082423182074812411.png)
(1)若图乙中示波器显示屏上横向的每小格对应的时间为1.00×10-2s(即乙图中t轴每大格表示5.00×10-2s),则圆盘的转速为________r/s。(保留3位有效数字)
(2)若测得圆盘直径为11cm,则可求得圆盘侧面反光涂层的长度为________cm。(保留3位有效数字)
(1)(1) 4.55 (3分) (2)1.57cm (3分)
试题分项:(1)由乙图可知,圆盘转动的周期
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231822293798.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231823619556.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231825397492.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231826848390.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231828205559.png)
(2)由乙图可知,反光涂层经过接收器的时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231829204587.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231830483407.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231831793663.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231833260600.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824231835007577.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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