题目内容
一端开口的粗细均匀的细玻璃管长30cm,管内有一段5cm长的水银柱,当玻璃管开口向上竖直放置时,管内水银上表面离管口2.5cm,如图所示,将管在竖直平面慢慢地沿逆时针方向转动,已知大气压强为75cmHg.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120132863377.gif)
问:
(1)转到水平位置时,管内空气柱长度是多少?
(2)转到什么位置时,水银恰好不致流出管口?
(角度可用反三角函数表示)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120132863377.gif)
问:
(1)转到水平位置时,管内空气柱长度是多少?
(2)转到什么位置时,水银恰好不致流出管口?
(角度可用反三角函数表示)
(1) 24cm
(2)管从水平位置继续向下转过arcsin0.6角时,水银恰不流出管口
(1)以管内封闭的空气柱为研究对象,在竖直位置时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120132941111.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120132957112.gif)
在水平位置时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120132988154.gif)
根据玻意耳定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120133003403.gif)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120133019111.gif)
(2)从(1)数据可知,管在水平位置时,水银不会流出.设管从水平位置继续向下转过
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412013306693.gif)
则此时空气压强
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120133081358.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120133159117.gif)
根据玻意耳定律
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120133191211.gif)
80×22.5=(75-5sin
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412013320692.gif)
sin
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412013320692.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412013320692.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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