题目内容
在做“用单摆测定重力加速度”的实验时,用摆长L和周期T计算重力加速度的公式是g=______。如果已知摆球直径为2.00cm,让刻度尺的零点对准摆线的悬点,摆线竖直下垂,如图所示,摆球最低点对应的度数为88.40 cm,那么单摆摆长是_____ cm 。如果测定了40次全振动的时间如图中秒表所示,那么秒表读数是_____s,此单摆的摆动周期是______s。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082413583551212602.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082413583551212602.png)
4π 2L/T2 , 87.40 , 75.2 , 1.88
根据
得
,摆长等于摆线长与球的半径之和,所以摆长为87.40cm,秒表读数为75.2,则周期为![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135835715829.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135835590763.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135835652768.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135835715829.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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