题目内容
机械手表的分针与秒针从重合至第二次重合,中间经历的时间为( )
A.![]() | B.1 min | C.![]() | D.![]() |
C
先求出分针与秒针的角速度:
ω分=
rad/s,ω秒=
rad/s
设两次重合时间间隔为△t,则有
φ分=ω分△t,φ秒=ω秒△t,φ秒一φ分=2π,
即 Δt=
=
(min)
故选项C正确
ω分=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824113742278455.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824113742309400.gif)
设两次重合时间间隔为△t,则有
φ分=ω分△t,φ秒=ω秒△t,φ秒一φ分=2π,
即 Δt=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824113742325520.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824113742263305.gif)
故选项C正确
![](http://thumb2018.1010pic.com/images/loading.gif)
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