物理功课！！！！帮帮忙

来源：百度知道 编辑：UC知道 时间：2024/05/15 11:15:42
Some people rejected the notion that the earth is rotating when it was first proposed. Since the earth is so large, points on the equator would be moving quite fast and it was thought that objects on the equator would be flung off into space. Show that the acceleration due to gravity is more than sufficient to keep this from happening through the following calculations.

1。Find the speed of a point on the equator.
2。How does this speed compare to the speed of sound in air?
3。Find the centripetal acceleration needed to remain on the equator.
4。How does the acceleration provided by gravity compare to the centripetal acceleration?

A cylindrical space station of diameter 500 m is set spinning to provide the sensation of normal earth gravity. Determine …

the speed of a point on the floor of the space station,

the period of one complete revolution, and

the number of revolutions per minute.

我就用汉语回答你吧，英语写得这么好，那汉语就更没有问题了。
解：（1）
地球赤道的周长为40075.24km,地球旋转一周的时间为24h，则赤道上一点的线速度（过该点与地球表面向切）为
40075.24/24=1669.80km/h=463.83m/s，所以这个速度是远小于空气中的光速（c=299792458m/s）的。
要想赤道上的物体与地球保持相对静止，则物体的向心加速度要大于该线速度所对应的向心加速度。问题要求这个临界向心加速度，a=线速度的平方/地球的赤道半径,(赤道半径为6356.9km)。
a=(463.83*463.83)/6356900=0.0338米/二次方秒，这个向心加速度是远小于地球提供的重力加速度的，所以不会出现物体脱离地球表面的情况！

（2）
由题目意思可以知道，这个空间站是失重的，也就是说地球提供的重力加速度刚好等于空间站作圆周运动的向心加速度，（重力加速度为g=9.80米／秒^2）。空间站底部的一点的线速度为
sqrt9.80*250=49.5m/s。求得空间站的轨道周长为(500*3.14)米，旋转一周的时间为500*3.14/49.5=31.7s，即周期为31.7s,那么1分钟=60s所转的圈数为60/31.7=1.9圈。

这是物理还是英语呀

既然英语这么好翻译下下面的吧
Dilute ferromagnetic oxides having Curie temperatures
far in excess of 300 K and exceptionally large ordered
moments per transition-metal cation challenge our
understanding of magnetism in solids. These materials
are high-k dielectrics with degenerate or thermally
activated n-type semiconductivity. Conventional superexchange
or double-excha