牛顿定理英文物理题You throw a ball of mass 1 kg straight up.You observe that it takes 3.2 s to go up and down,returning to your hand.Assuming we can neglect air resistance,the time it takes to go up to the top is half the total time,1.6 s.No
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/26 02:40:37
牛顿定理英文物理题You throw a ball of mass 1 kg straight up.You observe that it takes 3.2 s to go up and down,returning to your hand.Assuming we can neglect air resistance,the time it takes to go up to the top is half the total time,1.6 s.No
牛顿定理英文物理题
You throw a ball of mass 1 kg straight up.You observe that it takes 3.2 s to go up and down,returning to your hand.Assuming we can neglect air resistance,the time it takes to go up to the top is half the total time,1.6 s.Note that at the top the momentum is momentarily zero,as it changes from heading upward to heading downward.
Use the Energy Principle to determine the maximum height above your hand reached by the ball.
h=?m
牛顿定理英文物理题You throw a ball of mass 1 kg straight up.You observe that it takes 3.2 s to go up and down,returning to your hand.Assuming we can neglect air resistance,the time it takes to go up to the top is half the total time,1.6 s.No
这道题翻译下来就是:
把一个质量为1kg的小球竖直上抛,从出发到回到原点需要3.2秒.假设可以忽略空气阻力,则上升到顶部时间是1.6秒.由于它在最高点处掉头向下,动量为零,请使用能量原理来确定手到最高点的距离h.
选手的高度为零势能点,则小球从离手到最高点是动能全部转化为重力势能.
(1/2)mVo^2=mgh
Vo=gt
解得h=8m.
希望能帮到你,满意给个好评吧!谢谢!
题目大意:扔一个1kg的球,从离手到回到出发点共经历了3.2秒,如忽略空气阻力,则上升过程所花费的时间为1.6s。请利用能量原理确定小球达到的最高高度。
解题过程:s=1/2gtt
h=1/2×9.8×1.6×1.6=12.544米
由于Vo=gt,t=1.6s,则mgh=1/2mVo^2一0,因此h=Vo^2/2g=12.8m
h=1/2*a*t^2=12.8m
这里重力加速度我用的是10,如果要精确点用9.8也行
原题就是往上扔一个球,不考虑阻力,飞上去用1.6秒,问能飞多高。其实和质量没关系,高中物理学过s=1/2*a*t^2。 直接出结果。看来外国高中考试题,确实比国内简单的多啊...
全部展开
h=1/2*a*t^2=12.8m
这里重力加速度我用的是10,如果要精确点用9.8也行
原题就是往上扔一个球,不考虑阻力,飞上去用1.6秒,问能飞多高。其实和质量没关系,高中物理学过s=1/2*a*t^2。 直接出结果。看来外国高中考试题,确实比国内简单的多啊
收起