好难得物理题!用绳拴住木棒的AB的A端,使木棒在竖直方向上静止不动·在悬点A端正下方有一点C距A端0.8m.若把绳轻轻剪

一位老股名2022-10-04 11:39:543条回答

好难得物理题!
用绳拴住木棒的AB的A端,使木棒在竖直方向上静止不动·在悬点A端正下方有一点C距A端0.8m.若把绳轻轻剪断,测得木棒AB的长度为0.6m,重力加速度g=10ms2,求AB两端通过c点的时间差是多少?

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ff升级版 共回答了14个问题 | 采纳率92.9%
由s=1/2at^2得 t=√(s/(0.5g)).所以tA-tB=√(sAC/(0.5g))-√(sBC/(0.5g))=√(0.8/(0.5*10))-√((0.8-0.6)/(0.5*10))=0.4-0.2=0.2.(单位是秒)式子中间的单位自己加上去吧.
1年前
破车轮 共回答了1个问题 | 采纳率
B端通过C点时木棒下落0.2米,下落时间T1=根号(2X0.2/10),A端通过C点时木棒下落0.8米,T2=根号(2X0.8/10),时间差就是T2-T1。
望采纳!
1年前
我的屁有点怪味 共回答了6个问题 | 采纳率
S1=0.8m-0.6m=0.2m;
S2=0.8m;
1/2gt1^2=S1,带入数值求得t1=0.2s;
1/2gt2^2=s2, .....................t2=0.4s;
所以时间间隔t=t2-t1=0.2s。(S1指BC段距离,S2指AC段距离)
希望对你有用!
1年前

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