由D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体在A点时的速度是B点时的2倍;由B点在经过0.5S如图所示,一物体由低端D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体

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由D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体在A点时的速度是B点时的2倍;由B点在经过0.5S如图所示,一物体由低端D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体
由D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体在A点时的速度是B点时的2倍;由B点在经过0.5S
如图所示,一物体由低端D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体在A点时的速度是B点时的2倍;由B点在经过0.5S,物体滑到斜面最高点C时恰好速度为零,设AB=0.75M,求:(1)斜面的长度(2)物体由低端D点滑到B点时所需的时间

由D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体在A点时的速度是B点时的2倍;由B点在经过0.5S如图所示,一物体由低端D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体
设倾斜角为a,光滑斜面故加速度为g*sina,设B点速度为v,由物体在A点时的速度是B点时的2倍可得1.5v^2=gh(h为AB所对应的高度),则h=AB*sina.再由B点在经过0.5S,物体滑到斜面最高点C时恰好速度为零,可知加速度g*sina=2v.由以上三式可得倾斜角a,故可得加速度g*sina.此时可知运动所用时间t,长度为0.5g*sina*t^2(这是反着来看从C到D,即以匀加速g*sina运动ts).第二问利用前面的t减去0.5就是.

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由D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体在A点时的速度是B点时的2倍;由B点在经过0.5S如图所示,一物体由低端D点以4M/S的速度划上固定的光滑斜面,途径AB两点,抑制物体 如图所示,一物体由低端D点以4m/s的速度滑上固定的光滑斜面如图所示,一物体由底端D点以V0=4m/s的速度匀减速滑上固定的光滑斜面,途径A、B两点.已知物体在A点时的速度是B点时的2倍;由B点再 物体以4m/s的速度划上光滑的斜面,途径A、B两点,在A点时的速度是B点时的2倍,由B点在经过0.5S滑到斜面的顶点C,速度变为零,A、B相距0.75m,试求斜面长度及物体底端D滑到B时所用的时间. 如图所示,一物体由底端D点以V0=4m/s的速度匀减速滑上固定的光滑斜面,途径A、B两点.已知物体在A点时的速度是B点时的2倍;由B点再经过0.5s,滑到斜面最高点C时恰好速度为零.设SAB=0.75m,求:(1 物块以V0=4m/s的速度从D出发滑上光滑的斜面途经AB两点已知在A点时的速度是B点时的速度的两倍 由B点再经0.5s物块滑到斜面顶点C速度变为0 AB相距0.75m求斜面长度及物体由D运动到B的时间 物体以4m/s的速度滑上光滑斜面,途经A、B两点,已知A点速度是B点速度的两倍,由B点再经0.5s滑到顶点C,速度恰好为零,已知AB相距0.75m,试求斜面DC的长度,物体由底端D滑到B点所需时间. 物块以v0=4m/s的速度由光滑斜面底端D点沿斜面上滑,途经A,B两点已知在A点时的速度是B点时的速度的两倍 由B再经0.5s物块滑到斜面顶点C速度变为0 AB相距0.75m求斜面长度及物体由D运动到C的时间 物块以V0=4m/s 的速度滑至光滑的斜面底端D点,途径B,A两点(由上之下是BAD,)已知在A点的速...物块以V0=4m/s 的速度滑至光滑的斜面底端D点,途径B,A两点(由上之下是BAD,)已知在A点的速度是B 一物体以4m/s的速度滑上光滑的斜面,途以A、B两点,Va=2Vb,AB=0.75m,由B点经0.5s,物体到C速度刚为零一物体以4m/s的速度滑上光滑的斜面,途以A、B两点,已知物体在A点速度是B点速度的2倍,由B点再经0.5s, 1.物体以4M/S的速度滑上光滑的斜面,途径A,B两点,在A点是的速度是B点时的两倍,由B点再经过0.5S滑到斜面的顶点C,速度变为零.A,B相距0.75M,试求斜面长度及物体由底端D滑到B是所用的时间?(物体在 一物体以4M/S的速度滑上光滑斜面…………一物体以4M/S的速度滑上光滑斜面,途径A,B两点,已知它在A点时的速度是B点时的两倍,由B点再经0.5S物体就滑到斜面顶端C,速度恰好减至到0,A,B间相距0.75M, 高一物理 物体运动如图,一物体以4m/s的速度滑上光滑斜面,途经A、B两点,已知它在A点时的速度是B点时的2倍,由B点再经0.5s物体就滑到斜面顶端C,速度恰好减为零,A、B间相距0.75m.求斜面的长度及 如图,一物体以4m/s的速度滑上光滑斜面,途经A.B两点,已知它在A点时的速度是B点时的两倍.如图,一物体以4m/s的速度滑上光滑斜面,途经A.B两点,已知它在A点时的速度是B点时的2倍,由B点再0.5s物体 一物体以4m/s的速度滑上光滑斜面,途经A.B两点,已知它在A点的速度是在B点速度的两倍由B点在经过0.5s物体就滑到斜面的顶端C,速度恰好为0,已知A.B相距0.75m球斜面的长度及物体由斜面底端滑到B 一物体以4m/s的速度滑上光滑斜面,途经A.B两点,已知它在A点的速度是在B点速度的两倍由B点在经过0.5s物体就滑到斜面的顶端C,速度恰好为0,已知A.B相距0.75m球斜面的长度及物体由斜面底端滑到B 高一物理题你们快来我需要你们!如图所示,一物体以4m/s的速度滑上光滑斜面,途径A、B两点,已知它在A点时的速度是B点时的2倍,由B点再经0.5s物体就滑到斜面顶端C,速度恰好减至0,A、B间相距0.75m, 谁能解答这道高一物理题呢?谢谢~一质点以V0=4m/s的初速度从斜面底端滑上固定在水平地面上的光滑斜面,在斜面上做匀减速直线运动,运动过程中途经A、B两点,已知质点经过A点时的速度Va为经 一物体以4m/s的速度从D点滑上光滑的斜面,途径A、B两点,若物体在A点是速度是在B点是速度的2倍,且由B再经过0.5s物体滑至斜面的顶点C时速度恰好为零.已知AB=0.75m,求斜面长及物体由底端D点滑至B