求助，下面这个题，求过程
All right,we need to hurry.We have to do this soon.- There has to be enoughox-- Oxygen to ignite,yes.Yeah.Okay.Now,when I press that power button...The window will blow in,andwater will rush inwards.Yeah,which is gonna be like100 punches to the stomach,okay?The wind's gonna beknocked right out of us.Now,this is near empty,but I've rigged itTo let out a burst of very high pressure --Should force a breath into yourlungs,But you have to hold on to it,okay?Hold on tight.Should be enough to getyou up the 90 feet or so.One breath?But there's two of us.Yeah,I've done the math.That's why you're takingit.You're a better swimmer,anyway.- No.- Jemma.No.I'm not leaving you here.That's ridiculous.We need a new plan.We're not discussing it,okay?You're taking it-- end of story.I couldn't live if you didn't.Well,I feel the same way.There has to be another way.- You're taking it.- Why?Why would you make me do this?You're my best friend in the world!Yeah,and you're more than that,jemma.I couldn't find the courage to tell you.So,please......Let me show you.It's okay.No. No.Jemma. Jemma. Jemma.Jemma,we have to hurry.No. No.Take it,jemma.No.No.No!
求助物理高手，希望能把它解释清楚，谢谢
1.D第一类永动机违背了热力学第一定律,即能量守恒定律 一切自然现象都是遵循热力学第一定律的2.C由于是水平路面向心力只能是沿半径方向的力提供,且沿半径方向没有相对滑动,故只能是静摩擦力3.D实力物体只有地球,且重力充当向心力,物体处于失重状态,故只受重力
“上帝创造的八个真理”是什么?
真理就包含所有事与物的真理，真理就是真理没有个数之分
如何鉴别绿色歌曲
（1）A与C间的摩擦力为：f=μmg=0.5×20×10N=100N，B与C间摩擦为：f=μmg=0.25×40×10N=100N，推力F从零逐渐增大，当增大到100N时，物块A开始向右移动压缩轻弹簧（此时B仍保持静止），设压缩量为x，则力F=f+kx，当x=0.5m时，力F=f+f=200N，此时B将缓慢地向右移．B在移动0.5m的过程中，力F保持F=200N不变，弹簧压缩了0.5m，B离木板C的右端0.2m，A离木板C有左端1.0m．作出力F随A位移的变化图线如图所示．（2）在物块B移动前，力F作用于物块A，压缩弹簧使弹簧贮存了弹性势能E，物块A移动了s=0.5m，设力F做功为W，由能量守恒可得弹簧贮存的弹性势能大小为：E=W-fs=-100×0.5J=25J（3）撤去力F之后，AB两物块给木板C的摩擦力的合力为零，故在物块AB滑离木板C之前，C仍静止不动．物块AB整体所受外力的合力也为零，其动量守恒，可得 mv=mv由题可知，始终v：v=m：m=2：1 当物块B在木板C上向右滑动了0.2m，物块A则向左滑动了0.4m，但A离木板C的左端还有d=0.6m．可见，物块B先滑离木板C．并且两物体的相对位移△s=0.4m+0.2m=0.6m＞0.5m（弹簧的压缩量），弹簧储存的弹性势能已全部释放，由能量守恒定律有 12E=++f?△s由此求出物块B滑离木板C时A物块的速度为v=4m\\\/s 设此后A滑离木板C时，物体A的速度v′，木板C的速度v′，有动量守恒定律有 mv=mv′+mv′ 由能量守恒有fd=-（+）有将d=0.6m及有关数据代入上两式解得：v′=1m\\\/s，（v′=3m\\\/s，不合题意舍弃）答：（1）推力F随A位移的变化图线如图所示；（2）弹簧储存的弹性势能E的大小为25J；（3）最终C的速度是1m\\\/s．
