在某一真空空间内建立xoy坐标系,从原点O处向第Ⅰ象限发射一比荷=1×104c/kg的带正电的粒子(重力不计),速度大小v0=103m/s、方向与x轴正方向成30°角.                                                                    

                                       

(1)若在坐标系y轴右侧加有匀强磁场区域,在第Ⅰ象限,磁场方向垂直xoy平面向外;在第Ⅳ象限,磁场方向垂直xoy平面向里;磁感应强度均为B=1T,如图(a)所示.求粒子从O点射出后,第2次经过x轴时的坐标x1.                                                                                                                                 

(2)若将上述磁场改为如图(b)所示的匀强磁场.在t=0到t=×104s时,磁场方向垂直于xoy平面外;在t=×104s到t=×104s时,磁场方向垂直于xoy平面向里,此后该空间不存在磁场.在t=0时刻,粒子仍从O点以与原来相同的速度v0射入,求粒子从O点射出后第2次经过x轴时的坐标x2.                                

                                                                                                                                       


为了测量量程为3V的电压表V的内阻(内阻约2000Ω),实验室可以提供的器材有:                       

电流表A1,量程为0.6A,内阻约0.1Ω                                                                             

电压表V2,量程为5V,内阻约3500Ω                                                                             

变阻器R1阻值范围为0﹣9999Ω                                                                                      

变阻箱R2阻值范围为0﹣99.9Ω                                                                                       

滑动变阻器R3,最大阻值约为100Ω,额定电流1.5A                                                       

电源E,电动势6V,内阻约0.5Ω                                                                                    

单刀单掷开关K,导线若干.                                                                                          

                            

(1)请从上述器材中选择必要的器材,设计一个测量电压表V的内阻的实验电路,在图1方框中画出电路原理图  (    电路原理图中的元件要用题中相应的英文字母标注),要求测量尽量准确.                

(2)根据你所画的电路原理图在图2所给的实物图上画出连线.                                   

(3)说明实验所要测量的量:                                                                           .写出计算电压表V的内阻RV的计算公式为RV=                                                                                                               

                                                                                                                                       


在探究求合力的方法时,先将橡皮条的一端固定在水平木板上,另一端系上带有绳套的两根细绳,实验时,需要两次拉伸橡皮条,一次是通过两细绳用两个弹簧秤互成角度地拉橡皮条,另一次是用一个弹簧秤通过细绳拉橡皮条.                                                                                                                              

(1)实验对两次拉伸橡皮条的要求中,下列哪些说法是正确的                         (填字母代号).             

A.将橡皮条沿相同方向拉到相同长度                                                                             

B.将橡皮条拉伸相同长度即可                                                                                       

C.将弹簧秤都拉伸到相同刻度                                                                                       

D.将橡皮条和绳的结点拉到相同位置                                                                            

(2)同学们在操作过程中有如下议论,其中对减小实验误差有益的说法是              (填字母代号).             

A.两细绳必须等长                                                                                                         

B.弹簧秤、细绳、橡皮条都应与木板平行                                                                      

C.拉橡皮条的细绳要适当长些,标记同一细绳方向的两点要适当远些                            

D.用两弹簧秤同时拉细绳时两弹簧秤示数之差应尽可能大.                                         

                                                                                                                                       

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