大学【物理】- 搜题酱免费题库

在图示机构中，物体A质量为m1，放在光滑水平面上。均质圆盘C、B质量均为m，半径均为R，物块D质量为m2。不计绳的质量，设绳与滑轮之间无相对滑动，绳的AE段与水平面平行，系统由静止开始释放。试求物体D的加速度以及BC段绳的张力。

9.一台直流发电机,其端电压为 230V,内阻为 6Ω,输出电流为 5A,该发电机的电动势为( )。A. 230VB. 240VC. 260VD. 200V

121.[ 多选题 ] 影响电流对人体伤害程度的主要因素有哪些？A. 电流的大小B. 人体电阻C. 通电时间的长短D. 电流的频率

第11章恒定磁场11-1真空中有一电流元,在由它起始的矢径的端点处的磁感强度的数学表达式为。11-2如图所示,在真空中,几种载流导线在同一平面内,电流均为I,它们在O点的磁感强度的值各为多少11-3无限长直导线在P处弯成半径为R的圆,当通以电流I时,则在圆心O点的磁感强度大小等于[ ](A) . (B) .(C) 0. (D).11-4在一平面内,有两条垂直交叉但相互绝缘的导线,流过每条导线的电流i的大小相等,其方向如图所示.问哪些区域中有某些点的磁感强度B可能为零[ ](A) 仅在象限Ⅰ. (B) 仅在象限Ⅱ.(C) 仅在象限Ⅰ,Ⅲ. (D) 仅在象限Ⅰ,Ⅳ.(E) 仅在象限Ⅱ,Ⅳ.11-5取一闭合积分回路L, 使三根载流导线穿过L所围成的面. 现改变三根导线之间的相互间隔, 但不越出积分回路, 则[ ](A) 回路L内的I不变, L上各点的B不变(B) 回路L内的I不变, L上各点的B改变(C) 回路L内的I改变, L上各点的B不变(D) 回路L内的I改变, L上各点的B改变11-6若某空间存在两无限长直载流导线, 空间的磁场就不存在简单的对称性. 此时该磁场的分布[ ](A) 可以直接用安培环路定理来计算;(B) 只能用安培环路定理来计算;(C) 只能用毕奥–萨伐尔定律来计算;(D) 可以用安培环路定理和磁场的叠加原理求出。11-7在匀强磁场中,取一半径为R的圆,圆面的法线与成60°角,如图所示,则通过以该圆周为边线的如图所示的任意曲面S的磁通量_._____________________.11-8有一半径为R的无限长圆柱形导体, 沿其轴线方向均匀地通过稳恒电流I,如图所示.距轴线为r ( r>R )处的磁感应强度大小为 .11-9中所示的一无限长直圆筒,沿圆周方向上的面电流密度(单位垂直长度上流过的电流)为i,则圆筒内部的磁感强度的大小为B = ,方向 .11-10如图所示,一无限长载流平板宽度为a,线电流密度(即沿x方向单位长度上的电流)为,求与平板共面且距平板一边为b的任意点P的磁感强度。11-11所示为两条穿过y轴且垂直于x-y平面的平行长直导线的正视图,两条导线皆通有电流I,但方向相反,它们到x轴的距离皆为a。 (1) 推导出x轴上P点处的磁感强度B的表达式。(2)求P点在x轴上何处时,该点的B取得最大值。11-12有一同轴电缆,其尺寸如图所示,它的内外两导体中的电流均为I,且在横截面上均匀分布,但二者电流的流向正相反,则(3)在r < R1处磁感强度大小为 .(4)在r > R3处磁感强度大小为_______________ .11-13一根很长的铜导线载有电流10A,设电流均匀分布.在导线内部作一平面,如图所示.试计算通过S平面的单位长度的磁通量(沿导线长度方向取长为1m的一段作计算).铜的磁导率.11-14如图,无限长直载流导线与正三角形载流线圈在同一平面内,若长直导线固定不动,则载流三角形线圈将 [ ](A) 向着长直导线平移. (B) 离开长直导线平移.(C) 转动. (D) 不动.11-15流I的长直导线在一平面内被弯成如图形状,放于垂直进入纸面的均匀磁场中,求整个导线所受的安培力(R为已知)。11-16横截面为矩形的环形螺线管,圆环内外半径分别为R1和R2,芯子材料的磁导率为,导线总匝数为N,绕得很密,若线圈通电流I,求:(1) 芯子中的B值和芯子截面的磁通量.(2) 在r < R1和r > R2处的B值.11-17 边长为=的正三角形线圈放在磁感应强度=1T 的均匀磁场中,线圈平面与磁场方向平行.如题9-21图所示,使线圈通以电流=10A,求:(1)线圈每边所受的安培力;(2)对轴的磁力矩大小;11-18有一长直导体圆管,内外半径分别为R1和R2,如图,它所载的电流I1均匀分布在其横截面上.导体旁边有一绝缘“无限长”直导线,载有电流I2,且在中部绕了一个半径为R的圆圈.设导体管的轴线与长直导线平行,相距为d,而且它们与导体圆圈共面,求圆心O点处的磁感强度.11-19一根同轴线由半径为R1的长导线和套在它外面的内半径为R2、外半径为R3的同轴导体圆筒组成.中间充满磁导率为的各向同性均匀非铁磁绝缘材料,如图.传导电流I沿导线向上流去,由圆筒向下流回,在它们的截面上电流都是均匀分布的.求同轴线内外的磁感强度大小B的分布。11-20面积为S,截面形状为矩形的直的金属条中通有电流I.金属条放在磁感强度为的匀强磁场中,的方向垂直于金属条的左、右侧面(如图所示).在图示情况下金属条的上侧面将积累_._ ____电荷,载流子所受的洛伦兹力fm =_ _.(注:金属中单位体积内载流子数为n )11-21图中的三条线表示三种不同磁介质的关系曲线,虚线是=关系的曲线,试指出哪一条是表示顺磁质哪一条是表示抗磁质哪一条是表示铁磁质11-22磁介质有三种,用相对磁导率r表征它们各自的特性时,(A) 顺磁质r >0,抗磁质r <0,铁磁质r >>1.(B) 顺磁质r >1.抗磁质r =1,铁磁质r >>1.(C) 顺磁质r >1.抗磁质r <1,铁磁质r >>1.(C) A比B的动量增量大 (D) A与B的动量增量相等提示:动量定理:合外力的冲量等于动量的增量。2-6如图所示,一质量为、速率为10的小球,以与竖直墙面法线成角的方向撞击在墙上,并以相同的速率和角度弹回。已知球与墙面的碰撞时间为。求在此碰撞时间内墙面受到的平均冲力。

（）等于 1 的物体称为白体。A. 吸收率；B. 反射率；C. 透过率；D. 折射率

第五章静电场基本要求理解电荷、电量、电荷的量子化、以及电荷守恒定律。掌握电场强度、电势等基本物理量;理解电场强度E是矢量点函数,而电势V则是标量点函数。理解高斯定理和静电场的环路定理是静电场的两个重要定理。熟练掌握计算电场强度、电势的各种方法:用点电荷电场强度计算式和叠加原理求任意带电体系的电场强度;用高斯定理求电场对称分布的带电体系的电场强度;用电场强度和电势的关系求解较简单带电系统的电场强度;用点电荷电势计算式和叠加原理求任意带电体系的电势;用点电荷电势定义式求电场对称分布的带电体系的电势。理解电场强度通量、电势能的概念,以及电场强度与电势的关系。习题5-1.两块金属平行板的面积为S,相距为d(d很小),分别带电荷+q和-q,两板间为真空,则两板间的作用力由下式计算:(A) ;(B) ;(C) ;(D) 。5-2.关于高斯定理有下面的几种说法,其中正确的是:(A)如果高斯面上E处处为零,则该面内必无电荷;(B)如果穿过高斯面的电场强度通量为零,则高斯面上各点的电场强度一定处处为零;(C)高斯面上各点的电场强度仅仅由面内所包围的电荷提供;(D)如果高斯面内有净电荷,则穿过高斯面的电场强度通量必不为零;(E)高斯定理仅适用于具有高度对称性的电场。5-3.在某电场区域内的电场线(实线)和等势面(虚线)如图8-3所示,由图判断出正确结论为:(A),;(B),; 图;(C),;(D),。5-4.如图8-4所示,两块无限大平板的电荷密度分别为和,写出下列各区域内的电场强度(不考虑边缘效应):Ⅰ区:E的大小为 ,方向为 ;Ⅱ区:E的大小为 ,方向为 ; 图;Ⅲ区:E的大小为 ,方向为。b代表 BH曲线关系;c代表 BH曲线关系.7-19.如图所示,几种载流导线在平面内分布,电流均为I,它们在O点的磁感应强度各为多少?作业7-20.如图所示,一个半径为R的无限长半圆柱面导体,沿长度方向的电流I在柱面上均匀分布。求半圆柱面轴线OO的磁感强度.7-21.如图所示,一宽度为b的薄金属板,其电流为I.试求在薄板的平面上,距板的一边为r的点P的磁感强度.7-2..如图,在磁感强度为B的均匀磁场中,有一半径为R的半球面,B与半面的轴线夹角为。求通过该半球面的磁通量。7-23.电流I均匀地流过半径为R的圆形长直导线,试计算单位长度导线通过图中所示剖面的磁通量。7-24.如图所示,一根长直导线载有电流I =30A,矩形回路载有电流I =20A。试计算每个边的受力以及整个回路所受合力。已知7-25.利用霍尔元件可以测量磁场的磁感强度,设一霍尔元件用金属材料制成,其厚度为0.15mm,载流子数密度为10m,将霍尔元件放入待测磁场中,测得霍尔电压为42V,测得电流为10m A。求此时待测磁场的磁感强度。7-26.已知地面上空某处地磁场的磁感强度B=0.410T,方向向北。若宇宙射线中有一速率ms的质子垂直地通过该处,求:(1)洛仑兹力的方向;(2)洛仑兹力的大小,并与该质子受到的万有引力比较。7.27.在螺线环的导线内通有电流20.,环上绕有线圈400匝,环的平均周长是0.4m,测得环内磁感应强度是1T。求(1)磁场强度;(2)磁介质的相对磁导率。7-28.如图所示,一根长直同轴电缆,内、外导体之间充满磁介质,磁介质的相对磁导率为(<1),导体的磁化可以忽略不计。电缆沿轴向有稳恒电流I通过,内外导体上电流的方向相反。求:(1)空间各区域内的磁感应强度和磁化强度;(2)磁介质表面的磁化电流。7-29.在实验室为了测定某种磁性材料的相对磁导率μ,常将这种材料做成截面为矩形的环形样品,然后用漆包线绕成一个螺绕环,设圆环的平均周长为0.10m,横截面积为0.510m,线圈匝数为200匝,当线圈通以0.10A的电流时测得穿过圆环横截面积的磁通量为6.010Wb,求此时该材料的相对磁导率μ。

s时的速度和加速度分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:一质点自原点开始沿抛物线=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))运动,它在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))轴上的分速度为一恒量,其值为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),求质点位于=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))处的速度和加速度。题解:因vx = =dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))为一常数,故ax = 0。当t = 0时,x = 0,由=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))积分可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)) (1)又由质点的抛物线方程,有=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)) (2)由y方向的运动方程可得该方向的速度和加速度分量分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)) (3)=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)) (4)当质点位于x = m时,由上述各式可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:质点在Oxy平面内运动,其运动方程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。求:(1)质点的轨迹方程;(2)在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))到=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时间内的平均速度;(2)=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时的速度及切向和法向加速度。题解:(1)由参数方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))消去t得质点的轨迹方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))s到=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))s时间内的平均速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(3)质点在任意时刻的速度和加速度分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))则t1 = s时的速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))切向和法向加速度分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:质点的运动方程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),试求:(1)初速度的大小和方向;(2)加速度的大小和方向。题解:(1)速度分量式为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))当t = 0时,vx= 10=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),vy = 15=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),则初速度大小为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))设v与x轴的夹角为,则=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)加速度的分量式为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))则加速度的大小为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))设a与x轴的夹角为,则=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:一质点具有恒定加速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,其速度为零,位置矢量=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。求:(1)在任意时刻的速度和位置矢量;(2)质点在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))平面上的轨迹方程,并画出轨迹的示意图。题解:由加速度定义式,根据初始条件t0 = 0时v0 = 0,积分可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))又由=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))及初始条件t = 0时,r = (10 m)i,积分可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))由上述结果可得质点运动方程的分量式,即=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))消去参数t,可得运动的轨迹方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))这是一个直线方程,直线斜率=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。轨迹如图所示。题:飞机以=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))的速度沿水平直线飞行,在离地面高为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,驾驶员要把物品投到前方某一地面目标处。问:(1)此时目标在飞机下方前多远(2)投放物品时,驾驶员看目标的视线和水平线成何角度(3)物品投出=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))后,它的法向加速度和切向加速度各为多少?题解:(1)取如图所示的坐标,物品下落时在水平和竖直方向的运动方程分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))飞机水平飞行速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),飞机离地面的高度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))m,由上述两式可得目标在飞机正下方前的距离=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)视线和水平线的夹角为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(3)在任意时刻物品的速度与水平轴的夹角为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))取自然坐标,物品在抛出2 s时,重力加速度的切向分量与法向分量分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:一足球运动员在正对球门前=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))处以=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))的初速率罚任意球,已知球门高为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。若要在垂直于球门的竖直平面内将足球直接踢进球门,问他应在与地面成什么角度的范围内踢出足球(足球可视为质点)题解:取图示坐标系Oxy,由运动方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))消去t得轨迹方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))以x = m,v = =dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))及  y  0代入后,可解得º   ºº   º如何理解上述角度得范围?在初速度一定的条件下,球击中球门底线或球门上缘都将对应有两个不同的投射倾角(如图所示)。如果以或<踢出足球,都将因射程不足而不能直接射入球门;由于球门高度的限制,角也并非能取与之间的任何值。当倾角取值为 < < 时,踢出的足球将越过门缘而离去,这时也球不能射入球门。因此可取的角度范围只能是解中的结果。题:设从某一点=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))以同样的速率,沿着同一竖直面内各个不同方向同时抛出几个物体。试证:在任意时刻,这几个物体总是散落在某个圆周上。题证:取物体抛出点为坐标原点,建立如图所示的坐标系。物体运动的参数方程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))消去式中参数,得任意时刻的轨迹方程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))这是一个以=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))为圆心、vt为半径的圆方程(如图所示),它代表着所有物体在任意时刻t的位置。题:一质点在半径为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))的圆周上以恒定的速率运动,质点由位置=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))运动到位置=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))所对的圆心角为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。(1)试证位置=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))之间的平均加速度为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1));(2)当=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))分别等于=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))、=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))、=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,平均加速度各为多少?并对结果加以讨论。=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题解:(1)由图可看到=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),故=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))而=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))所以=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)将=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))分别代入上式,得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))上述结果表明:当=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,匀速率圆周运动的平均加速度趋于一极限值,该值即为法向加速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。题:一质点沿半径为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))的圆周按规律=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))运动,=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))、=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))都是常量。(1)求=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时刻的总加速度;(2)=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))为何值时总加速度在数值上等于=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(3)当加速度达到=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,质点已沿圆周运行了多少圈?题解:(1)质点作圆周运动的速率为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))其加速度的切向分量和法向分量分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))故加速度的大小为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))其方向与切线之间的夹角为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)要使=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))由=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(3)从t = 0开始到t = v/b时,质点经过的路程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))因此质点运行的圈数为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:碟盘是一张表面覆盖一层信息记录物质的塑性圆片。若碟盘可读部分的内外半径分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。在回放时,碟盘被以恒定的线速度由内向外沿螺旋扫描线(阿基米德螺线)进行扫描。(1)若开始时读写碟盘的角速度为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),则读完时的角速度为多少(2)若螺旋线的间距为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),求扫描线的总长度和回放时间。题分析:阿基米德螺线是一等速的螺旋线,在极坐标下,它的参数方程可表示为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),式中r为极径,r为初始极径,为极角,a为常量。它的图线是等间距的,当间距为d时,常量a = d/2。因此,扫描线的总长度可通过积分=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))得到。

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