In this lecture, we will particularize the conservation principles presented in the previous lecture to the case in which the system of particles considered is a 2D rigid body. Mass Moment of Inertia In the previous lecture, we established that the angular momentum of a system of particles relative to the center of mass, G, was

fz=f(x,y) 积为已知的立体的体积 y=y Dda (,y) D是矩形区域[a,b;cd Q(y)= fxyd I= avd, y d y (y) D b 问题:Q(y)是什么图形?是曲边梯形

第十二章祖国完全统一的构想习题答案 一、单项选择题 1b2B3c4b5a6c7d8c910c11c12d13a14B15C16D1718

第八章建设中国特色社会主义经济练习参考答案 编制整理刘荣材 一、单选题(共40道题) 1.A;2.a3.c4.d5.b6.c7.a8.c9.c;10.a11.c12.d13.b:14.a:15.d:16

In this lecture we will look at some applications of Newton's second law, expressed in the different coordinate systems that were introduced in lectures D3-D5. Recall that Newton's second law F=ma, (1) is a vector equation which is valid for inertial observers. In general, we will be interested in determining the motion of a particle given

一、定积分计算 1.设f(x)=,edx,求xf(x)d 2.设A=,试用表示:(1)B= 1t-a-1 (2 (1+t) 3.设feC,,证明:f(d=(x-x2)f(x)d 4.计算定积分xln(1+e)dx 二、定积分应用 1.设有曲线族y=kx2(k>0),对于每个正数k(k2),曲线y=kx2 与曲线y=sinx(0≤xs)交于唯一的一点(t,sint)(其中t=t()) 用S1表示曲线y=kx2与曲线y=sinx(0≤x≤)围成的区域的面积; S2表示曲线y=sinx,y=sint与x=围成的区域的面积求证在上述 曲线族中存在唯一的一条曲线L,使得S1+S2达到最小值 2.点A(3,1,-1)是闭曲面S1:x2+y2+z2-2x-6y+4z=10

Ⅰ、选择题(共10小题,每小题3分,共30分) 每题给出四个答案,其中只有一个是正确的,请将正确答案的标号(A或B或C或D) 写在题号前的横线上 1、积分∫es(-3)d等于 (A)0(B)1(C)e3OD)e3 2、下列等式不成立的是 (A)f0)6\(0)=f(0)6'(0 B)A06(0=f()5( (C)f)*6(0)=f(0 (D)f0)°(0)=f(0

1.以下程序的输出结果是 mair i int a=5, b=4, c=6, d printf(\%d\\n\, d=a>b? (a>c?a: c):(b)) 2.在C语言中,以下哪个函数用于将字符输出到标|two

一、选择题 1.烫发效果直接取决于对烫发工具和的选择 C A.长度深度 B.颜色长度 C.大小形状 D.深度颜色 2.为什么在烫发中需要烫发纸? D A.做好铺垫并保护易断裂的头发 B.更好的控制较短的头发 C.使整个卷很光滑 D.以上都是

第2章线性表 一.选择题 1.a2.b3.c4.a.d6.d7.d8.c9.b10.bc11.i11.2i11.3e