12.已知一质点的质量 =1kg, 其运动的位置矢量为-|||-overrightarrow (r)=-dfrac (6)(pi )(sin (dfrac (pi )(2)t)overrightarrow (i)+cos (dfrac (pi )(2)t)overrightarrow (j)) (SI制)-|||-试求:(1)第4秒时,质点的动量:(2)前4秒内,质点受到合力的冲量;③3(2)的计算,是否-|||-说明在(2)所指的过程中,质点的动量是守恒的?

=-sqrt (2)(sin dfrac (pi )(6)+icos dfrac (pi )(6))的指数表达式为（ ）。A =-sqr

[例2](1)(2020·全国卷Ⅲ)已知 sin theta +sin (theta +dfrac (pi )(3))=1,-|||-则 sin (theta

一质点在平面上运动,已知质点位置矢量的表示式为overrightarrow (r)=a(t)^2overrightarrow (i)+b(t)^2overrig

dfrac ({mu )_(0)I}(pi R)(dfrac (1)(2)+dfrac (pi )(6))-|||-dfrac ({mu )_(0)I}(pi

一质点在平面上运动，已知质点位置矢量的表达式为overrightarrow (r)=a(t)^2overrightarrow (i)+b(t)^2overrig

2.3.7 一质量为m的质点在xOy平面上运-|||-动,其位置矢量为-|||-r=acoswti+bsin wtj,-|||-求 t=0 到 =dfrac (

已知 omega gt dfrac (1)(4), 函数-|||-12. 已知 omega gt dfrac (1)(4), 函数-|||-(x)=sin (o

πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T

+sin dfrac (npi )(2)+... （ ）A 、是收敛的 ; B 、是收敛的，其和为sin dfrac (pi )(2)+sin dfrac (2

1-4 已知质点的运动方程-|||-=sqrt (3)cos dfrac (pi )(4)t, =sin dfrac (pi )(4)t-|||-(式中,x、y