[例题 -5 ]机构如图 7-9 所示,已知: =dfrac (4h)(9) ,=dfrac (sqrt {3)h}(3) ,轮E作纯滚-|||-动;在图示位置, varphi =(60)^circ ,bot OC ,杆AB的速度为v。试求此瞬时:(1)杆-|||-OC的角速度;(2)轮E的角速度。-|||-C、-|||-B 乙 A v-|||-30° 7777-|||-ve woc-|||-F-|||-UF-|||-WE-|||-C-|||-B A v-|||-777-|||-F-|||-R-|||-φ-|||-E-|||-77.-|||-UE 30° __ φ-|||-E-|||-p πT

^circ -dfrac (4)(2)-|||-:^circ +dfrac ({varphi )^1}(2) D. ^circ -dfrac ({varphi

11设 9.ipg-|||-OA dfrac (5)(9)-|||-CB dfrac (1)(9)-|||-oc dfrac (4)(9)-|||-dfrac

dfrac (sqrt {3)}(2)sin (15)^circ +dfrac (1)(2)cos (15)^circ 的值为( )A．dfrac

已知氢原子的 (varphi )_(2{p)_(2)}=dfrac (1)(4sqrt {2pi {{a)_(0)}^3}}(dfrac (r)({a)_(0)

2.求曲线 ^dfrac (2{3)}+(y)^dfrac (2{3)}=(a)^dfrac (2{3)} 在点 (dfrac (sqrt {2)}(4)a,d

2.求曲线 ^dfrac (2{3)}+(y)^dfrac (2{3)}=(a)^dfrac (2{3)} 在点 (dfrac (sqrt {2)}(4)a,d

[题目]-|||-求曲线 ^dfrac (2{3)}+(y)^dfrac (2{3)}=(a)^dfrac (2{3)} 在点 (dfrac (sqrt {2)

[题目]求椭圆 dfrac ({x)^2}(4)+dfrac ({y)^2}(9)=1 在点 (1,dfrac (3sqrt {3)}(2)) 处的法线方-||

19、单选 lim _-|||-A 1-|||-B) dfrac (4)(9)-|||-C dfrac (2)(3)-|||-D dfrac (9)(4)

设函数 (x)=dfrac (4sqrt {3)}(9)arctan (dfrac (2x+1)(sqrt {3)}) ,其反函数为φ(y),求φ(y).。。