已知正弦电压的振幅为10V,周期为100ms，初相位π/6为。试写出正弦量的函数表达式
未知类型：{'options': ['10sin(62.8t+[img=20x15]17e436158120103.jpg[/img])', ' 10[img=21x20]17e43630963ca84.jpg[/img]sin(62.8t+[img=20x15]17e436158120103.jpg[/img])', ' 10[img=21x20]17e43630963ca84.jpg[/img]sin(62.8t+[img=20x15]17e43aa0991ff38.jpg[/img])', ' 10cos(62.8t+[img=20x15]17e436158120103.jpg[/img])'], 'type': 102}
未知类型：{'options': ['10sin(62.8t+[img=20x15]17e436158120103.jpg[/img])', ' 10[img=21x20]17e43630963ca84.jpg[/img]sin(62.8t+[img=20x15]17e436158120103.jpg[/img])', ' 10[img=21x20]17e43630963ca84.jpg[/img]sin(62.8t+[img=20x15]17e43aa0991ff38.jpg[/img])', ' 10cos(62.8t+[img=20x15]17e436158120103.jpg[/img])'], 'type': 102}
举一反三
- 用戴维南定理求（a）中的电流向量I2[img=960x518]17e43aa0a588832.jpg[/img] 未知类型：{'options': ['2[img=21x20]17e43630963ca84.jpg[/img]∠-[img=20x15]17e43aa0ae535b8.jpg[/img]', ' 2.5[img=21x20]17e43630963ca84.jpg[/img]∠-[img=20x15]17e43aa0ae535b8.jpg[/img]', ' 2∠-[img=20x15]17e43aa0ae535b8.jpg[/img]', ' 2.5[img=21x20]17e43630963ca84.jpg[/img]∠[img=20x15]17e43aa0ae535b8.jpg[/img]'], 'type': 102}
- 正弦交流电流的频率为50Hz，有效值是1A，初相角90°，其瞬时值表达式为（）。 未知类型：{'options': ['i(t)=sin(314t+90°)A', ' i(t)=[img=21x20]17e43630963ca84.jpg[/img]sin(314t+90°)A', ' i(t)=cos(314t+90°)A', ' i(t)=[img=21x20]17e43630963ca84.jpg[/img]cos(314t+90°)A'], 'type': 102}
- 求不定积分[img=132x48]17da6537fc8dad6.png[/img]； （ ） A: -(4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) B: (4*(sin(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) C: (4*(cos(x/2)/2 + 2*sin(x/2)))/(17*exp(2*x)) D: (4*(cos(x/2)/2 + 2*cos(x/2)))/(17*exp(2*x))
- 已知[img=138x19]17e43f8966ee7a8.jpg[/img]A，[img=150x19]17e43f896f977b9.jpg[/img]A,则（ ）。 未知类型：{'options': ['17e43efebd80a36.jpg超前[img=10x17]17e438b794b2283.jpg[/img][img=20x15]17e43aa0991ff38.jpg[/img]', ' [img=10x17]17e43efebd80a36.jpg[/img]滞后[img=10x17]17e438b794b2283.jpg[/img][img=20x15]17e43aa0991ff38.jpg[/img]', ' 相位无法判断'], 'type': 102}
- 求微分方程[img=143x21]17da5f14490e50e.png[/img]的通解,实验命令为(). A: dsolve(D2y-2*Dy+5*y=sin(2*x),x)ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x) B: dsolve('D2y-2*Dy+5*y=sin(2*x)','x')ans =cos(2*x)*(sin(4*x)/17 - cos(4*x)/68 + 1/4) - sin(2*x)*(cos(4*x)/17 + sin(4*x)/68) + C1*cos(2*x)*exp(x) - C2*sin(2*x)*exp(x) C: dsolve(D2y-2*Dy+5*y=sin(2*x),'x','y')ans =exp(x)*sin(2*x)*C2+exp(x)*cos(2*x)*C1+1/17*sin(2*x)+4/17*cos(2*x)