已知,交流电压u1=10sin100πtV,u2=10sin(100πt-60°)V。则u=u1-u2的解析式为()。
A: u=10sin(100πt+60°)V
B: u=10sin(100πt-60°)V
C: u=8sin(100πt+30°)V
D: u=8sin(100πt-30°)V
A: u=10sin(100πt+60°)V
B: u=10sin(100πt-60°)V
C: u=8sin(100πt+30°)V
D: u=8sin(100πt-30°)V
举一反三
- 已知,交流电压U1=10sin100πtV,U2=10sin(100πt-60°)V,则u=u1-u2的解析式为() A: U=10sin(100πt+60°)V B: U=10sin(100πt-60°)V C: U=8sin(100πt+30°)V D: U=8sin(100πt-30°)V
- 已知,交流电压U1=10sin100πtV,U2=10sin(100πt-60°)V,则u=u1-u2的解析式为()
- 已知某正弦电压的频率f = 50Hz,初相角j=30°,在 t = 0.02 s时瞬时值u(0.02) = 100 V,则其瞬时值表达式可为( )。 A: u = 100 sin( 50t+30° ) V B: u = 141.4 sin( 50pt+30° ) V C: u = 200 sin( 100pt+30° ) V D: u = 282.8 sin( 100pt+30° ) V E: u = 282.8 sin( 50pt+30° ) V
- 已知某正弦电压的频率f =50 Hz ,初相角j=60°,有效值为50 V,则其瞬时表达式为 A: u = 100 sin( 50t+60°) V B: u = 70.7sin( 50pt+60°) V C: u = 200 sin( 100pt+60°) V D: u = 70.7 sin( 100pt+60°) V
- 画出下列扇形区域。[img=541x528]18030733b119884.png[/img] A: ParametricPlot[{v Cos[u],v Sin[u]},{u,0,3Pi/2},{v,1/2,1}] B: ParametricPlot[{v Cos[u],v Sin[u]},{v,1/2,1},{u,0,3Pi/2}] C: ParametricPlot[{v Cos[u],v Sin[u]},{u,-Pi/2,0},{v,1/2,1}] D: ParametricPlot[{v Cos[u],v Sin[u]},{v,1/2,1},{u,-Pi/2,0}]