CRH2型动车组车内信息显示器输入电压为() A: DC100×(l士10%)V B: DC11O×(l士10%)V C: DC100×(l土5%)V D: DCI10×(l士5%)V
CRH2型动车组车内信息显示器输入电压为() A: DC100×(l士10%)V B: DC11O×(l士10%)V C: DC100×(l土5%)V D: DCI10×(l士5%)V
砂轮 1—300×50×75—A 60 L 5 V—35m/s,表示磨料代号的是( )。 A: A B: 60 C: L D: V
砂轮 1—300×50×75—A 60 L 5 V—35m/s,表示磨料代号的是( )。 A: A B: 60 C: L D: V
【单选题】V g 和V l 分别为饱和蒸汽和饱和液体的体积,x为湿蒸汽干度,则湿蒸汽的体积V为() A. V= x(V g - V l ) + V g B. V= (V g - V l ) + x V l C. V= x(V g - V l ) +V l D. V=(1-x)V g + x V l
【单选题】V g 和V l 分别为饱和蒸汽和饱和液体的体积,x为湿蒸汽干度,则湿蒸汽的体积V为() A. V= x(V g - V l ) + V g B. V= (V g - V l ) + x V l C. V= x(V g - V l ) +V l D. V=(1-x)V g + x V l
船舶对水航速 V L ,对地航速 V G ,船速 V E ,如果 V L <V E ,而且 V G >V L ,则船舶航行在何种情况下。
船舶对水航速 V L ,对地航速 V G ,船速 V E ,如果 V L <V E ,而且 V G >V L ,则船舶航行在何种情况下。
对于饱和蒸汽进料,则有L’( )L ,V’( )V
对于饱和蒸汽进料,则有L’( )L ,V’( )V
精馏过程若为饱和液体进料,则()。 A: q = 1,L = L′ B: q = 1,V = V′ C: q = 1,L = V′ D: q = 1,L = V'
精馏过程若为饱和液体进料,则()。 A: q = 1,L = L′ B: q = 1,V = V′ C: q = 1,L = V′ D: q = 1,L = V'
A signal s(t)=3cos(20pt) (V) is quantized by a uniform quantizer, the step size of the quantizer is 0.1(V), then the minimal number of quantization level is L , the number of bits per sample is at least R , and the variance of the quantization error is T ([img=22x22]1803a28ec2869ac.png[/img]). A: L=60, R=5, T=0.000833 B: L=30, R=5, T=0.000833 C: L=60, R=5, T=0.001667 D: L=60, R=6, T=0.000833
A signal s(t)=3cos(20pt) (V) is quantized by a uniform quantizer, the step size of the quantizer is 0.1(V), then the minimal number of quantization level is L , the number of bits per sample is at least R , and the variance of the quantization error is T ([img=22x22]1803a28ec2869ac.png[/img]). A: L=60, R=5, T=0.000833 B: L=30, R=5, T=0.000833 C: L=60, R=5, T=0.001667 D: L=60, R=6, T=0.000833
A signal s(t)=3cos(20pt) (V) is quantized by a uniform quantizer, the step size of the quantizer is 0.1(V), then the minimal number of quantization level is L , the number of bits per sample is at least R , and the variance of the quantization error is T ([img=22x22]1802e166a76be1c.png[/img]). A: L=60, R=5, T=0.000833 B: L=30, R=5, T=0.000833 C: L=60, R=5, T=0.001667 D: L=60, R=6, T=0.000833
A signal s(t)=3cos(20pt) (V) is quantized by a uniform quantizer, the step size of the quantizer is 0.1(V), then the minimal number of quantization level is L , the number of bits per sample is at least R , and the variance of the quantization error is T ([img=22x22]1802e166a76be1c.png[/img]). A: L=60, R=5, T=0.000833 B: L=30, R=5, T=0.000833 C: L=60, R=5, T=0.001667 D: L=60, R=6, T=0.000833
精馏过程的操作线方程的斜率是()? V/R|V/L|L/V|R/V
精馏过程的操作线方程的斜率是()? V/R|V/L|L/V|R/V
某精馏塔操作时,若保持F、xF、q、D不变,增大回流比R,则L’、V’的变化趋势是( ) A: L’增大,V’减小 B: L’增大,V’增大 C: L’减小,V’减小 D: L’减小,V’增大
某精馏塔操作时,若保持F、xF、q、D不变,增大回流比R,则L’、V’的变化趋势是( ) A: L’增大,V’减小 B: L’增大,V’增大 C: L’减小,V’减小 D: L’减小,V’增大