1.单选题- (共8题)
A.楞次经过严密实验与逻辑推导,最终确认了电磁感应的产生条件:闭合线圈中磁通量变化,并找到了感应电流的方向的判断规律 |
B.安培发现了电流的磁效应并提出了分子电流假说 |
C.奥斯特发现了电流的磁效应并提出了分子电流假说 |
D.“闭合电路中感应电动势的大小,跟穿过这一电路的磁通量变化率成正比”,这是法拉第在对理论和实验资料严格分析后得出的法拉第电磁感应定律的内容 |

A.感应电流的方向始终是P→Q |
B.感应电流的方向先是由P→Q,后是由Q→P |
C.PQ受磁场力的方向垂直于杆向左 |
D.PQ受磁场力的方向先垂直于杆向右,后垂直于杆向左 |

A.铜盘转动将变慢 |
B.铜盘转动将变快 |
C.铜盘仍以原来的转速转动 |
D.铜盘的转动速度如何变化要由磁铁的上、下两端的极性来决定 |


A.导线框中的感应电动势为0.5V |
B.t=0.5s时刻,导线框中感应电流为零 |
C.导线框中产生俯视逆时针方向的感应电流 |
D.导线框一直静止在绝缘水平面上 |

A.![]() | B.mg | C.![]() | D.![]() |

A.![]() | B.![]() | C.![]() | D.![]() |

其中特高压输电是指输电电压为1000kV及以上的输电方式.由表中数据可以看出,当输送功率较大时,若输电电压较低则需要多组输电线路同时输送才可能完成.现欲将4400MW电功率输送相同距离,则用500 kV电压输电和1 000 kV电压输电,损失的电功率之比约为(若两种线路电缆材料相同,横截面积之比为1∶6)( )
A.24∶5 | B.24∶1 | C.5∶3 | D.2∶1 |

A.1∶1 | B.2∶1 | C.3∶1 | D.4∶1 |
2.选择题- (共3题)
阅读理解
A team of engineers at Harvard University has been inspired by Nature to create the first robotic fly. The mechanical fly has become a platform for a series of new high-tech integrated systems. Designed to do what a fly does naturally, the tiny machine is the size of a fat housefly. Its mini wings allow it to stay in the air and perform controlled flight tasks.
“It's extremely important for us to think about this as a whole system and not just the sum of a bunch of individual components (元件),” said Robert Wood, the Harvard engineering professor who has been working on the robotic fly project for over a decade. A few years ago, his team got the go-ahead to start piecing together the components. “The added difficulty with a project like this is that actually none of those components are off the shelf and so we have to develop them all on our own,” he said.
They engineered a series of systems to start and drive the robotic fly. “The seemingly simple system which just moves the wings has a number of interdependencies on the individual components, each of which individually has to perform well, but then has to be matched well to everything it's connected to,” said Wood. The flight device was built into a set of power, computation, sensing and control systems. Wood says the success of the project proves that the flying robot with these tiny components can be built and manufactured.
While this first robotic flyer is linked to a small, off-board power source, the goal is eventually to equip it with a built-in power source, so that it might someday perform data-gathering work at rescue sites, in farmers' fields or on the battlefield. “Basically it should be able to take off, land and fly around,” he said.
Wood says the design offers a new way to study flight mechanics and control at insect-scale. Yet, the power, sensing and computation technologies on board could have much broader applications. “You can start thinking about using them to answer open scientific questions, you know, to study biology in ways that would be difficult with the animals, but using these robots instead,” he said. “So there are a lot of technologies and open interesting scientific questions that are really what drives us on a day to day basis.”
2003春节晚会中,歌曲《让爱住我家》中唱道:“我爱我的家,弟弟爸爸妈妈,……让爱天天住你家,让爱天天住我家,不分日夜,秋冬春夏,全心全意爱我们的家。”可见作者认为家庭中最重要的是( )
3.多选题- (共6题)

A.磁感应强度B越大,上、下表面的电势差U越大 |
B.k越大,传感器灵敏度(![]() |
C.若图中霍尔元件是电子导电,则下板电势高 |
D.电流越大,上、下表面的电势差U越小 |

A.金属棒ab最终可能匀速下滑 |
B.金属棒ab一直加速下滑 |
C.金属棒ab下滑过程中M板电势高于N板电势 |
D.带电微粒不可能先向N板运动后向M板运动 |

A.电压表记录的电压为5 mV |
B.电压表记录的电压为9 mV |
C.河南岸的电势较高 |
D.河北岸的电势较高 |


A.通过R的电流方向为由外向内 |
B.通过R的电流方向为由内向外 |
C.R上产生的热量为![]() |
D.流过R的电荷量为![]() |

A.t=0时线圈的感应电动势为零 |
B.转过90°时线圈的感应电动势为零 |
C.转过90°的过程中线圈中的平均感应电动势为![]() |
D.转过90°的过程中线圈中的平均感应电动势为![]() |

A.原、副线圈中的功率之比为11:1 |
B.若滑动变阻器接入电路的阻值为10Ω,则1min内滑动变阻器产生的热量为1200 J |
C.若只将S从1拨到2,电流表示数减小 |
D.若只将滑动变阻器的滑片向下滑动,则两电表示数均减少 |
4.解答题- (共4题)


(1)发电机的输出电压为多少?
(2)变压器原副线圈的匝数比为多少?
(3)与变压器原线圈串联的交流电流表的示数为多少?

(1)此时通过电阻R上的电流;
(2)这一过程通过电阻R上的电荷量q;
(3)此时作用于导体棒上的外力F的大小.

(1)导线框刚下落时,bc边距磁场上边界MN的高度h;
(2)导线框离开磁场的过程中,通过导线框某一横截面的电荷量q;
(3)导线框穿越磁场的整个过程中,导线框中产生的热量Q.

(1)a匀速转动的方向;
(2)P、Q间电场强度E的大小;
(3)小球通过N点时对细线拉力FT的大小.
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【1】题量占比
单选题:(8道)
选择题:(3道)
多选题:(6道)
解答题:(4道)
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【2】:难度分析
1星难题:0
2星难题:0
3星难题:0
4星难题:4
5星难题:0
6星难题:5
7星难题:0
8星难题:7
9星难题:2