题目
题型:不详难度:来源:
(2)c、d两点间的距离L2=?
(3)外力F2的大小?
(4)金属棒从位置(I)运动到位置(Ⅲ)的过程中,电阻R上放出的热量Q=?
答案
E1=BL1 v1,,F安1···································①
F安1="4" N·································································②
根据牛顿第二定律得
F安1-F1 =ma·····························································③
a=" 1" m / s2·································································④
(2)设金属棒在位置(Ⅱ)时速度为v2,由运动学规律得
=-2a s1···························································⑤
v2=" 1" m / s·································································⑥
由于在(I)和(II)之间做匀减速直线运动,即加速度大小保持不变,外力F1恒定,所以AB棒受到的安培力不变即F安1=F安2
···························································⑦
m······················································⑧
(3)金属棒从位置(Ⅱ)到位置(Ⅲ)的过程中,做匀速直线运动,感应电动势大小与位置(Ⅱ)时的感应电动势大小相等,安培力与位置(Ⅱ)时的安培力大小相等,所以
F2= F安2="4" N······························································⑨
(4) 设位置(II)和(Ⅲ)之间的距离为s2,则
s2= v2t="2" m ································································⑩
设从位置(I)到位置(Ⅱ)的过程中,外力做功为W1,从位置(Ⅱ)到位置(Ⅲ)的过程中,外力做功为W2,则
W1= F1 s1="22.5" J ···························································11
W2= F2 s2="8" J······························································12
根据能量守恒得W1+ W2·······························13·
解得Q =" 38" J ·····························································14
解析
核心考点
试题【如图所示,两根不计电阻的金属导线MN与PQ放在水平面内,MN是直导线,PQ的PQ1段是直导线,Q1Q2段是弧形导线,Q2Q3段是直导线,MN、PQ1、Q2Q3相】;主要考察你对电磁感应等知识点的理解。[详细]
举一反三
均为 D=0.8m,轨道左端用阻值R=0.4Ω的电阻相连.水平导轨的某处有一竖直向上、磁感应强度B=0.06T的匀强磁场.光滑金属杆ab质量m=0.2kg、电阻r=0.1Ω,当它以5m/s的初速度沿水平导轨从左端冲入磁场后恰好能到达竖直半圆轨道的最高点P、Q.设金属杆ab与轨道接触良好,并始终与导轨垂直,导轨电阻忽略不计.取g=10m/s2,求金属杆:
(1)刚进入磁场时,通过金属杆的电流大小和方向;
(2)到达P、Q时的速度大小;
(3)冲入磁场至到达P、Q点的过程中,电路中产生的焦耳热.
A.S断开时,金属板沿斜面下滑的加速度 |
B. |
C.电阻R上产生的热量 |
D.CC′一定在AA′的上方 |
A.刚闭合S的瞬间,导体棒中的电流增大 |
B.刚闭合S的瞬问,导体棒两端电压增大 |
C.闭含S后,导体棒做减速运动直到停止 |
D.闭合S后,导体棒做减速运动直到再一次匀速运动 |
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