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2024年广东省初中学业水平模拟联考(三)物理试卷答案试卷答案

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2024年广东省初中学业水平模拟联考(三)物理试卷答案第一部分

到P和到Q的两段绳都是水平的,已知Q与P之间以及P与桌面之间的动摩擦因数都力1=0.4,物块P的重力为40N,物块Q的重力为20N.滑轮的质量,滑轮轴上的摩擦都不计。若用一水平向右的力F拉P使它做匀速运动,求:(1)物块Q所受的滑动摩擦力的大小和方向;(2)拉动物块P所需的力F的大小。

2024年广东省初中学业水平模拟联考(三)物理试卷答案第二部分（待更新）

如图甲15.(18分)如图甲所示,两根足够长、电阻不计且相距L=1m的平行金属导轨固定在倾角为L=1mθ=370-37的绝缘斜面上(斜面未画出),两导轨间有一磁感应强度大小B=1T、方向垂直斜面B=1T、m=1kg向上的匀强磁场,现将两根质量均为m=1kg、电阻均为R=10、长度均为L=1m的金属R=1ΩL=1m棒放置在导顶端附近,两金属棒与导轨接触良好,金属棒ab与导轨间的摩擦忽略不计,金属棒cd与导轨间的动摩擦因数为μ=0.75.在t=0时刻将金属棒ab由静止释放,此时金属=0.75.棒cd锁定在导轨上;t=t时刻将金属棒cd由静止释放,已知金属棒ab中的电流随时间变t=t1化的关系如图乙所示,重力加速度g取10m/s²,sin37°=0.6,cos37°=0.8,求10m/s^2,37^=0.6,37^=0.8,(1)t=t1(1)t=t时刻,金属棒ab的加速度大小a:及金属棒cd的加速度大小a2;a1a2;(20t1(2)若在0~t时间内,金属棒ab沿着斜面下滑的距离为x1=0.5m,求这段时间内金属棒x1=0.5m,ab产生的焦耳热Q;Q;t=t2(3)若t=t2时,金属棒ab的速度为v2=9m/s,求t=t2+1s时,金属棒cd的速度v.v2=9m/s,t=t2+1si/AI0.5t1t2乙甲