发布网友 发布时间:2024-10-24 02:42
共1个回答
热心网友 时间:2024-11-25 22:00
(1)由图示游标卡尺可知,主尺示数为2.9cm=29mm,游标尺示数为8×0.1mm=0.8mm,游标卡尺示数为29mm+0.8mm=29.8mm.
(2)①将小量程的电流表改装成电压表,电流表需要知道两个参数:量程和内阻,故电流表选A 2 .串联电阻阻值R= U- I g r g I g = 3 500×1 0 -6 Ω-1000Ω=5000Ω,定值电阻应选R 2 .
② R X R A1 =3.8~4.2, R V R X ≈31.6~28.6, R V R X > R X R A1 ,电流表应采用外接法,实验电路应选甲.
③电压表示数U=I 2 R= 500×1 0 -6 30 ×12×6000=1.2V,待测电阻阻值R X = U I 1 - U R =200.0Ω.
(3)计数点间的时间间隔t=0.02×5=0.1s,根据中点时刻的速度等于平均速度得:
v B = x AC 2t = 0.081-0.051 2×0.1 m/s=0.15m/s
v E = x DF 2t = 0.171-0.105 2×0.1 m/s=0.33m/s,
打纸带上B点到E点过程中小车重力势能的减少量:E P =mgx BE =0.072J;根据加速度的定义式得:
a= △v △t = v E - v B 3t = 0.33-0.15 3×0.1 =0.6m/s 2 ,根据牛顿第二定律得:mgsinθ-f=ma,
解得:f= 1 4 mg-0.6m=0.76N
所以克服阻力所做的功:W=-W f =0.76×(0.135-0.063)J=0.055J.
故答案为:(1)29.8;(2)①A 2 ,R 2 ;②甲;③1.20;,200.0;(3)0.15,0.072,0.055.