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2026 年初中毕业班(九年级)练习
物理(六) 参考答案
一、选择题(本大题共10小题,每小题2分,共20分)
1-5CABCB 6-10DADCD
二、非选择题(本大题共10小题,共40分)
11.(2分)流体中流速越大的地方,压强越小 静止
12.(2分)热传递 引力
13.(2分)相同路程比时间 8
14.(2分)2 2×10-4
15.(2分)二次 内
16.(5分)
(1)秒表、天平 a
(2)S 增强 变小
17.(6分)
(1)匀速直线 二力平衡(或平衡力)
(2)甲、乙 接触面粗糙程度
(3)没有控制压力一定
(4)不一定
18.(7分)
(1)如下图所示:
(2)B
(3)0.30(或0.3) 8.3
(4)不能
U R
【拓展】将滑动变阻器的滑片P移至a端 1 0
U U
2 1
19.(6分)
解:(1)作业本纸张密度:
m 2.25kg
ρ= = =0.5×103kg/m3·······························································2分
V 4.5103m3
物理练习(六) 参考答案 第 1 页 共 2 页(2)作业本的总重力:
G=mg=2.25kg×10N/kg=22.5N·········································································· 1分
1
作业本对桌面的压力等于自身的重力,即F=G=22.5N
1 1
作业本对桌面的压强为:
F 22.5N
p= 1 = =500Pa··············································································1分
1 S 4.5102m2
(3)此时桌面受到的压力:
F=pS=1500Pa×4.5×10-2m2=67.5N····································································1分
2 2
这盒器材的重力:
G=F-G=67.5N-22.5N=45N··········································································1分
2 2 1
20.(6分)
解:(1)当开关S、S 闭合,S、S 断开时,L和R 串联
2 1 3 1
电源电压为:U=IR=I(R+R)=0.4A×(6Ω+24Ω)=12V····························2分
1 L
(2)当所有开关都闭合,滑动变阻器的滑片移到最右端时,L和R 并联
2
U 12V
且小灯泡正常发光,流过L的电流为I = = =0.5A
L R 24Ω
L
U 12V
流过R 的电流为I= = =1.2A···································································1分
2 2 R 10Ω
2
电流表的示数I'=I+I=0.5A+1.2A=1.7A·····························································1分
L 2
(3)开关S、S、S 闭合,S 断开,小灯泡换为电压表时,R 和R 串联,电流表测通过电路的电流
2 3 1 1 2
1
题中要求的电流至少达到电流表测量范围的 ,电流表的测量范围为0~3A
3
1
则电路中的最小电流I = ×3A=1A
小 3
U 12V
电路中的最大电阻R'= = =12Ω
I 1A
小
滑动变阻器连入电路中的阻值R'=R'-R=12Ω-6Ω=6Ω······································1分
2 1
而此时R 两端的电压为U=I R'=1A×6Ω=6V>3V
2 2 小 2
又因为电压表测量范围为0~3V,所以当电压表满偏时R 连入电路的阻值最大
2
此时R 两端电压为U=U-U'=12V-3V=9V
1 1 2
U 9V
此时电路中的最小电流为I = 1 = =1.5A
最小 R 6Ω
1
U ' 3V
滑动变阻器R 连入电路的最大阻值R = 2 = =2Ω······································1分
2 2大 I 1.5A
最小
物理练习(六) 参考答案 第 2 页 共 2 页