文档内容
2026 年初中毕业班(九年级)练习
物理(五) 参考答案
一、选择题(本大题共10小题,每小题2分,共20分)
1-5CDABD 6-10BACCD
二、非选择题(本大题共10小题,共40分)
11.(2分)大 竖直向下
12.(2分)物体间力的作用是相互的 运动状态
13.(2分)增大 不变
14.(2分)升华 吸收
15.(2分)电 1.8×104
16.(5分)
(1)同一高度 缩小 左
(2)同种电荷相互排斥 绝缘体
17.(6分)
(1)标尺左端的零刻度线处 右
(2)27.0(或27)
(3)2.7×103
V
【拓展】 2 ρ 不变
V 水
1
18.(7分)
(1)如右图所示
(2)最大阻值(或最左端)
(3)滑动变阻器断路
(4)5
(5)保护电路
【拓展】电阻箱R和滑动变阻器 R-R
1 2
19.(6分)
解:(1)2600t=2.6×106kg
物理练习(五) 参考答案 第 1 页 共 2 页该盾构机主机的重力:
G=mg=2.6×106kg×10N/kg=2.6×107N························································2分
(2)该管片承受的压力:
F =pS=2×108Pa×5m2=1×109N····································································2分
压
(3)最大推进力做的功:
W=Fs=4.5×107N×0.3m=1.35×107J································································1分
最大推进力做功的功率为:
W 1.35107J
P= = =1.35×105W·····································································1分
t 100s
20. (6分)
U2
解:(1)根据P= 可知,只闭合开关S 时,打印笔处于低温挡,电路中的电流:
1
R
U 5V
I= = =1A···················································································1分
1 R 5Ω
1
(2)当开关S、S 均闭合,R 和R 并联时,打印笔处于高温挡
1 2 1 2
因为高温挡和低温挡的功率之比为3∶2
低温挡的功率为:P =UI=5V×1A=5W,P =P=5W······································1分
低 1 低 1
P (P P) (5WP) 3
即 高 = 1 2 = 2 =
P P 5W 2
低 1
所以R 的功率:P=2.5W··············································································· 1分
2 2
高温挡时通电时间t=1min=60s
R 产生的热量:Q=W=Pt=2.5W×60s=150J·····················································1分
2 2
(3)高温挡原来的功率P =P+P=7.5W································································1分
高 1 2
只要求将高温挡的功率提升20%,则改进后高温挡功率P '=P (1+20%)=7.5W×1.2=9W
高 高
由于只提高高温挡功率,则R 的阻值不变,改变R 的阻值
1 2
改变后,R 的功率:P'=P '-P =9W-5W=4W
2 2 高 低
U2 U2 (5V)2
由P= 可知,R 改变后的阻值:R'= = =6.25Ω······························ 1分
R 2 2 P' 4W
2
物理练习(五) 参考答案 第 2 页 共 2 页