八年级假日乐园快乐暑假答案(广西师范大学出版社)《所有》

如题所述

物理:
练习一:
一.选择题
1.C 2.D 3.B 4.D 5.B 6.D 7.B 8.D 9.B 10.A 11.A 12.D 13.A 14.A 15.D
二.填空与作图题
16.(1):kg;(2):g
17.5:1;5:1;1:1.
18.质量;绝缘。
19.较大 大 小
20. 66;60;1.1
21. 19.3×10³ ;金;0.29
22. 60;83;1200
23.大 ;大
24.
(1)主要器材:
密度表、小铁球、天平和量筒
(2)简要做法:
用天平称出小球质量、用量筒测出小球的体积.
(3)如何判断:
算出该小球的密度对照密度表进行判断:若ρ测<ρ铁,则小铁球是空心的;若ρ测=ρ铁,则小铁球是实心的.
25. 0.8;55
26.27;20;1.35×10³;相等
三、解答与探究题
27.(1).游码;平衡螺母(2)B、A、C(3)50(4)在操作过程中,手不能调节平衡螺母。
28.(1)62;(2)20;(3)3.1
29.(1)m=79g,v=75-60=15cm³,P=m/v=79/15≈5.3×10³kg/m³,因为5.3×10³kg/m³≠7.9×10³kg/m³,所以空心的。
(2)V1=m/P=79/7.9=10cm³,V2=15cm³,V=15-10=5cm³ 所以是5cm³
30.解:容积设为V,瓶的质量设为M,已知水的密度为1g/cm3,酒精密度为0.8g/cm3,有:
1V+M=400
0.8V+M=350
解出:V=250cm3,M=150g
答:空瓶的质量是150g,容积是250ml
31.
解:
汽车所用钢材的体积为: <?xml:namespace prefix = m />V钢=m钢/ρ钢=237/(7.9×10³)m³=3.0×10-2m³.
所用聚丙烯塑料的体积为V=V钢=3.0×10-2m3
所用聚丙烯塑料的质量为m=ρV=1.1×103×3×10-2Kg=33Kg
减轻质量为:△m=m钢-m=237-33=204Kg.
答:可减少质量204Kg.

语文:http://wenku.baidu.com/view/d8eb35f59e31433239689398.html

英语:楼上的答了

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第1个回答  2012-07-07
练习一

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