2019秋季人教新版八年级上学期 6.2 密度 同步练习试题卷 含详细答案

2019秋季人教新版八年级上学期   6.2  密度  同步练习试题卷   含详细答案
2019秋季人教新版八年级上学期   6.2  密度  同步练习试题卷   含详细答案

6.2 密度

一.选择题(共18小题)

1.对密度定义式ρ=的下列几种解释中,属于正确表达的是()A.质量越大的物体,密度越大

B.密度与物体的体积成反比

C.当一个物体的体积减少一半,其密度减少一半

D.密度与物体的体积、质量无关

2.以下关于密度的说法,正确的是()

A.密度是物质的特性,物质不同,密度一定不同

B.金属件在打磨过程中,质量和密度均变小

C.氢气球在升空胀大时,里面氢气的密度变小

D.温度计中的水银从室温放入开水中,密度不变

3.关于对密度公式ρ=的理解,下列说法正确的是()

A.某种物体的质量越大,密度越大

B.某种物质的体积越大,密度越小

C.某种物质的密度与质量、体积都有关

D.密度是物质的一种特性,与质量和体积无关

4.甲、乙两种物质的质量和体积关系如图所示,由图象可知()

A.ρ甲>ρ乙B.ρ甲<ρ乙C.ρ甲=ρ乙D.无法判断

5.将一根粗细均匀的铁丝截去一段后,不变的物理量是()

A.重力B.质量C.体积D.密度

6.甲、乙两种物质的质量之比是5:2,密度之比是10:7,则两种物质的体积比是()A.50:14 B.9:5 C.7:4 D.4:7

7.关于物质的密度理解正确的是()

A.物质的密度与质量成正比,密度与体积成反比

B.不同种类的物质,物质的密度一定是不相同的

C.不同种类的物质,单位体积的质量一般不同的

D.物质密度不随状态、位置、形状的变化而变化

8.用不同材料做成的甲乙两个实心球,已知ρ甲>ρ乙,则甲乙两球体积与质量之间比较不可能的关系是……()

A.甲的体积较小,质量较大

B.甲的体积较大,质量较大

C.乙的体积较小,质量较大

D.乙的体积较大,质量较大

9.如图甲所示,质量相同的a、b、c三种液体分别装在三个相同的玻璃杯中,如图乙是它们的质量与体积的关系图象,三个杯子从左至右依次装的液体种类是()

A.bca B.acb C.abc D.bac

10.氧气瓶中的氧气密度为ρ,用去一半质量的氧气后,瓶中氧气的密度为()A.2ρB.ρC.ρD.无法判定

11.甲乙两物体是实心正方体,放在水平桌面上,对桌面压强相等,压力之比为9比4,则两物体的密度之比是()

A.2:3 B.3:2 C.1:1 D.4:9

12.一个由甲、乙两种金属做成的零件,其体积为61.3cm3.若用与零件质量相同的纯金属甲和纯金属乙做此零件,其体积分别为96.5cm2和52.5cm3.则原零件中金属甲的质量和金属乙的质量之比为()

A.l:2 B.1:3 C.1:4 D.1:5

13.分别由甲、乙两种物质组成的物体,其质量与体积的关系如图所示。分析图象可知两种物质的密度之比ρ甲:ρ乙为()

A.8:1 B.2:1 C.4:1 D.1:2

14.关于物质的密度,下列说法中正确的是()

A.不同物体的密度不相同

B.不同种类的物质,其质量与体积的比值一般不同

C.由公式ρ=可知,物质的密度跟质量成正比,跟体积成反比

D.密度是物质的特性,其大小不随温度、形状、状态的变化而变化

15.下列关于密度及其公式(ρ=)理解的说法中,正确的是()A.同一种物质的密度一定是不变的

B.不同种物质的密度一定是不同的

C.不同种物质的密度一般是不同的

D.物体的密度与其质量成正比、与其体积成反比

16.有一瓶食用油用掉一半,则剩下的半瓶油()

A.密度为原来的一半

B.质量为原来的一半

C.质量与体积的比值为原来的一半

D.质量、体积和密度都为原来的一半

17.有位同学想估测自己的体积,她只知道自己的身高为165cm,体重50kg,那么,她的体积大约为()

A.0.005m3B.0.05m3C.0.33m3D.1.65m3

18.在国际单位中,密度的单位是()

A.℃B.kg C.m/s D.kg/m3

二.计算题(共6小题)

19.一把汤勺的质量为10.8g,它的体积为4cm3,则它的密度为多少g/cm3?合多少kg/m3?

20.测得一木块的质量是10.8g,体积是24cm3,木块的密度是多少kg/m3?

21.一个空玻璃杯的质量50g,装满水时总质量为550g,装满某种液体时总质量为450g,求液体的密度。

22.一个质量为0.5kg的塑料瓶,用它装满水后,总质量为3kg,用它装满某液体后,总质量为2kg,求:

(1)该塑料瓶的容积为多少?

(2)这种液体的密度为多少?

23.一个瓶子的质量是0.2kg,装满水时是0.7kg装满另一种液体时总质量是0.8kg,该液体的密度多大?

24.如图所示,一只容积为3×10﹣4 m3的瓶内盛有0.2dm3的水,一只口渴的乌鸦每次将一块质量为0.01kg的小石子投入瓶中,当乌鸦投了25块相同的小石子后,水面上升至瓶口,求:

(1)瓶内石块的总体积;

(2)石块的密度。

三.解答题(共1小题)

25.一石块的体积为20m3,为了获知它的质量,某同学在石块上取下了样品,用仪器测得样品的质量为1×10﹣2kg,体积为4×10﹣6m3,求:

(1)石块的密度ρ石;

(2)石块的质量m石。

参考答案与试题解析

一.选择题(共18小题)

1.【解答】解:某种物质单位体积的质量叫做这种物质的密度,所以密度就是质量与体积的比值,不同物质密度一般不同,所以不同种类的物质,单位体积的质量一般不同;

密度是物质的一种属性,与质量大小、体积大小无关,故ABC错误,D正确。

故选:D。

2.【解答】解:A、密度是物质的特性,物质不同,密度一般不同,也有相同的,例如酒精的密度与煤油的密度相同,故A错误;

B、金属件在打磨过程中,金属件的质量减小,由于组成金属件的物质没有发生变化,其

密度不变,故B错误;

C、氢气球在升空胀大时,体积变大,而氢气质量不变,根据ρ=可知,球内氢气的

密度变小,故C正确;

D、温度计是利用液体的热胀冷缩的性质制成的,温度计中的水银从室温放入开水中,水

银的体积变大,质量不变,故密度变小,故D错误。

故选:C。

3.【解答】解:

密度是物质本身的一种特性,与物质的种类和状态有关,与质量和体积无关;同种物质的质量和体积成正比,可以利用ρ=计算物质的密度,故ABC错、D正确。

故选:D。

4.【解答】解:

由图象知,在两种物质的体积相同时(如都取V=1cm3),甲的质量大于乙的质量,由公式ρ=可知,甲的密度大于乙的密度。

故选:A。

5.【解答】解:一根粗细均匀的铁棒,截去一段后,体积、质量、重力都变小了,但是密度是物质的一种特性,密度大小与物体的质量和体积无关,所以密度没有变。

故选:D。

6.【解答】解:

由题知,m甲:m乙=5:2,ρ甲:ρ乙=10:7,

由ρ=可得体积之比:

V甲:V乙=:==7:4。

故选:C。

7.【解答】解:

A、某种物质单位体积的质量叫做这种物质的密度,其质量与体积的比值是一个常量,密

度不随质量和体积的变化而变化,故A错误;

B、密度是物质的一种特性,物质不同,密度一般也不同;但不同物质,密度也可能相同,

故B错误;

C、某种物质单位体积的质量叫做这种物质的密度,所以密度就是质量与体积的比值;不

同物质密度一般不同,所以不同种类的物质,单位体积的质量一般不同,故C正确;

D、物质的密度不随位置、形状的变化而变化,当状态发生变化时,体积会发生变化,密

度也会改变,故D错误。

故选:C。

8.【解答】解:A、已知ρ甲>ρ乙,根据公式m=ρV可知,存在甲的体积较小,质量较大的情况,故A可能存在;

B、已知ρ甲>ρ乙,根据公式m=ρV可知,甲的体积大,质量大,故B可能存在;

C、已知ρ甲>ρ乙,根据公式m=ρV可知,乙的体积较小,则乙的质量也会小,故C

不可能存在;

D、已知ρ甲>ρ乙,根据公式m=ρV可知,若乙的体积远远大于甲的体积,则乙的质

量可能会大于甲的质量,故D可能存在。

故选:C。

9.【解答】解:

(1)如图乙,取相同体积的三种液体V,可得质量关系m a>m b>m c,由ρ=可知ρa >ρb>ρc;

(2)在图乙中,三种液体的质量相同,而ρa>ρb>ρc,由V=可知V a<V b<V c,所以左边烧杯液体为b,中间为c,右边为a,

三个杯子从左至右依次装的液体种类:b、c、a。

故选:A。

10.【解答】解:

设原来氧气的质量为m,体积为V,原来氧气的密度为ρ=;

用掉一半后,氧气的质量变为m,

因为氧气的体积不变,

所以此时瓶中氧气的密度:ρ′==×=ρ。

故选:C。

11.【解答】解:由它们对桌面的压强相等,可知=,

压力之比为9:4,可得,=,

则==,

因为是甲乙两个实心正方体,

所以=,

=,

因为静止在水平面上的物体,此时的压力即为重力,

所以,====,

所以==×=×=2:3。

故选:A。

12.【解答】解:

设零件的质量为m,

由题可知,纯甲金属和纯乙金属的质量相等,则两纯金属的质量也为m,

已知纯甲金属和纯乙金属的体积分别为96.5cm3和52.5cm3,

所以,甲的密度ρ甲=,乙的密度ρ乙=,

已知零件的体积为V=61.3cm3,设零件中含甲金属的体积为V甲,

则零件中乙金属的体积:

V乙=V﹣V甲=61.3cm3﹣V甲﹣﹣﹣﹣①

零件的质量:m=ρ甲V甲+ρ乙V乙,

即:m=V甲+V乙﹣﹣﹣﹣②

②式两端同时除以m,并把①式代入可得:

1=V甲+(61.3cm3﹣V甲),

解得:V甲=19.3cm3,

则V乙=V﹣V甲=61.3cm3﹣19.3cm3=42cm3,

零件中甲金属的质量:m甲=ρ甲V甲=×19.3cm3,

零件中乙金属的质量:m乙=ρ乙V乙=×42m3,

所以,零件中甲乙两金属的质量之比:

m甲:m乙=×19.3cm3:×42m3=1:4。

故选:C。

13.【解答】解:由图象可知,当m甲=20g时,V甲=5cm3,m乙=10g时,V乙=20cm3,则甲、乙的密度分别为:

ρ甲===4g/cm3,ρ乙===0.5g/cm3,

所以,甲、乙的密度之比:

ρ甲:ρ乙=4g/cm3:0.5g/cm3=8:1。

故选:A。

14.【解答】解:A、不同物体密度一般不同,但个别不同物质密度是相同的,如煤油和酒精。故A错误;

B、某种物质单位体积的质量叫做这种物质的密度,所以密度就是质量与体积的比值,不

同物质密度一般不同,所以不同种类的物质,其质量与体积的比值一般不同。故B正确;

C、某种物质单位体积的质量叫做这种物质的密度,其质量与体积的比值是一个常量,密

度不随质量和体积而变化。故C错误;

D、当温度变化时,质量不变,但体积变化,所以,密度会变化。故D错误。

故选:B。

15.【解答】解:密度是物质的一种特性,不同物质密度一般不同,密度与物质种类和状态有关,即物质在不同温度下或不同状态时,密度不同,不能说物质的密度跟它的质量成正比,跟它的体积成反比,故C正确,ABD错误。

故选:C。

16.【解答】解:一瓶食用油用掉一半,所含物质变少,质量变为原来的一半,体积也减小为原来的一半,但密度与原来相同,保持不变,故ACD错误,B正确。

故选:B。

17.【解答】解:该同学的质量m=50kg,人的密度约为ρ=ρ水=1.0×103kg/m3,由ρ=可得,她的体积约为:

V===0.05m3。

故选:B。

18.【解答】解:

A、摄氏度(℃)是温度的单位,故A不符合题意;

B、千克(kg)是质量的单位,故B不符合题意;

C、米每秒(m/s)是速度的单位,故C不符合题意;

D、千克/m3(kg/m3)是密度的单位,故D符合题意。

故选:D。

二.计算题(共6小题)

19.【解答】解:汤勺的密度:

ρ===2.7g/cm3,

因1g/cm3=1×103kg/m3,所以2.7g/cm3=2.7×1×103kg/m3=2.7×103kg/m3。

答:汤勺的密度为2.7g/cm3,合2.7×103kg/m3。

20.【解答】解:木块的密度:

ρ===0.45g/cm3=0.45×103kg/m3。

答:木块的密度是0.45×103kg/m3。

21.【解答】解:

由题知,空玻璃杯的质量m0=50g,装满水时总质量m总水=550g,装满某种液体时总质量m总液=450g,

则玻璃杯内水和某种液体的质量分别为:

m水=m总水﹣m0=550g﹣50g=500g,

m液=m总液﹣m0=450g﹣50g=400g,

因同一玻璃杯装满不同液体时液体的体积相等,

所以,由ρ=可得,容器的容积:

V容==,

则液体的密度:

ρ液=ρ水=×1.0g/cm3=0.8g/cm3。

答:液体的密度为0.8g/cm3。

22.【解答】解:

(1)水的质量:

m水=m总1﹣m瓶=3kg﹣0.5kg=2.5kg,

由ρ=得塑料瓶的容积:

V=V水===2.5×10﹣3m3;

(2)某液体的质量:

m液=m总2﹣m瓶=2kg﹣0.5kg=1.5kg,

液体的体积V液=V=2.5×10﹣3m3,

这种液体的密度:

ρ液===0.6×103kg/m3。

答:(1)该塑料瓶的容积为2.5×10﹣3m3;

(2)这种液体的密度为0.6×103kg/m3。

23.【解答】解:瓶子装满水时水的质量:

m水=m总﹣m瓶=0.7kg﹣0.2kg=0.5kg,

由ρ=可得,水的体积:

V水===5×10﹣4m3,

则瓶子的容积:

V=V水=0.5×10﹣3m3;

瓶子装满另一种液体时液体的质量:

m液=m总′﹣m瓶=0.8g﹣0.2kg=0.6kg,

液体的体积:

V液=V=5×10﹣4m3,

则液体的密度:

ρ液===1.2×103kg/m3。

答:这种液体的密度是1.2×103kg/m3。

24.【解答】解:

(1)由题意可知,瓶内石块的总体积:

V石=V容﹣V水=3×10﹣4m3﹣0.2×10﹣3m3=1×10﹣4m3;

(2)石块的总质量:

m石=0.01kg×25=0.25kg,

石块的密度:

ρ石===2.5×103kg/m3;

答:(1)瓶内石块的总体积为1×10﹣4m3;

(2)石块的密度为2.5×103kg/m3。

三.解答题(共1小题)

25.【解答】解:(1)石块的密度:

ρ石=ρ样品===2.5×103kg/m3;

(2)石块的质量:

m石=ρ石V石=2.5×103kg/m3×20m3=5×104kg。

答:(1)石块的密度为2.5×103kg/m3;

(2)石块的质量为5×104kg。

人教版八年级初二英语期末考试习题及答案

欢迎共阅人教版八年级英语期末考试题及答案 八年级上学期期末测试英语试卷 一、单项填空(共25小题,每题1分,计25分) 1.What____you______atthistimeyesterday? A.did;do B.are;doing C.were;doing 2.Shedidn’tgohome.Ididn’tgohome,_______. A.too B.also C.either 3.Heaskedme_____areportcard. A.write B.writing C.towrite 4.TheteachersaidIcoulddobetter_____English. A.in B.at C.for 5.Thankyoufor______mewithmyhomework. A.help B.helps C.helping 6.____iseasyforachildtowakeupandknowwheretheyare. A.It B.That C.This 7.I’msurehewill_____passtheexam. A.can B.isableto C.beableto 8.____eyeexercises____good_____youreyes. A.Doing;are;for B.Do;is;for C.Doing;is;for 9.Theriverisverydirty.______peoplewanttoswiminit.

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