DE THI HSG LY 8

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Trờng THCS Xuân Dơn= g Đề thi học= sinh giỏi tuyến trờng lớp 8

Năm h= ọc 2009-2010

Môn: Vật lý 8 = Thời gian: 60 phút

Họ và tên: ………̷= 0;……… Lớp 8…

Điểm

Lời

phê

của

giáo

viên

Đề bài:

Câu 1(2đ):Một ô tô chuyển động từ địa điểm A &re= g;ến địa điểm B cách nhau 180km.

Trong n&= ouml;a đoạn đờng đầu xe đi v&i= acute;i vận tốc v1=3D 45 km/h, nửa đoạn đờng còn lại

xe chuyển động với vận tốc v2 =3D 30 km/h

a)&n= bsp; Sau bao lâu xe đến B?

b)&n= bsp; Tính vận tốc trung b&time= s;nh của xe trên cả đoạn đờng AB

Câu 2(2,5đ): Một cục n= ớc đá có thể tích V =3D 500 cm3 n&ae= lig;i trên mặt nớc Tính thể tích của phần ló ra khỏi mặt nớc b= iết khối lợng riêng của nớc đ&ced= il; là 0,92 g/cm3, trọng lợng riêng của nớc dn =3D 10000 N/m3.

Câu 3(2đ): Một ngời kéo một vật từ giếng sâu 12m lên đều trong 24 gi&cop= y;y Ngời ấy phải dùng một lực F=3D 150= N

Tính công và công su&Ec= irc;t của ngời kéo.

Câu 4(2,5đ): Ngời ta thả một thỏi đồng nặng 0,4 kg ở nhiệt độ 800C vào 0,25 kg nớc ở nhiệt độ 180C Hãy xác định nhiệt độ

khi có cân bằng nhiệt Cho nhiệt dung riêng của đồng= C

1 =3D 400 J/kg.độ, của n&ia= cute;c C2 =3D 4200 J/kg.độ.

Câu 5(1đ): Trớ= ;c mặt em là một lon nớc ngọt và một cục đá lạnh Em phải đặt nh thế nào? lon nớc tr d= f;n cục đá hay cục ∨

đá trên lon n&iacu= te;c để có thể lon nớc lạnh đi nha= nh nhất.

Bài làm:

……… 8230;………= ………̷= 0;……

……… 8230;………= ………̷= 0;……

……… 8230;………= ………̷= 0;……

……… 8230;………= ………̷= 0;……

……… 8230;………= ………̷= 0;……

……… 8230;………= ………̷= 0;……

……… 8230;………= ………̷= 0;……

……… 8230;………= ………̷= 0;……

……… 8230;………= ………̷= 0;……

a)&n= bsp; Thời gian xe đi nửa đo&su= p1;n đờng đầu: t 1 =3D giờ = &nb= sp; (0,75đ) Thời= gian xe đi nửa đoạn đờng sau := t 2 =3D giờ = &nb= sp; (0,75đ) Thời= gian xe đi cả đoạn đờng : t =3D t 1 + t 2 =3D 2+ 3=3D 5 giờ (0,5đ) b)&n= bsp; Vận tốc trung bình c&ntild= e;a xe : v =3D km/h Câu 2(2,5đ): Khối lợng của cục nớc đá : m =3D D.V =3D 500 0,92 =3D 460 g =3D 0,46 kg = &nb= sp; (0,5đ) Trọng lợng của cục nớc đá : P =3D 10 m = =3D 10 0,46 =3D 4,6 N = (0,5đ) Khi c&o= circ;c đá nổi trọng lợng của cục nớc đá bằng đúng trọng lợng của nớc bị chiếm chỗ tức bằng lực đẩy Acsimet = &nb= sp; = &nb= sp; = &nb= sp; (0,5đ) Thể tích ph&C= cedil;n bị chìm trong n ớc : V/ =3D (d là trọng l ợng riêng c&ntild= e;a n ớc)

V / =3D m 3 =3D460 cm 3 = = &nb= sp; = (0,5đ)

Thể tích ph&C= cedil;n cục đá nhô ra khỏi mặt n ớc là:

V =3D V =3D 500- 460 =3D 40 cm / 3 = &nb= sp; = (0,5đ= ;)

Vậy thể tich ph= ần cục đá nhô ra khỏi mặt n ớc là 40 cm 3

Câu 3(2đ)

Công thực hiện : A =3D F s =3D 150.12 = =3D1800 (J) = (1đ)

Công= suất : P =3D (w) = &nb= sp; (1đ)

= ;

Câu 4(2,5đ):

Gọi m 1 , m 2 là khối l ợng của thỏi đồng và của n ớc

t( 0 C) là nhiệt độ khi có cân bằng nhiệt

Nhiệt l ợng đồng toả ra: Q 1 =3D m 1 C 1 (t 1 – t) = (0,5đ)

Nhiệt l ợng của n ớc thu và= o Q 2 =3D m 2.C2(t - t2) = (0,5đ)

= ;

Khi có cân bằng nhiệt : Q1 =3D Q 2 = &nb= sp; = &nb= sp; (0,5đ)

ú m 1 C 1 (t 1 –= ; t)=3D m 2 C 2 (t - t 2 ) = &nb= sp; = (0,5đ)

= =3D>t =3D =3D =3D 27 C = 0 = (0,5đ)

Câu 5(1đ): = Nên dặt cục đá lên trên lon n ớc Thực vậy nếu đặt lon n ớc lên cục đá thì chỉ có lớp

n

ớc thấp nhất bị lạnh đi, còn những phần ở trên vẫn bị lớp không khí lạnh bao xung quanh, lon n

ớc sẽ lâu lạnh hơn (0,5đ)

Nếu cục đ&= cedil; phía trên lon n ớc th&tim= es; lớp n ớc phía trên lon n ớc lạ= ;nh đi rất nhanh và chìm xuống và lớp n ớc ch a lạnh ở d ới xẽ lên = thay thế (do hiện t ợng đối l u) Mặt khác không khí lạnh xung quanh mặt n ớc cũng đi xuống và bao bọc lon n ớc làm lon n ớc lạnh đi nhanh hơ= n (0,5đ)

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