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Fizik - Tingkatan 5 - Soalan Kuiz
 Online Tutoring : Forum : Soalan Kuiz (Tingkatan 5) : Fizik - Tingkatan 5 - Soalan Kuiz
Topic Topic: Prinsip Archimedes: Daya Julangan = Berat Cecair Tersesar Post ReplyNew Question
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Sistem
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Joined: 26 June 2005
Posts: 2755
Posted: 13 July 2005 at 11:57pm | IP Logged Quote Sistem

SOALAN KUIZ #249
Objek M berbentuk kubus dengan panjang setiap sisinya ialah 10 cm.  Ketumpatan objek M ialah 5 g/cm.  Objek diletakkan atas penimbang dan kedua-duanya ditenggelamkan dalam air. Dengan menganggap ketumpatan air sebagai  1 g/cm  dan pecutan graviti sebagai  10 m/s ,  berapakah berat objek M yang akan diukur oleh penimbang tersebut?
    A.   10 N
    B.   20 N
    C.   30 N
    D.   40 N
    E.   50 N
    F.   60 N
 SOALAN KUIZ #249 digubah oleh Space pada 13 July 2005 pukul 9:09pm

Jawapan betul: D    Klik betul: 33.94%  Klik salah: 66.06%
 
    A.   [22 klik 6.73%] 
 
    B.   [34 klik 10.40%] 
 
    C.   [62 klik 18.96%] 
 
    D.   [111 klik 33.94%] 
 
    E.   [84 klik 25.69%] 
 
    F.   [14 klik 4.28%] 
 
Sila bincang soalan kuiz ini.
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Fiyasakura
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Joined: 17 November 2005
Location: Malaysia
Posts: 47
Posted: 25 November 2005 at 11:12pm | IP Logged Quote Fiyasakura

tak faham laa....

 

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Iz_aan
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Joined: 24 November 2005
Location: Malaysia
Posts: 129
Posted: 29 November 2005 at 6:41pm | IP Logged Quote Iz_aan

hai safiah...

nie aku faham...

Isipadu objek = 1000 cm     

Berat objek = Vg

                 =
[5g cm 1000cm]
[1000]
10

                 = 50 N

Kehilangan berat ketara = tujah ke atas

                                    = Vg

                                    =
[1g cm-31000cm]
[1000]
10

                                    = 10 N

Berat objek dalam air = berat sebenar objek - kehilangan berat ketara

                              = 50 N - 10 N

                              = 40 N

senang je nie....

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