Soal un-ipa-smp-kode-ipa sp-38-volumekuate28093ese2809380f_2
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Transcript of Soal un-ipa-smp-kode-ipa sp-38-volumekuate28093ese2809380f_2
Dipindai oleh http://urip.wordpress.com Kode IPA_SP_38 (VolumeKuat–Es–80F) Halaman 3
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Besaran pokok
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Besaran pokok
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Konsep : Tenggelam jika massa jenis benda > massa jenis zat cair Terapung jika massa jenis benda < massa jenis zat cair
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Jawaban : D
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TF = (TC x 1,8) + 32 = (80 x 1,8) + 32 = (144) + 32 = 176 F
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= 180 N
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300 x 3 5
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=
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F = W.h s
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Jadi jawabannya : A
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Ep1 : Ep2 m1gh1 : m2gh2 2.10.3 : 3.10.4 60 : 120 1 : 2
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Jadi jawabannya : D
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a = 8/2 = 4
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F = 6 + 5 - 3 = 8
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a= 3/2 = 1,5
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F = 5+4-6 = 3
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a = 7/2 = 3,5
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F = 4+6-3 = 7
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a = 5/2 = 2,5
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F = 6 + 3 - 4 = 5
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:
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Persamaan
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a = F/m
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Q(A-B-C) = Q (A-B) + Q (B-C) = 63000 + 1008000 = 1071000 Joule = 1071 kJ
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Q(B-C) = m.L = 3 . 336000 = 1008000
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Q(A-B) = m. c. dt = 3. 2100. 10 = 63000
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Konsep : Tekanan berbanding lurus dengan gaya, dan berbanding terbalik dengan luas penampang alas (P = F/A)
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Semakin besar luas alas, maka tekanannya akan semakin kecil
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8 cm^2
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6 cm^2
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4 cm^2
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4 cm^2
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Jadi jawabannya : A
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satu gelombang 2 m satu gelombang ditempuh dalam 0,5 s jadi dlm 1 sekon menghasilkan panjang gelombang 4 m maka kecepatannya adalah 4 m/s
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W = P x t = (5x20x10) + (1x60x8) +(1x150x24) = (1000) + (480) + (3600)
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= 5080
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besar energi yg digunakan dalam 1 bulan adalah : 5080 x 30 = 152400 Wh
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