# Kunci Jawaban Fisika Klas x Bab 1 Martin Kanginan1

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Bab 1Besaran Fisika dan Satuannya Ayo Uji Pemahaman Anda 1.(13,35 0,05) cm 2.(a) (1,670 0,005) cm (b) (6,230 0,005) cm 3.(a) 6,5 + 43 x 0,01 = (6,930 0,005) mm (b) 4,0 + 11 x 0,01 = (4,110 0,005) mm 4.(a) 4200 m = 4,2 103 m Bilangan penting = 4,2 Orde besar = 103 (b) 5807,6 m = 5,8076 103 m Bilangan penting = 5,8076 Orde besar = 103 (c) 200 300 000 m = 2,003 108 m Bilangan penting = 2,003 Orde besar = 108 (d) 0,007 kg = 7 10-3 kg Bilangan penting = 0,007 Orde besar = 10-3 (e) 0,006 300 kg = 6,300 10-3 kg Bilangan penting = 6,300 Orde besar = 10-3 (f) 0,000 000 54 kg = 5,4 10-7 kg Bilangan penting = 5,4 Orde besar = 10-7 5.(a) 4 angka penting (b) 5 angka penting (c) 2 angka penting (d) 2 angka penting 6.(a) 43,35 (b) 88,0 (c) 0,090 (d) 225 7.(a) 24,286 + 2,343 + 3,21 = 30,24 m (b) 3,67 x 104 + 2,54 x 103 = 36,7 x 103 + 2,54 x 103 = (36,7 + 2,54) x 103

= 39,2 104 g (c) 297,15 13,5 = 283,7 m(d) 6,35 x 103 5665 = (6,35 5,665) x 103 = 0,69 103 m (e) 0,012 kg + 30 g = 12 g + 30 g = 42 g 8.(a) 2,5 m x 3,14 m = 7,9 m2

2 AP 3 AP2 AP;(AP = angka penting) (b) 2,5 m x 4,20 m x 0,305 m = 10,5x 0,3052 m3 = 3,2 m3 2 AP 3 APhasil antara dari 2 AP 4 AP2 AP (c) 323,75 N : 5,0 m2 = 65 N/m2 5 AP 2 AP 2 AP (d) 3 cm 5, 2 cm 15,6cm2,10 cm 2,10= =7,4 cm = 7,0 cm 9.massa = 2,1 x 25 g = 525 g = 530 g = 5,3 x 102 g 10. suhu rata-rata = 21 21, 2 21,113+ += 21 11. R1 = 36 5%; R2 = 75 5%;(a)R1 = 536100 = 1,8 R2 = 575100 =3,75 (b) R1 seri dengan R2; R = R1 + R2 = 36 + 75 = 111 R = R1 + R2 = 1,8 = 3,75 = 5,55Hambatan total = R R = (111 5,55) = (111 6) Ketidakpastian R total = 6 Persen = 6100% 5%111 =12. L = (90,0 0,1) cm = (90,0 0,1) x 10-2 m T = (3,00 0,05) s 22 22L L 4 LT 2 ;T 4 ;gg g T= = =Karena L dan T dalam 3 AP, maka ambil = 3,142 (4 AP) 224 LgT=2 224(3,14) (90,0x10 )g 3,949(3,00)= = m/s2 (4 AP; hasil antara) 22 22g 4 L L Tg 4 LT ; 2T g L T = = = +g = 220,1x10 0,052 0,0344 3, 44%90,0x10 3,00| |+ = = |\ Sesuai persamaan (1-5), 3,44% dekat dengan 1%, sehingga berhak atas 3 angka penting. g =( )3, 443,949 0,1358100=m/s2 Jadi, g = (3,949 0,1358) m/s2 = (3,95 0,14) m/s2 dalam 3 AP 13.L = 100,00 cm (sL = 0,04 cm);T = 2,00 s (sT = 0,05 s) 224 LgT= g = 42LT-2;2 2 2 2L Tg s s 0,04 0,051 2 2g L T 100,00 2,00 | | | | | | | |= + = + ||||\ \ \ \ gg= 0,05000 = 0,05000 100% = 5% dekat dengan 1%, sehingga berhak atas 3 angka penting. g = 4(3,142)2(100,00 cm)(2,005)-2 = 987,2164 cm s-2 = 9,87 m s-2 g = 5%(9,87 m s-2) = 0,4935 m s-2 Jadi, g = (9,87 0,49) m s-2 14. (a). 45 000 mg = 45 000 10-6 kg = 0,045 kg (b) 200 dm3 = 200 10-3 m3 = 0,200 m3 (c)0,8 3gcm= (8 10-1) 103 3kgm= 8 102 3kgm= 800 3kgm 15. (a) [Luas] = [p][] = L.L = L2 (b) [p] = [m][v] = [M][L.T-1] = MLT-1 (c)[p] = 21 22[F] M.L.TML T[A] L = =(d) [s] = 22 23[w] M.L.TML T[V] L = =16. v = P + Qt + Rt2; [v] = [P] + [Q].T + [R].T2 [P] = [v] = L.T-1 = m s-1 [Q] = [v].T-1 = [L.T-1][T-1] = L.T-2 satuan m s-2 [R] = [v].T-2 = [L.T-1][T-2] = L.T-3 satuan m s-3 17. [p] = [m][v] = [M][L.T-1] = MLT-1 [I] = [F][t] = [MLT-2][T] = MLT-1 [p] = [I]. Jadi, momentum dan impuls adalah besaran vektor yang setara. 18. (a) a = mF; LT-2 = 2 1 22M;LT L TMLT ; tidak sama, sehingga persamaan pasti salah (b) v2 = vo2 + 2as; (L.T-1)2 =(L.T-1)2 + [L.T-2][L] L2T-2= L2T-2 + L2T-2 sama, sehingga persamaan mungkin benar 19. 2 2 21 3 2 1 221 2m m Fr (M.L.T ).LF G ; G M L Tr m m M.M = = = =20. perpindahan; s = kax.ty;L = [L.T-2]x .[T]y

2,6 106 s Jawab A 6.Jarak = 20 000 tahun cahaya; 1 tahun = 365 hari Jarak = [(2 104)(365)(24)(3 600) s] 3 108 m/s = (1,89 106) 1011 km 2 1017 km Jawab B 7.Ek = mv2 kg(m/s)2 = kgm2s-2 Jawab A 8.Dimensi momentum, p = mv = MLT-1 Dimensi gaya, F = MLT-2 Dimensi daya, ( )( )2MLT LF.sPt T= = =ML2T-3 Jawab C 9.Dimensi daya, ML2T-3 (cara sama dengan nomor 8) Jawab D 10. Dimensi momentum, MLT-1 (cara sama dengan nomor 8) Jawab C 11. ( ) ( ) ( )x y zy x z 1 2 3 2m Mkp A ; ML T ML Lt T = =M1T-1L0 = Mx + yT-2xL-x 3y + 2z Pangkat T; -1 = -2x; x = Pangkat M; 1 = x + y; 1 = + y; y = Pangkat L; 0 = -x 3y+ 2z; 0 = - - 1 + 2z; z = 1 Jawab B 12. sudah jelas Jawab E 13. x = xo xxo = (0,2 + 5 0,01) cm = (0,2 + 0,05) cm = 0,25 cm x = xo x = (0,250 0,005) cmJawab D 14. mikrometer sekrup, x = 0,01 mm = 0,005 mm x = 5,5 mm + (37 x 0,01 mm) = 5,87 mm x relatif = 0,005x100%5,87= 0,085% Jawab A 15. 0,07060 m memiliki 4 angka penting Jawab C 16. Luas = 12,61 x 5,2 m2 = 65,572 m2 = 66 m2 (2 AP) Jawab E 17. R1 = 400 1%; R1 = 1%(400) = 4 R2 = 600 1%; R2 = 1%(600) = 6 R3 = 100 0,5%; R3 = 0,5%(100) = 0,5 Rx = 1 23R RR; Rx = ..? Rx = 400(600)2400100= Rx = R1R2R3-1; 3 x 1 2x 1 2 3R R R RR R R R = + +xxR1% 1% 0, 5% 2, 5%R= + + = ; Rx = 2,5%(2400 ) = 60 Jawab B 18. Nilai benar g 9,80 m/s2 Murid D menghasilkan pengukuran berulang yang baik yang berarti tepat (presisi), tetapi nilainya jauh dari nilai benar, g = 9,80 m/s2 (berarti tidak teliti atau tidak akurat) Jawab D 19.T = 5,00 s (sT = 0,10 s);L = 100,0 cm (sL = 3,0 cm) T = 2 2L2 ;g 4 LTg = 2 2 2 2L Tg s s 0,10 3,01 2 2 0,05 5%g L T 5,00 100,0 | | | | | | | |= + = + = = ||||\ \ \ \ Jawab E 20. F = 1 22q q 14 r| | |\ sumbu tegak F dan sumbu mendatar r-2 tan = 1 2q q4, dari grafik tan = ( )( )21100 500 13(r ) 400 200 = = Jadi, 3 = 1 2q q4; ( )( )( )6 612 10 1 262,6x10 106, 2x10q q177x10 1,8x1012 12 3,14 = = = = Jawab C 21. D + A + B = C Jawab B F3x = 50 cos 143 = 50 (-cos 37) = 50(-0,8) = -40 N F2y = 50 sin 143 = 50 (sin 37) = 50(0,6) = 30 N 22. F Fx = F cos ; Fy = F sin P = 10 N Px = 0; Py = -10 Rx = F cos Ry = F sin 10 Diketahui R = 20 N arah OA atau = 0, artinya Rx = R = 20 N dan Ry = 0 Jadi, F sin 10 = 0; F sin = 10 F cos = 20 F si n 10 1F cos 20 2= =; tan Jadi, sin = 1 155 5 =Jawab C 23. A = 40; B = 20; = 60 cos 60 = |A B| = 2 2 2 21A +B - 2AB cos40 +20 - 2(40)(20)2| | = |\ = 2 220 2 +1 - 2 20 3 =Jawab B 24. P = D;(P, Q) = 90 tan = oQ1; 45P = =tan = oQ1; 45P= =(P + Q, P - Q) = + = 45 + 45 = 90 Jawab D 25. Diketahui: P dan Q dengan P = Q = x. Misal(P, Q) = 122=+P QP Q; 2 22 2P Q 2PQcos22P Q 2PQcos+ =+ + 22 2 22 2 2x x 2x cos 2x x 2x cos 2| | + = |+ + \ ; ( )( )222x 1 cos2 12x 1 cos 4 2 = =+ 1 + cos = 2 2 cos ; 3 cos = 1; cos = 13 Jawab A 26. 60 + 120 + 180 240; 360 240. Jadi, mungkin Jawab C 27. Diketahui: 1 petak = 1 N F1x = 6; F1y = 0; F2x = 2; F2y = 6 Fx = 6 + 2 = 8; Fy = 0 + 6 = 6 2 2 2 2x yF F +F 8 +6 10 = = = N Jawab C 28. Diketahui A = 8 km; C = 6 km R = A + C; R = ? Ax = 0; Ay = -8; Cx = 6 cos 30 = 33; Cy = 6 sin 30 = 3 Rx = 0 + 33 = 33; Ry = -8 + 3 = -5 ( )( )222 2x yR R +R 3 3 + -5 27+25 2 13 = = = = km Jawab C 29. R = F1 + F2 + F3 F1x = -2; F1y = 7 F2x = -6; F2y = 8 F3x = 3; F3y = -3 Rx = -5; Ry = 12 ( )22 2 2x yR R +R 5 +12 13 = = =N Jawab D 30. Diketahui: F1 = 20 N; 0; F1x = 20; F1y = 0 F2 = 203 N; 90; F2x = 0; F2y = 203F3 = 30 N; 60; F3x = 15; F3y = 153 Fx = 20 + 0 + 15 = 35; Fy = 0 + 203 + 153 = 353 Besar ( )( )222 2x yF F +F 35 + 35 3 70 = = =N Arah tan = yxF35 33F 35= = = 60 Jawab B II

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