Bab 1 Besaran 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 AP
2 AP;
(AP = angka penting)
(b) 2,5 m x 4,20 m x 0,305 m = 10,5 2 AP
3 AP
hasil antara dari 2 AP
x 0,3052 m3 = 3,2 m3 4 AP
(c) 323,75 N : 5,0 m2 = 65 N/m2 5 AP (d)
2 AP
2 AP
3 cm × 5,2 cm 15,6 = cm = 7,4 cm = 7,0 cm 2,10 cm 2,10
9. massa = 2,1 x 25 g = 525 g = 530 g = 5,3 x 102 g 10. suhu rata-rata =
21 + 21, 2 + 21,11 = 21º 3
11. R1 = 36 Ω ± 5%; R2 = 75 Ω ± 5%; (a) ∆R1 = ∆R2 =
5 × 36 Ω = 1,8 Ω 100
5 × 75 Ω = 3,75 Ω 100
(b) R1 seri dengan R2; R = R1 + R2 = 36 + 75 = 111 Ω ∆R = ∆R1 + ∆R2 = 1,8 = 3,75 = 5,55 Hambatan total = R ± ∆R = (111 ± 5,55) Ω = (111 ± 6) Ω Ketidakpastian R total = 6 Ω Persen =
6 × 100% = 5% 111
12. L = (90,0 ± 0,1) cm = (90,0 ± 0,1) x 10-2 m
2 AP
T = (3,00 ± 0,05) s L 2 L 4π2L ;T = 4π2 ;g = 2 g g T
T = 2π
Karena L dan T dalam 3 AP, maka ambil π = 3,142 (4 AP) g=
4π2L T2
g=
4(3,14)2 (90,0x10−2 ) 2 = 3,949 m/s (4 AP; hasil antara) 2 (3,00)
g=
∆g ∆L 4π2L ∆T = 4π2LT −2 ; = +2 2 T g L T 0,1x10−2
0,05
∆g = +2 = 0,0344 = 3, 44% −2 3,00 90,0x10 Sesuai persamaan (1-5), 3,44% dekat dengan 1%, sehingga berhak atas 3 angka penting. ∆g =
3, 44 ( 3,949) = 0,1358 m/s2 100
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) g=
4π2L → g = 4π2LT-2; T2 2
∆g s 0,05 0,04 s = 1× L + −2 × T = + −2 × 2,00 g L T 100,00 2
2
2
∆g = 0,05000 = 0,05000 × 100% = 5% dekat dengan 1%, sehingga berhak atas 3 g
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
g kg kg kg = (8 × 10-1) ×103 3 = 8 × 102 3 = 800 3 3 cm m m m
15. (a) [Luas] = [p][ℓ] = L.L = L2 (b) [p] = [m][v] = [M][L.T-1] = MLT-1 (c) [p] =
[F] M.L.T −2 = = ML−1T −2 2 [A ] L
(d) [s] =
[w ] M.L.T −2 = = M L−2T −2 [V] L3
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 =
m M ; LT-2 = ;LT −2 ≠ L−1T 2 ; tidak sama, sehingga persamaan pasti salah F M LT −2
(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. F = G
m 1m 2 Fr 2 (M.L.T −2 ).L2 ;G = = = M −1L3T −2 2 r m 1m 2 M.M
20. perpindahan; s = kax.ty; L = [L.T-2]x .[T]y L1T0 = Lx.T-2x + y Pangkat L: 1 = x; x = 1 Pangkat T: 0 = -2x + y; 0 = -2(1) + y; y = 2 Jadi, persamaannya adalah s = kat2 21. Tekanan hidrostatik; p = kρxgyhz; ML-1T-2
= (ML-3)x.(LT-2)y.(L)z
M+1L-1T-2 = MxL-3x+y+z.T-2y Pangkat M; +1 = x; x = +1 Pangkat T; -2 = -2y; y = 1
Pangkat L; -1 = -3x + y + z; -1 = -3 + 1 + z; z = 1 Jadi, persamaannya adalah p = kρgh 22.
23.
24.
25. (a)
(b)
(c)
26. (a)
(b)
(c)
27. (a)
(b)
28.
S = F1 – F2 = F1 + (-F2) |F2| = F2, sehingga S2 = F12 + F22 + 2F1F2 cos (180 – α) S2 = F12 + F22 - 2F1F2 cos α S=
F12 +F22 - 2F1F2 cos α (terbukti)
29. Diketahui: A = 5; B = 12; ∠ (A, B) = α; cos α = cos 120º = -½ Ralat soal F1 diganti A F2 diganti B (a) R = A + B
R2 = A2 + B2 + 2AB cos 120º = 52 + 122 + 2(5)(12)(-½) R2 = 109 R = √109 Arah β B R = ; sin β = sin β sin 60o
(b) S = A - B
1 12 3 2 = 0,9954 → β = 84,5º 109
S2 = A2 + B2 + 2AB cos 60º = 52 + 122 + 2(5)(12)(½) R2 = 229 R = √229 Arah γ B S = ; sin γ = sin γ sin 120o
1 12 3 2 = 0,68674 → β = 43,4º 229
(c) T = B - A
T2 = A2 + B2 + 2AB cos 60º = 52 + 122 + 2(5)(12)(½) R2 = 229 R = √229 Arah x 1 5 3 B T ; sin x = 2 = 0,2861→ x = 16,6º = sin x sin 120o 229
Jadi, arah = 120 + 16,6º = 136,6º terhadap A 30. (a) 2 m + 8 m ≥ 10 benar, sehingga mobil mungkin dapat kembali ke titik awal keberangkatan. (b) 5 m + 14 m ≥ 20 m salah, sehingga mobil tidak mungkin dapat kembali ke titik awal keberangkatan.
(c) 10 m + 10 m ≥ 10 m benar, sehingga mobil mungkin dapat kembali ke titik awal keberangkatannya. (d) 150 m + 250 m ≥ 450 m salah, sehingga mobil tidak mungkin dapat kembali ke titik awal berangkatnya. 31. s = 8 m arah 143° terhadap sumbu X+ (arah mendatar) sx = 8 cos 143° = 8(-cos 37°) = 8(-0,8) = -6,4 m sy = 8 cos 143° = 8(sin 37°) = 8(0,6) = 4,8 m 32. (a) A = (b) F =
32 + 42 = 5 cm; tan α =
( −40 3 )
2
Ay Ax
=
4 (Kuadran I) → α = 53,13º 3 Fy
+ (−40)2 = 80 N; tan α =
Fx
=
−40 (Kuadran III) −40 3
→ α = (180 + 30)° = 210º (c) B = 122 + (−13)2 = 313 m; tan α =
By Bx
=
−13 (Kuadran IV) 12
→ α = -47,29 (d) P = (−20)2 + 202 = 20 2 N; tan α =
Py Px
=
20 (Kuadran II) −20
→ α = (180 – 45)º = 135º 33. F1x = +5 N; F1y = 0; F2x = +3 N; F2y = 6 N Fx = 5 + 3 = 8 N; Fy = 0 + 6 = 6 N F=
82 + 62 = 10 N
Arah tan α = 34.
Fy Fx
=
6 (Kuadran I); α = 37º 8
F1 = 100 N; α1 = 90º F2 = 120 N; α2 = 0º F3 = 50 N; α1 = 143º Ralat soal Orang ketiga menarik ke arah 53º barat dari selatan dengan gaya 50 N.
F1x = 0 N; F1y = 100 N F2x = 120 N; F2y = 0 F3x = 50 cos 143º = 50 (-cos 37º) = 50(-0,8) = -40 N F2y = 50 sin 143º = 50 (sin 37º) = 50(0,6) = 30 N Fx = 0 + 120 – 40 = 80 N; Fy = 100 + 0 + 30 = 130 N F=
802 + 1302 = 10 64 + 169 = 10 233 N
Arah tan α =
Fy Fx
=
130 (Kuadran I); α = 1,02º 80
Uji Kompetensi Bab 1 I. Pilihan Ganda 1. Besaran pokok: panjang, massa, waktu, kuat arus listrik, suhu, jumlah zat, intensitas cahaya. Jawab C 2. 1 m3 = 1000 dm3 = 1000 L Jawab C 3. 500 cc = 500 cm3 = 500 × 10-6 m3 = 5 × 10-4 m3 Jawab D 4. 2,47 × 102 km = 2,47 × 102(103 m) = 2,47 × 105 m Jawab E 5. 1 bulan = 30 hari = 30(24 jam) = 720(3 600 s) = 72(36)103 s = 2592 × 103 ≈ 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 = ½mv2 → kg(m/s)2 = kgm2s-2 Jawab A 8. Dimensi momentum, p = mv = MLT-1 Dimensi gaya, F = MLT-2 F.s ( MLT Dimensi daya, P = = t T
−2
) ( L ) = 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 z m M = kp x ρy A z ; = ( M L−1T −2 ) ( M L−3 ) ( L2 ) t T
M1T-1L0 = Mx + yT-2xL-x – 3y + 2z Pangkat T; -1 = -2x; x = ½ Pangkat M; 1 = x + y; 1 = ½ + y; y = ½ Pangkat L; 0 = -x – 3y+ 2z; 0 = -½ - 1½ + 2z; z = 1 Jawab B 12. sudah jelas Jawab E 13. x = xo ± ∆x xo = (0,2 + 5 × 0,01) cm = (0,2 + 0,05) cm = 0,25 cm x = xo ± ∆x = (0,250 ± 0,005) cm Jawab D 14. mikrometer sekrup, ∆x = ½ × 0,01 mm = 0,005 mm x = 5,5 mm + (37 x 0,01 mm) = 5,87 mm ∆x relatif =
0,005 x100% = 0,085% 5,87
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 Ω R2 = 600 ± 1%; ∆R2 = 1%(600) = 6 Ω R3 = 100 ± 0,5%; ∆R3 = 0,5%(100) = 0,5 Ω Rx =
R1R2 ; ∆Rx = …..? R3
Rx =
400(600) = 2400 Ω 100 ∆Rx ∆R1 ∆R 2 ∆R3 = + + Rx R1 R2 R3
Rx = R1R2R3-1;
∆Rx = 1% + 1% + 0,5% = 2, 5% ; ∆Rx = 2,5%(2400 Ω) = 60 Ω Rx
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) L ;g = 4π2LT −2 g
T = 2π
2
2
∆g s 3,0 0,10 s = 1× L + −2 × T = + 2× = 0,05 = 5% g L T 100,0 5,00 2
2
Jawab E q q 1
20. F = 1 2 2 sumbu tegak F dan sumbu mendatar r-2 4πε r tan θ =
q1q 2 ∆1 ( 1100 − 500) , dari grafik tan θ = = =3 −2 4πε ∆(r ) ( 400 − 200)
Jadi, 3 =
( 62,6x10−6 )(106, 2x10−6 ) = 177x10−12 = 1,8x10−10 q1q 2 qq ; ε= 1 2 = 12π 12 ( 3,14) 4πε
Jawab C 21. D + A + B = C Jawab B F3x = 50 cos 143º = 50 (-cos 37º) = 50(-0,8) = -40 N F2y = 50 sin 143º = 50 (sin 37º) = 50(0,6) = 30 N 22. F → Fx = F cos θ; Fy = F sin θ P = 10 N → Px = 0; Py = -10 Rx = F cos θ Ry = F sin θ – 10 Diketahui R = 20 N arah OA atau α = 0º, artinya Rx = R = 20 N dan Ry = 0 Jadi, F sin θ – 10 = 0; F sin θ = 10 F cos θ = 20 F sin θ 10 1 = = ; tan ½ F cosθ 20 2
Jadi, sin θ =
1 1 5 = 5 5
Jawab C 23. A = 40; B = 20; θ = 60º → cos 60º = ½ |A – B| =
1 A 2 +B2 - 2A B cos θ = 402 +202 - 2(40)(20) 2
= 20 22 +12 - 2 = 20 3 Jawab B 24. P = D; ∠ (P, Q) = 90º
tan α =
Q = 1; α = 45o P
tan β =
−Q = 1; β = 45o P
∠ (P + Q, P - Q) = α + β = 45º + 45º = 90º
Jawab D 25. Diketahui: P dan Q dengan P = Q = x. Misal ∠ (P, Q) = θ P−Q 1 2; = P+Q 2
P2 + Q 2 − 2PQ cosθ P + Q + 2PQ cosθ 2
2
=
2 2
2 x 2 + x 2 − 2x 2 cosθ 2 2x ( 1 − cos θ ) 2 1 = ; = = x 2 + x 2 + 2x 2 cos θ 2 2x 2 ( 1 + cosθ ) 4 2
2
1 + cos θ = 2 – 2 cos θ; 3 cos θ = 1; cos θ =
1 3
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 F = Fx2 +Fy2 = 82 +62 = 10 N
Jawab C 28.
Diketahui A = 8 km; C = 6 km R = A + C; R = ……? Ax = 0; Ay = -8; Cx = 6 cos 30º = 3√3; Cy = 6 sin 30º = 3
Rx = 0 + 3√3 = 3√3; Ry = -8 + 3 = -5
(3 3)
R = R2x +R2y =
2
+ ( -5) = 27+25 = 2 13 km 2
Jawab C 29.
R = F1 + F2 + F3 F1x = -2; F1y = 7 F2x = -6; F2y = 8 F3x = 3; F3y = -3 Rx = -5; Ry = 12 R = Rx2 +R2y =
( −5)2 +122 = 13 N
Jawab D 30. Diketahui: F1 = 20 N; 0º; F1x = 20; F1y = 0 F2 = 20√3 N; 90º; F2x = 0; F2y = 20√3 F3 = 30 N; 60º; F3x = 15; F3y = 15√3 Fx = 20 + 0 + 15 = 35; Fy = 0 + 20√3 + 15√3 = 35√3 Besar F = Fx2 +Fy2 =
2 ( 35) + ( 35 3 ) = 70 N 2
Arah tan θ =
Fy Fx
Jawab B
=
35 3 = 3 → θ = 60º 35
II. Esai Tingkat 1 Pengaplikasikan Skill A. Pengukuran 1. L = x + ∆x = (11,45 ± 0,05) cm 2. L = x + ∆x = (1,680 + 0,005) cm 3. mikrometer sekrup; ∆x = 0,005 mm (a) 3,5 mm lebih 45 × 0,01 mm = 0,45 mm Jadi, L = (3,5 + 0,45) mm ± 0,005 mm = (3,950 ± 0,005) mm (b) 1,5 mm lebih 17 × 0,01 = 0,17 mm Jadi, L = (1,5 + 0,17) mm ± 0,005 mm = (1,670 ± 0,005) mm (c) 5 mm lebih 40 × 0,01 = 0,40 mm Jadi, L = (5 + 0,40) mm ± 0,005 mm = (5,400 ± 0,005) mm 4. x =
11, 38 + 11,28 + 11,32 + 11, 42 + 11, 30 = 11,34 5
1 sx = N
N Σx i2 − ( Σx i )
2
N −1
; N = 5; N – 1 = 4
Σx 2i = 11,382 + 11, 282 + 11, 322 + 11, 422 + 11,302 = 642,992
( Σx )
2
i
= ( 11,38 + 11, 28 + 11, 32 + 11, 42 + 11,30) = 3214,89 2
Jadi, sx =
1 5( 642,992) − 3214,89 = 0,0265 5 4
sx 0,0265 × 100% = × 100% = 0,2% x 11,34
0,2% dekat dengan 0,1%, sehingga berhak atas 4 angka penting. Hasil diameter = x ± sx = (11,34 ± 0,03) mm 5. Hasil pengukuran selang waktu dari 20 kali ayunan adalah 20T = 40,0; 40,1; 39,8; 39,8; 39,9 T = 2,000; 2,005; 1,990; 1,990; 1,995 ΣTi 9,980 = = 1,996 5 5
T=
ΣTi2 = 2,0002 + 2,0052 + 1,9902 + 1,9902 + 1,9952 = 19,9203
( ΣT ) i
2
= ( 2,000 + 2,005 + 1,990 + 1,990 + 1,995) = ( 9,980) 2
2
1 sT = N
N ΣTi2 − ( ΣTi )
2
N −1
1 5( 19,9203) − ( 9,980) = = 0,003317 5 4 2
sT 0,003317 × 100% = × 100% = 0,17% T 1,996
0,17% dekat dengan 0,1%, sehingga berhak atas 4 angka penting. Jadi, periode bandul T = T ± sT = (1,996 ± 0,003) detik 6. Temannya ragu karena dua nol dalam 500 g tidak terdefinisi dengan jelas, apakah termasuk angka penting atau bukan. Untuk mengatasi masalah ini, maka harus ditulis dengan menggunakan notasi ilmiah. 500 g = 5,00 × 102 g berarti 3 angka penting 500 g = 5,0 × 102 g berarti 2 angka penting 500 g = 5 × 102 g berarti 1 angka penting 7. (a) 125,97 g + 8,15 g = 134,12 g (b) 112,6 m + 8,005 m + 13,48 m = 134,085 m = 134,1 m (c) 78,05 cm2 – 32,046 cm2 = 46,004 cm2 = 46,00 cm2 (d) 1,6 kg + 23 kg – 0,005 kg = 24,595 kg = 25 kg (e) 14,0 cm × 5,2 cm = 72,80 cm2 = 73 cm2 (f) 0,1682 m × 8,2 m = 1,379 m2 = 1,4 m2 (g)
54, 5 J = 78,75 = 79 J/s 1, 2 s
(h)
180 N × 35 m N.m 3 = 1050 = 1000 N m/s = 1 × 10 N m/s 6s s
(i)
7,500 × 103 kg
( 5,0 × 10
1
m ) × 5,0 m × 4 m
2
3
= 0,075 × 10 kg/m
= 0,1 × 102 kg/m3 = 1 × 101 kg/m3 8. (a) (2,46 + 5,4) × 103 g = 7,86 × 103 g = 7,9 ×103 g (b) (5,80 × 109) + (3,20 × 109) = (5,80 + 0,320) × 109 = 6,12 × 109 s (c) (5,87 × 10-6) + (2,83 × 10-6) = 3,04 × 10-6 m (d) (8,12 × 107) + (6,30 × 106) = (8,12 – 0,630) × 107 = 7,49 × 107 g (e) (5,60 × 10-7 m) : (2,8 × 10-12 s) = 2,0 × 105 m/s (f) (9,2 × 10-4 km)(1,5 × 10-3 km) = 13,8 × 10-7 km2 = 1,4 × 10-6 km2
9. 90 lembar = 1,35 cm; tebal kertas =
1 -2 × 1, 35 cm = 0,015 cm = 1,50 × 10 m 90
10. Sisi persegi, a = 15,300 cm Keliling persegi = 4a = 4(15,300 cm) = 61,200 cm Luas persegi = a2 = (15,300 cm)2 = 234,09 cm2 11. ρ =
m 4,500 × 103 g 1 3 3 = = 0,64 × 10 g/cm = 6,4 g/cm V 7,0 × 102 cm 3
12. massa air dalam tangki = 51,7 kg – 3,66 kg = 48,04 kg = 48,0 kg 13. siswa yang hasil pengukuran panjangnya 63,58 cm Data: 63,65 cm; 63,64 cn; 63,66 cm; 63,66 cm L=
ΣL i 63,65 + 63,64 + 63,66 + 63,66 = = 63,65 cm N 4
14. (a) ∆t = ½(0,5) = 0,25ºC (b) t = 100ºC sehingga
∆t 0,25 = × 100% = 0, 25% t 100
(c) ketelitian = 100% - 0,25% = 99,75% 15. ∆t = ½(0,1) 0,05 sekon (a) ketelitian 96% berarti kesalahan relatif Jadi, (b)
∆t × 100% = 4% t
0,05 × 100 = 4 ; t = 1,25 s t
∆t × 100% = 1% (dari 100% - 99% = 1%) t
0,05 × 100% = 1% ; t = 5,00 s t
16. Ketelitian 99% berarti kesalahan relatif
∆d × 100% = 1% d
d kira-kira 25 mm → ∆d × 100 = 25 mm; ∆d = 0,25 mm mistar, ∆L = 0,5 cm = 5 mm jangka sorong, ∆L = 0,5 mm Jadi, jelas mistar mauoun jangka sorong tidak dapat digunakan untuk mengukur tebal papan. 17. R1 = (28,4 ± 0,1); R2 = (4,25 ± 0,01); R3 = (56,605 ± 0,001); R4 = (90,75 ± 0,01) Empat resistor tersebut dirangkai secara seri
Ditanya: Rtotal berikut ketidakpastiannya = ….? R = R1 + R2 + R3 + R4 = 180,005 Ω N = 4, N - 1 = 3 ∆R = ∆R1 + ∆R2 + ∆R3 + ∆R4 = 0,121 Ω ∆R 0,121 = × 100% = 0,07% dekat dengan 1%, sehingga berhak atas 4 angka R 180,005
penting. Rtotal = R + ∆R = (180,0 ± 0,121) Ω 18. Bola, diameter, d = (10,00 ± 0,05) mm (a) V =
4 3 4 d 3 1 3 πr = π( ) = πd ; d = diameter 3 3 2 6
∆V ∆d 0,05 =3 = 3 = 0,0150 = 1,5% V d 10,00
(b) Ketidakpastian relatif 1,5% dekat dengan 1% sehingga berhak atas 3 angka penting. (c) V =
1 (3,142)(10,00)3 = 523,7 mm3 6
∆V = 0,0150V = 0,0150(523,7) = 7,855 mm3 Jadi, V = (524 ± 8) mm3 dalam 3 angka penting 19. T = 2π (a) T 2 =
m ; T = (0,0825 ± 0,0025) s; m = (15,02 ± 0,05) kg k
4π2m 4π2m ;k = k T2
∆k ∆m ∆T = + 2. k m T ∆k 0,05 0,0025 → = + 2. = 0,0639 = 6, 39% = 6, 4% k 15,02 0,0825
(b)
∆k = 6,4% dekat dengan 10%, sehingga berhak atas 2 angka penting. k k = 4π2mT −2 = 4(3,14)2 (15,02)(0,0825)−2 = 87032 N/m
∆k = 6,4%k = 6,4%(87032) = 5570 N/m Jadi, k = (87032 ± 5570) N/m
= (87000 ± 5600) N/m = (8,7 ± 0,6) × 104 N/m dalam 2 angka penting. 20. R = 10 Ω; L1 = 60,52 cm; L2 = 39,49 cm; sL = 0,08 cm x=
L1 R , dengan R dianggap konstanta. L2 2
2
2
2
s ∆x s 0,08 0,08 −3 = 1× L + −1× L = + 39, 49 = 2, 4 × 10 x L L 60,52 1 2 ∆x = 2, 4 × 10−3 × 100% = 0,24% dekat dengan 0,1% sehingga berhak atas 4 angka x
penting. x=
60, 52 (10 Ω) = 15,33 Ω 39, 49
∆x = 2,4 × 10-3(15,33) = 0,037 Ω Jadi, x = (15,33 ± 0,04) Ω dalam 4 angka penting.
B. Besaran dan Satuan 21. x → meter (m); t → sekon (s); v dalam m/s; a dalam m/s2 (a) (b)
v 2 (m / s)2 = = m / s2 x m x m = = s2 = s 2 a m/ s
(c) ½at2 = ½
m 2 ( s) = m s2
22. (a) 5,2 ton = 5,2 × 103 kg = 5200 kg (b) 150 mg = 150 × 10-6 kg = 1,5 ×10-4 kg (c) 2500 g = 2500 × 10-3 k = 2,5 kg (d) 0,5 hm2 = 0,5 × 104 m2 = 5 × 103 m2 (e) 400 cm2 = 400 × 10-4 m2 = 4 × 10-2 m2 (f) 24 000 mm2 = 24 000 × 10-6 m2 = 2,4 × 10-2 m2 (g) 5000 dm3 = 5000 × 10-3 m3 = 5 m3 (h) 7500 cm3 = 7500 × 10-6 m3 = 7,5 × 10-3 m3
10−3 kg 3 = 13,6 × 103 kg/m −6 3 10 m
23. (a) 13,6 g/cm3 = 13,6
(b) 0,5 g/cm3 = 0,5 × 103 kg/m3 = 500 kg/m3 1000 m = 25 m/s 3600 s
(c) 90 km/jam = 90
5m = 37, 5 m/s 18 s
(d) 135 km/jam = 135
1N 2 = 800 N/m −4 2 10 m
(e) 0,08 N/cm2 = 0,08
1N 2 = 7 × 106 N/m −4 2 10 m
(f) 700 N/cm2 = 700
(g) 700 N cm = 700(1 N × 10-2 m) = 7 N m (h) 1 kW jam = 1(1000 W)(3600 s) = 3,6 × 106 Ws 24. (a) 1 tahun = 1(365)(24)(3600) s = 3,15 × 107 s (b) lama =
6,2 × 1023 molekul NA 16 = = 1,97 × 1016 tahun ≈ 2 × 10 tahun 7 v 3,15 × 10 s molekul 1 s 1 tahun
25. (a) [Ep] = [massa][percepatan][tinggi] = M(LT-2)(L) = ML2T-2 [gaya][ jarak] ( MLT (b) [Daya] = = [w aktu] T
−2
) ( L ) = ML T 2
−3
(c) [Ek] = ½[massa][kelajuan]2 = M(LT-1)2 = ML2T-2 (d) [momen inersia] = [massa][jarak]2 = M(L)2 = ML2 26. v = Pt + Qt2 +Rt3; LT-1 = [P]T + [Q]T2 + [R]T3 [P] =
LT −1 -2 = LT −2 satuan m s T
[Q] =
LT −1 -3 = LT −3 satuan m s 2 T
[R] =
LT −1 -4 = LT −4 satuan m s T3
27. (a)
[volume] L3 = = L3T −1 [menit] T
(b)
[perpindahan] L = = T2 [percepatan] LT −2
(c) [gaya][volume] = ( MLT −2 )( L3 ) = M L4T −2 (d) [massa jenis][kelajuan] 2 = ( M L−3 )( LT −1 ) = M L−1T −2 2
28. Dimensi T = [T] 1
g LT −2 2 −1 T = 2π → [T] …. ;T ≠ T (salah) L L 1
L L 2 T = 2π → [T] …. −2 ;T = T (benar) g LT
Jadi, rumus yang benar adalah T = 2π
L . g
29. Dimensi vo = LT-1 1
(a) v o =
2g LT −2 2 −1 −1 → (LT-1) … ;LT ≠ T (salah) h L 1
2h L 2 → (LT-1) … −2 ;LT −1 ≠ T (salah) (b) v o = g LT 1
(c) v o = 2gh → (LT-1) … ( LT −2 (L)) 2 ;LT −1 ≠ LT −1 (benar) Jadi, rumus yang benar adalah v o = 2gh . 30. T = 2π [k] =
31. (i)
4π2m 4π2m m → T2 = ;k = k k T2
[m] M = 2 = M T −2 2 [T] T
(ii) (a)
(b)
(c)
(d)
(e)
(f)
32. (a) C + B = A, karena vektor A yang menutup (b) Ralat soal: vektor diberi nama
G + D + E = F, karena vektor F yang menuturp. (c) L + M + N + P = Q, sebab Q yang menutup. (d)
Mulai dari A memutar searah dengan jarum, lalu kembali ke titik A W+V–U+T+S+X=0 U=W+V+T+S+X 33. (a) Ax = 20 cos 37º = 20(0,8) = 16 m; Ay = 20 sin 37º = 20(0,6) = 12 m (b) Bx = 30 cos 60º = 15 m; By = 30 sin 60º = 15√3 m (c) Cx = 40 cos 150º = 40(-½√3 ) = -20√3 m; Cy = 40 sin 150º = 40(0,5) = 20 m (d) Dx = 10 cos 217º = 10 cos (180 + 37) = 10(-cos 37º) = 10(-0,8) = -8 m Dy = 10 sin 217º = 10 (-sin 37º) = 10(-0,6) = -6 m (e) Ex = 50 cos (-45º) = 50(½√2) = 25√2 m Ey = 50 sin (-45º) = 50(-½√2) = -25√2 m (f) Fx = 36 cos (330º) = 36 cos (360 - 30) = 36 cos 30º = 36(½√3) = 18√3 m Fy = 36 sin (330º) = 36(-sin 30º) = 36(-½) = -18 m 34.
(a) ∠ (C, D) = θ = 120º; cos θ = cos 120º = -½ Besar
R=C+D R=
1 42 +52 + 2(4)(5)(− ) = 21 2
Arah C R 4 21 = → = sin α sin D sin α sin 60 1 4 3 2 =2 3= 2 sin α = 21 21 7
→ α = 49º Arah terhadap sumbu X+, yaitu θR = 270 – α = 270 – 49 = 221º (b) S = C – D; ∠ (C, -D) = 60º;
S=
1 42 +52 + 2(4)(5)( ) = 61 m 2
C R 4 61 = → = sin α s sin ( 180 − 60)o sin α s 1 3 2 sin α s =
(2 3) → α 61
s
= 26o
Arah θs = 90º + αs = 90 + 26 = 116º (c) T = D + C Dengan cara yang sama dengan (a) dan (b): T = √61 m; θs = 296,33º 35. A = 3 m; B = 4 m R = A + B; besar dan arah R (a) α = 0º → R = A + B = 7 m, arah θ = 0º (b) α = ∠ (A, B) = 60º; cos α = cos 60° = ½ R=
A 2 +B2 + 2A Bcos α
R=
1 32 +42 + 2(3)(4)( ) = 37 2
1 4 3 B R Bsin α 4sin 60 2 → θ = 35o = → sin θ = = = sin θ sin α R 37 37 o
(c) Dengan cara yang sama, R = 5 m; θ = 53º (d) Dengan cara yang sama, R = √13 m; θ = 74º (e) Dengan cara yang sama, R = -1 m; θ = 180º 36. B = 2A; |A + B| = ½√6|A - B|; cos θ = ….? A 2 +B2 + 2A Bcos α = ½√6 A 2 +B2 − 2A Bcosα A 2 +4A 2 + 2A(2A )cos α =
6 2 ( A +4A 2 − 4A 2 cosα ) 4
3 2 A ( 5 - 4 cosθ ) 2 10 + 8 cos θ = 15 − 12 cosθ
A 2 ( 5 + 4 cosθ ) =
20 cos θ = 5; cos θ = ¼ 37. Misalnya dua vektor tersebut adalah A dan B, maka A = B = 5 m; θ = …? (a) R = 5√3 m R=A+B R2 = A 2 +B2 + 2A Bcosα
25(3) = 52 + 52 + 2(5)(5) cos α; cos α =
75 − 50 1 = → α = 60º 50 2
(b) R = 5 m, dengan cara yang sama, α = 120° (c) R = 3√10 m, dengan cara yang sama, α = 37° (d) R = 5√2 m, dengan cara yang sama, α = 90° 38. R = P + Q, S = P – Q, Buktikan bahwa R2 + S2 = 2(P2 + Q2) Misalkan P = Px, Py
Q = Qx, Qy Rx = Px + Qx; Ry = Py + Qy; Sx = Sx - Sx; Sy = Sy - Sy R2 + S2 = (Rx2 + Ry2) + (Sx2 + Sy2) = (Px + Qx)2 + (Py + Qy)2 + (Px - Qx)2 + (Py – Qy)2 = (Px2 + Qx2 + 2PxQx) + (Py2 + Qy2 + 2PyQy) + (Px2 + Qx2 - 2PxQy) + (Py2 + Qy2 - 2PyQy) = (Px2 + Py2) + (Qx2 + Qy2) + (Px2 + Py2) + (Qx2 + Qy2) = P2 + Q2 + P2 + Q2 R2 + S2 = 2(P2 + Q2) terbukti 39. Misal, ketiga vektor yang besarnya 3 N itu adalah A, B, dan C. Jadi, A = B = C = 3 N. R = A + B + C (a) R terbesar, jika A, B, dan C searah, sehingga Rmaks = 3 + 3 + 3 = 9 N
(b) Nilai terkecil, Rmin = 0, yaitu jika ∠ (A, B) = ∠ (B, C) = ∠ (C, A) = 120º
(c) R = 3, jika dua vektor dan vektor ketiga berlawanan arah
40. A = 30 N; B = 40 N; R = A + B (a) Rmaks = 30 + 40 = 70 N (b) Rmin = 40 - 30 = 10 N (c) (i) 30 N + 5 N ≥ 40 N, salah → tidak dapat seimbang
(ii) 30 N + 10 N ≥ 40 N, benar → dapat seimbang (iii) 30 N + 30 N ≥ 40 N, benar → dapat seimbang (iv) 30 N + 40 N ≥ 50 N, benar → dapat seimbang (v) 30 N + 40 N ≥ 70 N, benar → dapat seimbang (vi) 30 N + 40 N ≥ 75 N, salah → tidak dapat seimbang 41. (a) Besar, A =
A x2 + A 2y = −62 + 82 = 10 m
Ay
Arah, tan θ =
Ax
=
8 (Kuadran II) −6
θ = (180 – 53) = 127° (b) Dengan cara yang sama, B = 10 m; θ = 330° (c) Dengan cara yang sama, C = 16 m; θ = 60° (d) Dengan cara yang sama, D = 6√2 m; θ = 225° 42. Lihat gambar di soal; Ax = -12 m; Ay = 0 Bx = 18 cos 37° = 18(0,8) = 14,4 m; By = 18 sin 37° = 18(0,6) = 10,8 m (a) R = A + B; Rx = -12 + 14,4 = 2,4 m; Ry = 0 + 10,8 = 10,8 m Besar, R =
Rx2 + Ry2 =
Arah, tan α =
2 2 ( 2, 4) + (10,8) = 1,2 85 m
10,8 (Kuadran I)→ α = 77,5° 2, 4
(b) S = A - B; Sx = -12 - 14,4 = -26,4 m; Sy = 0 - 10,8 = -10,8 m Besar, S =
Sx2 + S2y =
Arah, tan α =
( −26, 4)2 + ( -10,8)2 = 1, 2 565 m
−10,8 (Kuadran III)→ α = (180 + 22,2)° = 202,2º −26, 4
43. Harus menggunakan metode vektor komponen (a) A = 8 N; ∠ A = 90° → Ax = 0; Ay = -8 B = 10 N; ∠ B = 0° → Bx = 10; By = 0 C = 20 N; ∠ C = 53° → Cx = 20 cos 53° = 12 N; Cy = 20 sin 53° = 16 N R=A+B+C Rx = 0 + 10 + 12 = 22 N; Ry = -8 + 0 + 16 = 8 N R=
R2x + R2y =
( 22)2 + ( 8)2 = 2 137 N
tan θR =
8 (Kuadran I)→ θR = 20° 22
(b) Dengan cara yang sama (a) A = 18 N; ∠ A = 0° → Ax = 18 N; Ay = 0 B = 30 N; ∠ B = 143° atau 37° (Kuadran II) → Bx = -30 cos 37° = -(30)(0,8) = -24 N; By = 30 sin 37° = (30)(0,6) = 18 N C = 24 N; ∠ C = 270° → Cx = 0; Cy = -24 N Rx = 18 + -24 + 0 = -6 N Ry = 0 + 18 + -24 = -6 N R=
Rx2 + Ry2 =
tan θR =
( −6)2 + ( -6)2 = 6 2 N
−6 (Kuadran III)→ θR = (180 + 45)° = 225° −6
(c) A = 7 N; ∠ A = 180° → Ax = -7 N; Ay = 0 B = 18 N; ∠ B = 60° → Bx = 7 cos 60° =
7 7 3 N; By = 7 sin 60° = N 2 2
C = 7 N; ∠ A = -60° → Cx = 7 cos (-60°) = Rx = -7 + Ry = 0 +
7 7 3 N; Cy = 7 sin (-60°) = N 2 2
7 7 + =0 2 2 7 3 7 3 +=0 2 2
R=0
(d) Dengan cara yang sama pada (a), (b), dan (c), R = 80√3 N; θR = 0º 44.
Ax = 40 cos 60° = 20 km; Ay = 40 sin 60º = 20√3 km Bx = 10 km; By = 0 Cx = 0; Cy = -10√3 km Rx = 20 + 10 + 0 = 30 km Ry = 20√3 + 0 - 10√3 = 10√3 km R=
Rx2 + R2y =
tan θR =
2 ( 30) + (10 3 ) = 20 3 km 2
10 3 3 (Kuadran I)→ θR = 30° = 3 3
45.
R=A+B+C+D Ax = 100 m; Ay = 0 Bx = 120 cos 60º = 60 km; By = 120 sin 60° = 60√3 m Cx = 100 cos 120° = -50 m; Cy = 100 sin 120° = 50√3 m Dx = -60 cos 60° = -30 m; Dy = -60 sin 60° = -30√3 m Rx = 100 + 60 + (-50) + (-30) = 80 km Ry = 0 + 60√3 + 50√3 + (-30√3) = 80√3 m Besar, R =
Rx2 + Ry2 =
2 ( 80) + ( 80 3 ) = 160 m 2
Arah, tan θR =
80 3 (Kuadran I)→ θR = 60° 80
Tingkat II Soal Tantangan 46. t = 1 tahun = 3,15 x 107 s (baca jawaban nomor 24) 1 tahun cahaya = (3,15 x 107 s)(kecepatan cahaya) = (3,15 x 107 s)( 3 x 108 m/s) = 9,45 x 1015 m Jarak Gomeisa = 250 tahun cahaya = 250(9,45 x 1015)(10-3 km) = 2362,5 x 1012 km = 2,36 x 1012 km 47. R = 6,38 x 106 m sehingga ambil π = 3,142 V=
4 3 4 18 3 πR = (3,142)(6, 38 × 106 )3 = 1087 x 10 m 3 3
= 1,09 x 1021 m3 48. 1 kmol = 6 x 1026 atom; massa 1 atom karbon = 2,0 × 10-26 kg Massa 5 mol atom karbon = (5 mol)
12,0 kg -3 = 60 x 10 1 kmol
= 6,0 x 10-2 kg (dalam 2 angka penting) Nm 2 ( MLT )( L = 2 M2 kg −2
49. vL = GxMyRz; [G] =
2
) =M
−1 3
LT-1 = (M-1L3T-2)xMyLz M0L1T-1 = M-x + yL3x + zT-2x Pangkat T → -1 = -2x → x = ½ Pangkat M → 0 = -x + y → 0 = -½ + y → y = ½ Pangkat L → 1 = 3x + z; 1 =
3 + z; z = -½ 2
Persamaannya yaitu vL = G½M½R-½ = (GMR-1)½ vL = k
GM R
L T −2
50. µ =
massa M = = M L−1 ; panjang kawat, L panjang L
F = gaya tegang = MLT-2; Periode, To To = kµxLyFz T = (ML-1)xLy(MLT-2)z M0L0T1 = Mx + zL-x + y + zT-2z Pangkat T → 1 = -2z → z = -½ Pangkat M → 0 = x + z → 0 = x - ½ → x = ½ Pangkat L → 0 = -x + y + z; 0 = -½ + y - ½; y = 1 Persamaan periode, To = kµ½L1F-½ → To = kL
µ F
−1 F M LT −2 m M = = M L T −2 ; [ρ] = = 3 = ML−3 2 A L V L
51. v = kpxρy; [v] = LT-1; [p] = LT-1 = (ML-1T-2)x(ML-3)y M0L1T-1 = Mx + yL-x – 3yT-2x
Pangkat T → -1= -2x → x = ½ Pangkat M → 0 = ½ + y → y = -½ Persamaannya adalah v = kp½ρ-½ → v = k
p ρ
52. Lihat gambar di soal Dx = 0; Dy = 3; 3 3 3 ; Cy = 3 sin 30º = 2 2
Cx = 3 cos 30º = 3(½√3) = (a) R = 2C + D
Rx = 2Cx + Dx = 3√3 + 0 = 3√3 Ry = 2Cy + Dy = 3 + 3 = 6
(3 3)
R = R2x +R2y =
2
+ ( 6) = 3 7 m 2
(Kunci jawaban diralat) Arah tan θ =
Ry Rx
=
6 3 3
=
2 → θ = 49º 3
(b) S = 2C – D Sx = 2Cx - Dx = 3√3 - 0 = 3√3 Sy = 2Cy - Dy = 3 - 3 = 0 Besar S=
(3 3)
Sx2 +Sy2 =
Arah, tan θ =
Sy Sx
=
2
+0 = 3 3
0 → θ = 0° 3 3
(c) T = C + 2D Tx = Cx + 2Dx =
3 3 3 -0= 3 2 2
Ty = Cy + 2Dy =
3 15 +6= 2 2 2
2
3 15 T = Tx2 +Ty2 = 3 + = 3 7 (kunci jawaban diganti) 2 2
15 5 → θ = 71° Arah, tan θ = = 2 = Tx 3 3 3 2 Ty
(d) U = C – 2D Ux = Cx - 2Dx =
3 3 3 -0= 3 2 2
Uy = Cy - 2Dy =
3 9 -6=2 2 2
2
3 9 U = U x2 +U y2 = 3 + − = 3 3 2 2
9 −3 −1 Arah, tan θ = = 2 = = (Kuadran IV) θ = -60° 3 Ux 3 3 3 2 Uy
−
53. R = P + Q; 2R = P + 2Q; 2R = P - Q Bukti, ∠ (P, Q) = ∠ (P, 2Q) = α R=P+Q R2 = P2 +Q 2 +2PQ cos α …………………………(1)
2R = P + 2Q
2 2 ( 2R ) = P2 + ( 2Q ) +2P ( 2Q ) cos α
4R2 = P2 + 4Q 2 +4PQ cos α ……………………..(2)
Eliminasi persamaan (1) dan (2) 4R2 = P2 + 4Q 2 +4PQ cos α (dikalikan 1) → 4R2 = P2 + 4Q 2 +4PQ cos α R2 = P2 +Q 2 +2PQ cos α
(dikalikan 2) → 2R2 = 2P2 +2Q 2 +4PQ cos α
-
2R2 = -P2 + 2Q2 …………………(3) 2R = P – Q 2 2 ( 2R ) = P2 + ( Q ) - 2PQ cos α
4R2 = P2 + Q 2 - 2PQ cos α ……………………….(4)
Eliminasi antara persamaan (4) dan (1) 4R2 = P2 + Q 2 - 2PQ cos α R2 = P2 +Q 2 + 2PQ cos α
+
5R2 = 2P2 + 2Q 2 …………………………………….(5)
Eliminasi antara persamaan (5) dan (3) 5R2 = 2P2 + 2Q 2
2R2 = -P2 + 2Q2
-
3R2 = 3P2
R=P Substitusi ke persamaan (3) 2R2 = -P2 + 2Q2 ; 2R2 = -R2 + 2Q2; 3R2 = 2Q2; Q = Jadi, P : Q : R = R :
R 3 :R=1: 2
R 3 2
3 : 1 = √2 : √3 : √2 (ralat kunci jawaban) 2
54. Ambil sudut terhadap sumbu X+ A = 100 ∠ A = 0° → Ax = 100; Ay = 0 B = 300 ∠ B = -90° → Bx = 0; By = -300 C = 150 ∠ C = (180 + 30)° → Cx = -150 cos 30° = -75√3; Cy = -150 sin 30° = -75
D = 200 ∠ D = (180 – 60)° → Dx = -200 cos 60° = -100; Ay = +200 sin 60° = 100√3 Resultan R=A+B+C+D S = -R dengan S adalah perpindahan kelima yang mengembalikan ke titik berangkatnya. Rx = 100 + 0 - 75√3 – 100 = -75√3 Ry = 0 - 300 - 75 + 100√3 = -375 + 100√3 R = Rx2 +R2y =
( −75 3 ) + ( -375+100 3 ) 2
2
= 16(875) + (−201,8)2 = 240 m
S = R = 240 m Sx = -Rx = 75√3; Sy = -Ry = 375 - 100√3 Arah, tan θ =
Sy Sx
=
375 − 100 3 → θ = 57° 75 3
55. Ambil sudut terhadap sumbu X+ sebagai acuan A = 6 N ∠ A = 60° → Ax = 6 cos 60° = 3 N; Ay = 6 sin 60° = 3√3 N B = 3 N ∠ B = -θ → Bx = 3 cos θ; By = -3 sin θ C = 4 N ∠ C = -90° → Cx = 0; Cy = -4 N D = 7 N ∠ D = 180° → Dx = -7 N; Dy = 0 (a) R = A + B + C + D; θR = 180º; sin θ = ….? Rx = 3 + 3 cos θ + 0 + (-7) = -4 + 3 cos θ Ry = 3√3 – 3 sin θ - 4 + 0 = 3√3 – 4 – 3 sin θ
Jadi, 3√3 – 4 – 3 sin θ = 0; sin θ =
3 3−4 4 = 3 − (ralat kunci jawaban) 3 3
(b) R = ….? Di sini θ telah ditetapkan sebagai sudut lancip, sehingga 2
4 cos θ = + 1 − sin 2 θ = 1 − 3 − = 0,917
3
Rx = -4 + 3(0,917) = -1,25 Resultan R = |Rx| = 1,25 N (ralat kunci jawaban)
