Penyelesaian Fisika Kelas Xi [PDF]

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PENYELESAIAN BUKU FISIKA KLAS XIA (2019- 2020)



BUKU PINTAR BELAJAR FISIKA



SMA/MA XI A



SEMESTER Ganjil 2019/2020



SAGUFINDO KINARYA Kunci Jawaban:an Fisika XIA



1



Sagufindo Kinarya



10 N



5. UJI KOMPETENSI KESETIMBANGAN DAN DINAMIKA ROTASI



4N



SOAL PILIHAN 9N



1. Dengan menentukan momen gaya yang terbesar Jawaban : D



Ʃ 𝜏 = -10 . 5 . 10-2 - 5√2 . 5√2 . 10-2 - 10 . 5 . 10-2 – 4 . sin-2 + 9 . 5 . 10-2 = - 50 . 10-2 – 50 . 10-2 – 50 . 10-2 - 20 . 10-2 + 45 . 10-2 = - 170 . 10-2 + 45 . 10 -2 = - 125 . 10-2 = - 1,25 Nm = 1,25 Nm (Arah berlawanan jarum jam)



3. Ʃ 𝜏 = F2 (3+x) – F1 . 3 9,6 = 8 (3+x) – 8 . 3 9,6+24 = 24+8x 9,6 = 8x X= 1,2 m Jawaban : D



Jawaban : E F2 = 10√2 N



4. FA = 10N A



45 ̊



30 ̊



Poros



A 53 ̊



B



B



C



10 cm



D 10 cm Poros



FB = 20N



F3 = 20N



AB = 2AC = 1,2 m AC = 2AP CB = AC



Ʃ 𝜏 D = - F1 sin53 . 40 . 10-2 + F2 sin45 . 20 . 10-2 + F3 . 10 . 10-2 4 1 = - 10 . 5 . 0,4 + 10√2 . 2√2 . 0,2 20 . 0,1 16 =+2–2 5 = - 3,2 Nm (arah berlawanan dengan putaran jarum jam) Jawaban : D



Ʃ 𝜏 = FB . BP + FA . AP – FC sin30 . CP = 20 . 0,9 + 10 . 0,3 – 20 . ½ . 0,3 = 18 + 3 – 3 = 18 Nm



Jawaban : C Kunci Jawaban:an Fisika XIA



20 cm



6.



FC = 20N C



10 N



5√2



2. Karena gaya di P dan di R sama besar serta arahnya berlawanan. Jawaban : B



P



45 ̊



1



Sagufindo Kinarya



7.



10. Jawaban : Karena besar gaya tidak ada, maka tidak bisa dikerjakan.



F = 280 N 0 20 cm



20 N



30 N



Jawaban : D



∑ 𝜏𝑜 = 30.0,2 + 20.0,4 – 280.0,05 = 0 Jawaban : A FA = 20N A



F2 = 5 N



A



30 ̊ 1



1m



C



B



Poros



1m



F1 = 2 N



FB = 10N



D



C



B



2



1



E 30̊ 1m



1m



F



1m



Poros



τD = −τA + τB − τF



AP : AC : AB = 1:2:4 AB = 1,2 m ∑ 𝜏𝑝 = FA . 0,3 + FB . 0,9 – FC sin30 . 0,3 1 = 20 . 0,3 + 10 . 0,9 – 20 . 2 . 0,3 = 12 Nm



= − FA . r A + FB . r B − FF . r F = -2.3 + 5.2 – 1 . 2 = 2 N.m Jawaban : A 13. Jawaban : Karena mA dan mB tidak diketahui, maka tidak bisa dikerjakan.



Jawaban : A 9.



F3 = 2 N



12.



FC = 20N



P



a



4 . 10 . 0,12 = = 0,2 kg. m2 2



40 cm



8.



m . g . r2



11. I =



5 cm



F1 = 20N



100 cm



14. Jawaban : tidak ada jawabannya. Karena jawabannya 254 kgm2.



O 40 cm



F2 = 10N



15. 𝐼 = 2 ∙ 0,22 + 3 ∙ 0,32 = 0,35 𝑘𝑔. 𝑚2 Jawaban : E



F3 = 40N



∑ τo = 20.0,7 + 40.0,7 – 10.0,3 = 39 Nm



16. 𝐼 = 𝑚𝑎2 + 𝑚. (2𝑎)2 = 5𝑚𝑎2 Jawaban : C



Jawaban : B Kunci Jawaban:an Fisika XIA



2



Sagufindo Kinarya



17. I = 4. 22 + 4. 12 + 2. 22 + 2. 42



23. τ = I.α a τ = I.R I = τ . R a-1 Jawaban : Tidak ada di option



I = 16 + 4 + 8 + 32 = 60 kg. m2 Jawaban : E 18. Pada buku paket, berada di nomor 17 kedua



24.



1



F = 21 N



f



∑ 𝜏 = 𝐼. 𝛼 𝑎



∑ 𝜏 = 𝑚𝑅2 . 2 𝑅 𝑎 = 20 𝑚. 𝑠 −2 Jawaban : C



τ



= I.α 1



a



2



R



f . R = mR2 1



f =2ma



19. 0,2 kg



l1 4m



l2



F – f = m.a



0,6 kg



F-



8m



1 2



a = ∑ 𝐼 = 0,2 . 42 + 0,6 . 82 = 0,2 . 16 + 0,6 . 64 = 3,2 + 38,4 = 41,6 kgm2 Jawaban : B



2F 3m



=



2.21 3.5



= 2,8 m/s2



Jawaban : Tidak ada di option IA A = (IA + IB) 



25.



A



20.  = I .  F.R = β.α 𝛽. 𝛼 𝐹= = β . α . R−1 𝑅 Jawaban : D



=



IA . ωA I A  IB



Jawaban : A 26. 𝐼1 . 𝜔1 = 𝐼2 . 𝜔2 6.9 = 𝐼2 . 12 𝐼2 = 4,5 𝑘𝑔. 𝑚2 Jawaban : D



21. Jawaban : E 22. ∑ τC = FA . AC − FB . BC = 10 .0,4 − 20 .0,2 = 0 Jawaban : A Kunci Jawaban:an Fisika XIA



m a = m.a



27. t 0  3



= o +  t = 20 +  . 10 = - 2 put/s Sagufindo Kinarya







32. o = 20 rad/s θ = 100 rad t = 60 rad/s I = 4 kg m2 τ = ...... ? t2 = o2 + 2  θ 3600 = 400 + 2  . 100  = 16 rad/s2 τ = I .  = 4 . 16 = 64 Nm Jawaban : D



= - 2 . 2π rad/s = - 4 π rad/s =I. = 5 . 10-3 . 4π = 2π .10-2 Nm







Jawaban : B 28. Jawaban : D 29. Erotasi = ½ I ω2 =



1 1 · · m r2 2 2 2



33. I = β τ = I .α F.R=β.α α = F . R . β-1 Jawaban : D



2



 32 . 2π  1 1  = · ·0,4.(12.10-2)  2 2  60  = 1,6 . 10-2 Joule Jawaban : Tidak ada di option 30. EmA = EmB EkR + Ek + m g h = C



34. m = 8 kg v = 15 m/s Vb = 5 m/s EK tot = ?



EK tot = E K rot + KK t 1 1 = 2 . I . ⍵2 . + 2 m v2



Va = 10 m/s



1 1



1



𝛼



1



= 4 . mv2 + 2 . mv2 3



= 4 . mv2



½Iω2 + ½mv 2 + m g h = C 2 vA = vB2 + gh 102 = 52 + 10 ℎ h = 7,5 h 7,5 s= = = 12,0 m sin α 0,6 Jawaban : C



3



= 4 . 8 . 152 = 1350 J Jawaban : B 35. F = 20 N m = 0,5 kg R = 20 . 10-2 m  = ......?  = I. = m r2 . 



31. Soal salah



Kunci Jawaban:an Fisika XIA



1



= 2 . 2 . mr2 . ⍵2 + 2 . mv2



s



4



Sagufindo Kinarya



=



1 a · (20 . 10-2)2 · 2 R



38. IA = 8 kg m2 m = 2 kg I0 = ......? 1 IA = 3 mR2



F 1 m -4 = · 400 · 10 · 2 R 20 = 200 . 10-4 . 1 2 2 . 20 .10



8.3



R = √ 2 =√12 m 1



= 0,4 Nm Jawaban : C



1



I0 = 12 mR2 = 12 2(√12)2 = 2 kgm2 Jawaban : A



36. F . R = I . α SOAL URAIAN 1. τ = F1 ℓ1 – F2 Sin 30o.ℓ2



F = I . α . R-1 F = β . α . R-1



= 16 . 5 – 16 ·



Jawaban : D ( R kurang pangkat -1)



1 ·2 2



= 64 N/m 37.



2. τ 1



Mk =4kg τ2



m2 =2kg m1 = 4kg



τ3



( 𝑚1 − 𝑚2 ) 𝑔 1 𝑚1 + 𝑚2 + 2 𝑚𝑘 (4 − 2) 10 = 1 4+ 2 + 2.4 20 = 8



= F 1 ℓ1 =6.4 = 24 Nm, berlawanan putaran jarum jam =6.3 = 18 Nm, searah putaran jarum jam =6.0 = 0 Nm



𝑎=



τ4



3.



τ



= F4 4  3 =6.5 = 30 Nm, searah putaran jarum jam 2



2



= F . Cos 60o . R



= 2,5 m/s2 Jawaban : D Kunci Jawaban:an Fisika XIA



5



Sagufindo Kinarya



=5·



1 · 0,6 2



m2 = 2 kg



m1 = 11 kg



= 1,5 Nm



( 𝑚1 – 𝑚2 ) 𝑔 1 1 𝑚1 + 𝑚2 + 2 𝑚𝑘1 + 2 𝑚𝑘2 (11 − 2) 10 𝑎= 1 1 11 + 2 + 2 . 2 + 2 . 2 90 = = 6 𝑚⁄𝑠 2 15 𝑎 =



4.  τ o



= F1 R – F2 R = 10 . 0,3 – 6 . 0,3 = 1,2 Nm



5. a. I



= m . R2 = 1 . 0,92 = 0,81 kg m2



b. τ



6. I =



= F . R = 0,08 . 0,9 = 0,072 Nm



9. Mk = 1 kg Mb = 2 kg rk = 50 cm = 0,5 m ⍺ =?



1 m R2 2 1 = . 80 . 4 . 10-2 = 16 kg m2 2



𝑚𝑏 . 𝑔 1 𝑚𝑏 + 2 𝑚𝑘 2 . 10 𝑎 = 1 2+ 2 . 1 = 8 𝑚⁄𝑠 𝑎 =



7.  = 6t rad/s r = 4 cm a. aT =  . R = 6t . 4 = 6 . 2 . 4 = 48 cm/s2 b. v =  . R = ∫  . dt . R = 3 t2 R = 3 . 22 . 4 = 48 cm/s c. S



8.



𝑎



⍺ =𝑅 =



8 0,5



= 16 𝑟𝑎𝑑⁄𝑠 2



= r . θ = R ∫  dt = 4 ∫ 3 t2 dt = 4t3 = 32 m



10. A. 𝑎 = mK2 = 2 kg



𝑚𝑏 . 𝑔 . sin 37 1 2



𝑚𝑏 + 𝑚 𝑘



mK1 = 2 kg



=



Kunci Jawaban:an Fisika XIA



6



1 . 10 . 1



3 5



1+ 2 𝑅



Sagufindo Kinarya



=



6 1+



0,8 0,22



=



6 21



𝑚⁄ 2 𝑠



I=



=1,25 kg m2



B. T = m . g . sin37 – m . a = 1 . 10 . ⅗ - 1 . 6⁄21 = 6 – 6⁄21 = 120⁄21 = 5,7 N 11.



L I. 12 . 



mR 2 Δv  13. a. τ = I .  = R Δt = 1. 0,6



= L' = I' . ' = 8 . '



b. F =



8 ' 12 2 = ' 3







=



EK   EK



18  12 = 12 Nm 0,3



σ 12  = 20 N R 0,6



c. Ianguler =L' – L= I . ' – I .  = I (' – ) = I 



 Δv    R 



= m . R2 . 



I ω 2 12 2 3  2   1 I ω 2 8 3 2 1



10 8



2



2



= m . R . v = 1 . 0,6 . (18 – 12) = 3,6 kg m2 rad/s



12. a.



T1 – W1 Sin 37o = m1 . a T1 – 150 . 0,6 = 15 . 2 T1 = 30 + 90 = 120 N



14. v =



b. W2 – T2 = m2 . a 200 – T2 = 20 . 2 T2= 200 – 40 = 160 N



a.



( T2 – T1 ) R = I ·



a R



=



4 . 9,8 . 14,70 . Sin 30 3



=



96,04 = 9,8 m/s



v = a=



2 (160 – 120) 0,25 = I · 0,25 10 = I . 8 Kunci Jawaban:an Fisika XIA



4 gh 3



96,04 = 3,27 m/s 2 .14,70



a. t = 7



2as



2s  a



2 .14,7 = 9 sekon 3,27 Sagufindo Kinarya



1



b. ER =



2



1



. I . 2 =



V2 2 .I. R2



V2 = 2 . 2 mR . 2 R 1 = mR 1



1



2



4



1 . 0,3 . 96,04 4 = 7,203 Joule



=



15. V 2 = h = = =



2. g. h 1+k 2



=



2 . 10 . h 1+



2 5



7 V  5 20 7 202  5 20



7 . 400 100



= 28 m



Kunci Jawaban:an Fisika XIA



8



Sagufindo Kinarya



T1



UJI KOMPETENSI KESEIMBANGAN DAN TITIK BERAT



𝑇



sin127o T1 0,8



100



2 = 𝑠𝑖𝑛143 𝑜 = sin90o



T



= 0,62 =



100 1



𝑇1 = 80 N dan T2 = 60 N SOAL PILIHAN



Jawaban : A



1. Sumbu x



3.



∑ Fx = 0



T



T 60o



60o



∑ Fx = −T1 . cos 30 + T2 . cos 60 0 = −T1 . ½√3 + T2 . ½ Sumbu y



W=10 N



∑ Fy = 0



Gaya dalam arah sumbu Y : ∑ 𝐹𝑦 = 0 2Tsin60𝑜 = W 1 2T2 √3 = 10 T = 5,8 N Jawaban : C



∑ Fy = T1 . sin 30 + T2 . sin 60 − w 0 = T1 . ½ + T2 . ½√3 − 100 0 = T1 + T2 √3 − 200 Jawaban : E 2.



4. 502 = 502 + 502 + 2 . 50 . 50 . cos𝚹 502 = 502 + 502 + 2 . 502 . cos𝚹 502 - 502 - 502 = 2 . 502 . cos𝚹 - 502 = 2 . 502 . cos𝚹 1 = cos 𝚹 2 𝚹 = 120o Jawaban : Tidak Ada di Option



100N



Kunci Jawaban:an Fisika XIA



9



Sagufindo Kinarya



5. m1



200 cm



7. m2



Z



o



90 o



150



50 cm 50 cm C



80 cm



120o



20 kg



WC . 0,5 = 200 . 0,2 40 WC = 0,5



m1 m2  o sin 120 sin 150o



WC = 80 N 80 MC = 10 = 8 kg



m1 m  2 1 1 3 2 2 m2 1  m1 3 Jawaban : A



Jawaban : B 8. ∑ τ𝐴 = 0 W.ZA – Wc.AB = 0 100 . 4 – Wc . 5 = 0



6.



T sin 37o



T



Wc = 80 N



0,6 m



Jawaban : C



0,8 m



3 = WB · 1 2 3 mA · g = mB · g 2 2 mA = · mB 3 2 = · 45 = 30 kg 3



0,2 m



W b = 15N



9. WA ·



2 kg



∑𝜏 = 0



T sin 37 . 0,8 – 20 . 0,6 – 15 . 0,4 = 0



T.



3 5



.



8 10



= 12 + 6



T=



18 . 50 24



=



3 . 50 4



= 37,5



Jawaban : A



Jawaban : E



Kunci Jawaban:an Fisika XIA



10



Sagufindo Kinarya



10.  τ B = 0 NA.30 = WT . 20



20 = 10000 N 30



NA = 15000· Fy NA + NB 10000 + NB NB



13. T sin 30o . AB = 80. AB T = 160 N Jawaban : D 14. 50.x = 30 (160 – x) X = 60 cm Jawaban : D



=0 = WT



15. m batang . g . 1= T . 2 = 10 . g . 2 m batang = 20 Kg Jawaban : D



= 15000 = 5000 N



Jawaban : B



16. −𝑇 sin 𝜃 . 𝐴𝐵 + 𝑊. 𝐴𝐵 + 1 𝑊𝐵 . 𝐴𝐵 = 0



11.



2



100 cm



30o



1



𝑊 + 2 𝑊𝐵 = 𝑇 sin 𝜃 1



30 + 2 . 18 = 𝑇 . 0,6 𝑇 = 65 𝑁 Jawaban : D



30o C 30 N



20 N



17.



μ



1 1 3   2 tg θ 2 4 3 8



Jawaban : C



AD = 100sin30o = 50 cm 18.



AC = 50sin300 = 25 cm



C



NC



AE = 100cos30o = 50√3 cm ∑ τ𝐴 = 0 30 . AD + 20 . AC = Tsin30o . AB



WO WT



30 . 0,5 + 20 . 0,25 = T . 0,5 . 1 15 + 5 = 0,5 T



B



20



T = 0,5 = 40 N Jawaban : C







WO fg . A



A



A



Fy = 0 NA = Wo + WT = 700 + 300 = 1000 N A = 0 NC.5sin=Wo.3cos+WT.2,5cos



12. ~T sin 30o . l = 8.10. ½.l T = 80 N Jawaban : B Kunci Jawaban:an Fisika XIA



N



11



Sagufindo Kinarya



21. Jawaban : B



3 3 + 300.2,5. 5 5 NC = 427,5 N  . N A = NC 427,5  = = 0,4275= 0,43 1000 Jawaban : B NC .4 =700.3.



22. Dari gambar sudah jelas Z 0 = (4;3) Jawaban : D 23. Jawaban : Tidak bisa dikerjakan karena AD tidak diketahui 24. 𝑦1 = 5 𝑐𝑚 𝑦2 = 5 + 10 = 15 𝑐𝑚 𝐴1 = 200 𝑐𝑚2 𝐴2 = 300 𝑐𝑚2 𝑦 .𝐴 +𝑦 .𝐴 𝑦0 = 1 1 2 2



19. m . g = 5 . sin30o m . 10 = 5 . 0,5 m = 0,5 . 0,5 m = 0,25 kg Jawaban : D



𝐴1 +𝐴2 5.200+15.300



𝑦0 = 200+300 Jawaban : C



20.



= 11 𝑐𝑚



25. I B



II 6



AB = BC =√13



 Z2 (2 ; 2) A2 = 16



= = =



tinggi



Y1 . A1 + Y2 .A2



2



=6



1



Y1 = 6 + 3.3 = 7 , Y2 = 3



20



= 2,6



 Z0 = (2 ; 2,6)



Yo =



𝐴1.𝑌1 +𝐴2 𝑌2 𝐴1 +𝐴2



=



6.7+24.3 6+24



= 3,8



Jawaban : C



Jawaban : Tidak ada di option Kunci Jawaban:an Fisika XIA



4.3



AACDE = A2 = 4 . 6 = 24



4 + 16 20 + 32



20



= √13 − 22 = 3



AABC = A1 =



A1 + A2 5 . 4 + 2 . 16



52



D 4



E 0



 Z1 (2 ; 5) A1 = 4



 Y0 =



C



A



12



Sagufindo Kinarya



2 . 16 – 2 . 4 =2 12 2 . 16 − 1 . 4 1 x0 = = 23 12 Jawaban : E



26. Jawaban : A 27. Y1 = ½ . 6 = 3 A1 = 6 . 6 = 36



y0 =



Y2 = 3 + 23 .3 = 5 A2 = ½ . 6 . 3 = 9 Yo =



31. Dari gambar sudah jelas 1 𝑍0 = (2,12)



𝐴1.𝑌1 +𝐴2 𝑌2 𝐴1 +𝐴2



3 . 36 − 5 .9 = 2,33 36 − 9 Jawaban : A



𝐉𝐚𝐰𝐚𝐛𝐚𝐧 ∶ 𝐂



y0 =



32. Z1 = (3 ; 1,5) cm A1 = 18 cm2 Z2 = (4,5 ; 4) cm A2 = 4,5 cm2 1,5 . 18 + 4 . 4,5 𝑦0 = 18 + 4,5 27+18 = 22,5



28. 𝑥0 = 2 𝑐𝑚 𝑦1 = 3 𝑐𝑚 𝑦2 = 6 + 2 = 8 𝑐𝑚 𝐴1 = 24 𝑐𝑚2 𝐴2 = 6 𝑐𝑚2 𝑦 .𝐴 +𝑦 .𝐴 𝑦0 = 1 𝐴1 +𝐴2 2 1



2



3.24+8.6



𝑦0 = 24+6 = 4 𝑐𝑚 Jawaban : A



= 2 𝑐𝑚 Jawaban : Option jawaban tidak lengkap 33. 𝑦1 = 2 𝑐𝑚 𝑦2 = 2 𝑐𝑚 𝐴1 = 18 𝑐𝑚2 𝐴2 = 6 𝑐𝑚2 𝑦 .𝐴 −𝑦 .𝐴 𝑦0 = 1 𝐴1 −𝐴2 2



29. Z1 = (3;2) Z2 = (5;6) 2 . 24 − 6 . 8 32 8 . (6 + 6) = 8 . (3 + 1) = 3 𝑐𝑚 Jawaban : D 𝑦0 =



1



2



2.18−2.6



𝑦0 = 18−6 = 2 𝑐𝑚 Jawaban : A



30. Z1 = (2,2) cm A1 = 16 cm2 Z2 = (1,2) cm A2 = 4 cm2 Kunci Jawaban:an Fisika XIA



13



Sagufindo Kinarya



SOAL URAIAN 3. Jarak Z ke tumpuan 1 Jarak anak ke tumpuan 2 Jarak tali ke tumpuan 3 .∑ 𝜏 = 0 𝑇. 3 = 𝑊𝐴 . 2 + 𝑊𝐵 . 1 𝑇. 3 = 420.2 + 60.1 𝑇 = 300 𝑁 𝑚𝐵 = 30 𝑘𝑔



1.



T2 T1 T3 T1 T2  o sin 75 sin 150o 20 T  2 0,97 0,5 T2 = 10,35 N



4. fg=  . N = 0,4 . 7,5 . 10 = 30 N



fg W  o sin 120 sin 150o 30 W  1 1 3 2 2



T3 T1  o sin 75 sin 135o



T 20  3 0,97 0,71



W = 10



T3 = 14,64 N



m=



2. = F . r . sin  a.  = 1,5 . 40 = 60 Nm b.  = 1,5 . 40 . sin 37o = 60 . 0,6 = 36 Nm c.  = 40 . 0 = 0 Nm d.  = 20 . 2 + 40 . 2 – 10 . 4 = 80 Nm



3



3 kg x



5.



A



B



WA=150 WBAT =300 WB=200 N N N WA . x + WBAT (x–15)= WB (3–x) 150 . x + 300 (x–15)= 200(3–x) 150 x + 300 x – 450 = 600 –200x 650 x = 1050 x = 1,61 m



Kunci Jawaban:an Fisika XIA



14



Sagufindo Kinarya



6.



yo 



C 𝑇 sin 𝜃



T 𝑇 cos 𝜃



yo 



𝜃



A



10 .13 π r  3.12 π r  - 2,5.10 π r  13 π r  12 π r  10 π r



130  36  25 141   4,03 cm 13  12  10 35



B 1m



3m



9.



Titik pusat keping 1 (besar) sebagai acuan



𝑊𝐵



xo 



∑𝜏 = 0 𝑇 sin 𝜃 . 𝐴𝐵 − 𝑊𝐵 . 3 = 0 𝑇. 0,6 − 100.3 = 0 𝑇 = 125 𝑁



0  r .πr r2 π r2  2 2 22  2 2 2 π r1  π r2 π . 2 . r2  π r2 2



7. Z1 = (2,5 ; 5) cm A1 = 30 cm2 Z2 = (5 ; 1) cm A2 = 20 cm2 2,5 . 30 + 5 . 20 𝑦0 = 30 + 20



2



2







r2 π r2 25  = 5 cm 4  1 π r22 5



Jadi letak titik berat gabungan 5 cm di kanan pusat lingkaran besar.



= 3,5 𝑐𝑚 10.



8.



yo 



y1 A 1  y 2 A 2 A1  A 2



2R . 8R 2  5R . π R 2 8R 2  π R 2



I. Kerucut







II. Silinder



 16  5π  R . R 2  8  π  R 2 



III. Setengah bola



yo 



x 1 A1  x 2 A 2 A1  A 2



16  5π  R 8 π



y1 A 1  y 2 A 2  y 3 A 3 A1  A 2  A 3



Kunci Jawaban:an Fisika XIA



15



Sagufindo Kinarya



UJI KOMPETENSI ELASTISITAS 1. Tetapan pegas: k



F







Δx



25 0,05



 500 N



6. Tetapan pegas: k



2. 𝐼1 = 𝐼2



Δx



Δx 2



0,002



w



Δx







10 0,02



=



0,16.10



 500 N



m



= 40



x 0,04 80.40 3200



x=k =



120



0,16.10 160 6



T



=



160 6



= 0,06 m



Jawaban : C 8. 𝑘𝑝 = 2𝑘 1 𝑘𝑠



1



1



1



1



=𝜋+𝜋+𝜋 𝑘



𝑘𝑠 = 3



1



1



1



1



= 2𝜋 + 𝜋 + 𝜋 + 𝜋



𝑘𝑠 =



2𝑘 7 1



1



𝐸𝑝 = 2 𝑘𝑥 2 = 7 𝑘𝑥 2 Jawaban : D m



9.



F 7 -2  2  1,4 . 10 m k 500



= 1,4 cm Jawaban: B Kunci Jawaban:an Fisika XIA



mn



w



𝑘𝑠



 500 N







k T = 80+40 =



4. Tetapan pegas: 1



Δx



7. k =



3. Jawaban : B







F



Jawaban : B



l A 3  l B 1 dA 1 A 1   A  dB 2 BB 4 EA  .............. ? EB E A ( F lo A l ) A AA l 2   E B ( F lo A l ) B AB l1 1 4 =4x  3 3 Jawaban: A



F



𝐾



K terbesar bila T terkecil Jawaban : C



m



Jawaban: B



k



𝑀



5. T 2 = 4𝜋 2 . 4𝜋 2 𝑚 𝐾= 𝑇2



16



1



1



F . x = 2 m . v2 2 F . x = m . v2 20 . 0,2 = 0,25 . v 2 4 = 0,25 . v 2 4 v2 = = 16 0,25 Sagufindo Kinarya



v = 4 m⁄s Jawaban : E



15. k p = k + k = 2k 1 ks



1



1



𝑠



3𝑘 2 1



𝐸𝑃 = 2 𝑘𝑥 2



1 3(1600)



= 2k



1 2k



3𝐾



17. Ks = 3+1 = 1200 N/m F = Ks . x = 1200 . 0,05 = 60 N Jawaban : A



12. 𝐹 = 𝑘. ∆𝑥 𝐹 4 𝑘= = ∆𝑥 0,1 = 40 𝑁. 𝑚−1 Jawaban : Tidak ada di option



18. E = 4 x 106 N/m2 A = 20 x 10-4 m2  =5m l l o  F E  A   Δl lo A . l 1600 N/m Jawaban : C



13. Ep1 = ½. F. ∆x Ep1 = ½. 50.2 = 50 J Ep2 = ½. 100.4 = 200 J Ep3 = ½. 150.6 = 450 J Jawaban : C 14. Energi potensial: Ep = luas grafik



k = ....?



19. 𝜏 = 2 x 106 N/m E = 2,5 x 108 N/m l = ........?  = 4 meter    . lo E    Δl lo l



 1,6 Joule



2



atau energi potensial: Ep = ½.k.x2 = ½.F.x = ½ (40)(0,08) = 1,6 Joule



 . lo



2 x10 6.4   0,032 m l = E 2,5 x108 Jawaban : C



Jawaban : D



Kunci Jawaban:an Fisika XIA



1



F 30 = 1500 N/m  x 2 x 10 -2 F = k .x =1500 x 5,4.10-2= 81 N Jawaban : C



11. Jawaban : B



40 x 0,08



+ 2k



16. k =



−2 )2



= ∙ ∙ (7. 10 2 2 = 5, 88 𝑗𝑜𝑢𝑙𝑒 Jawaban : Tidak ada di option



=



1



k s = 2 k = 100 N/m 0,4 . 10 ∆x = = 0,04 m 100 Jawaban : B



1



10. 𝑘 = 3𝑘 + 3𝑘 𝑘𝑠 =



=



17



Sagufindo Kinarya



20. 𝑘𝑡𝑜𝑡𝑎𝑙 = 1



3𝑘



F 3 = 75 N/m  x 4 x 10 -2 Jawaban : B



24. K =



2



𝐸𝑝 = 2 𝑘. ∆𝑥 2 1 3 𝐸𝑝 = ∙ ∙ 200 ∙ (10=1 )2 2 2 = 1,5 𝑗𝑜𝑢𝑙𝑒 Jawaban : C 𝐹



25. k1 = 400 N/m k2 = 400 N/m diperoleh : kp = 400 + 400 = 800 N/m x = 5 x 10-2 meter F = .......? F = k . x = 800 . 5x 10-2 = 40 N Jawaban : B



200



21. 𝑘 = ∆𝑥 = 8.10−2 = 2500 𝑁. 𝑚−1 𝑘𝑝 = 2500 + 2500 = 5000 𝑁. 𝑚−1 ∆𝑥 = 8. 10−2 1 𝐸𝑝 = 2 𝑘. ∆𝑥 2 1 𝐸𝑝 = ∙ 5000 ∙ (8. 10−2 )2 2 = 16 𝑗𝑜𝑢𝑙𝑒 Jawaban : C



26. Jawaban : C 27. K =



22. w1 = w kp = 2k x 1 = x jika diseri, maka : 1 1 1 2    ks k k k 1 ks = k 2 w2 = .......? 2 w1 12 k P x P 2k . x2   4 1 k . x2 1 k x 2 w2 2 2 s s w2 = ¼ w 1 = ¼ w Jawaban : A



=



F ∆X 88 0,11



= 800 N⁄m



Jawaban : A 28. Ep = ½. F. ∆x 0,4 = ½. 40. x X = 0,02 m F 40 k = X = 0,02 = 2000 N/m Jawaban : C 29. F = k . x 20 = k . 0,05 20 K = 0,05 = 400 N⁄m  Ep =½ . k . x2



F 1,2 = 200 N/m  x 6 x 10 -3 Jawaban : A



23. k =



=½ . 400 . (0,1)2 = 200 . 0,01 = 2 j Jawaban : A



Kunci Jawaban:an Fisika XIA



18



Sagufindo Kinarya



30. Jawaban: A











E l  l0



Karena identik pertama bukan panjang pegas sama.







800  0,032 2,5 x 10 4



40 x10 2 l0    1,25 meter  0,032 l



31. Jawaban: E ktotal = k + k + k + k



4. 1 = 0,5 % F1 = F



= 4k



2 = 2% F2 = ........?



 FA F   E E AE  F  1 F1 0,5% F     2 F2 2% F2 F2 = 4F 



SOAL URAIAN 1. F1 = 100 N x 1 = 2 cm = 2x10-2 meter x 2 = 5 cm = 5x10-2 meter F2 = ......? F 100 k = = 5000 N/m  x 2 x 10 -2 F2 = k . x 2 = 5000 . 5x10-2 = 25



5. 1 = 4x105 meter l2 = 0,02 meter l1 = 1 x 10-3 meter 2 = ........? 1  (silahkan buktikan) l  1  l1 W2    2  l2 W1



2.  A E F  



= 2 meter = 10 mm2 = 10 x 10-6 m2 = 2,5 x 1011 N/m2 = 8 x 105 N = .......? = .......? F 8 x105  0,8 x1011 N / m2 a   -6 A 10 x 10  0,8x1011  0,32 b.    E 2,5 x 1011



4 x10 5



0,02 2 1x10 3 2 = 2 x 104 N/m2 6. x1 = 4 x10-2 meter x2 = 2 x10-2 meter W = 0,16 joule F2 = .........? W= ½ F . x 2 2W 2.0,16 F   8N x 4 x 10 -2 F  x



3.  = 800 N/m2 l = 40 x 10-2 meter E = 2,5 x 104 N/m2 l0 =.........? Kunci Jawaban:an Fisika XIA







19



Sagufindo Kinarya



F  x F1 x1  F2 x2 8 4 x10 2  2 F2 2 x10 2 F2 = 4N 7. k = 400 N/m g = 10 m/s2 m = 2 kg W = ............? x = F/k = 20/400 = 0,05 meter W = ½ . F. x = ½ . 20 . 0,05 = 5 joule 8.



w 20  = 400 N/m x 5 x10  2 Ep (system)  dengan acuan adalah saat pegas sedang ada beban, sehingga : Ep = ½ (400) (5 x 10-2)2 = 0,5 joule Ep (pegas)  dengan acuan saat pegas tanpa beban, sehingga : Ep = ½ (400) (10 x 10-2)2 = 2 joule k =



Kunci Jawaban:an Fisika XIA



20



Sagufindo Kinarya



0,04 0,1 vB  8000 10000 = 5.10-6 m3 =1.10-5 m3 m  mB 0,04  0,1 ρ A  v A  vB 5 .106  1.10 5 = 9.300 kg/m3 Jawaban : D



4. v A 



UJI KOMPETENSI FLUIDA TAK BERGERAK SOAL PILIHAN 1. v = 1000 ℓt = 1000 dm3 = 1 m3 m = 789 kg m 789   = = 789 kg/m3 v 1 Jawaban : B



5. P = ρ g h = 1000 . 10 . 4,8 = 48.000 Pa = 48 kPa Jawaban : C



2. m =  . v =.Ah m =  . π r2 . h 314= 1 . π r2 . 10 314 r2 = = 3,16 cm 10 Jawaban : B



6. P = ρ ∙ g ∙ h = 1000 . 10 . 2,5 = 25.000 N/m2 Jawaban : C 7. P = ρ ∙ g ∙ h = 1200 . 10 . 12 = 144.000 Pa = 144 Kpa Jawaban : C



3.  air = 1gr/cm3  x = ......? m air = 300 gr m x = 270 gr m=v



ρ1 m  1 ρ2 m2



8. P = ρ ∙ g ∙ h = 1000 . 10 . 2,5 = 25.000 N/m2 Jawaban : C



1 300  ρ2 270 2 = 0,9 gr/cm3 Jawaban : E m



v



Kunci Jawaban:an Fisika XIA



21



Sagufindo Kinarya



9. a . ha = ρm . hm



V1 = V2 𝜋r1 2 . h1 = 𝜋r2 2 . h2



1000 . 8 ℎ𝑚 = = 10 cm 800 Jawaban : D



802 . h1 = 2002 . 1 h1 = 6,25 m Jawaban : A



10. a . ha = ρm . hm 0,92 . 10 ℎ𝑚 = = 1,53 cm 6 Jawaban : D



11. ρm =



1000 . 7,2 7,2+42



15. PA = Po + 16 cmHg = 76 + 16 = 92 cmHg Jawaban : A



= 600 kg. m−3



16.



Jawaban : D 12. pb =



m .g π . r2



22 . 9,8



pb =



22 7



.



0,72



= 0,014. 105 N. m−2



= 𝜋.202



F1 = 1 N Jawaban : B



14. F1 = = =



A1 A2



. F2



𝜋r1 2 𝜋r2 2 802 2002



Kunci Jawaban:an Fisika XIA



vb



= 960 kg/m3



18. F = W - FA = mg - 𝜌𝑧𝑒 . g . Vb = 4500 . 10 – 1025 . 10 . 3 = 45000 – 10 . 250 . 3 = 45000 – 30 . 750 = 14.250 Jawab : A



400



𝜋.12



= 1200 . 0,8



17. P = 1,06 . 10-3 . 10 . 1,7 = 18.020 Pa = 18 . 103 Pa Jawaban : B



F1 F  2 A1 A2 𝐹1



ρc g v c = ρc v c =



Jawaban : D



ρm = 1 + 0,01 = 1,01 bar Jawaban : C



13.



ρb g vb ρb vb ρb vb ρb



19. Pada cairan A 𝜌𝐵 = 900 𝑘𝑔. 𝑚−3 2 𝑉𝐵 = 3 𝜌𝐴 . 𝑉𝐴 = 𝜌𝐵 . 𝑉𝐵



. F2 . 2000 = 320 N 22



Sagufindo Kinarya



2



24. FA = ρ v g



𝜌𝐴 . 1 = 900. 3 𝜌𝐴 = 600 𝑘𝑔. 𝑚−3



= 800 · m · g ρ



Pada cairan B 𝜌𝐵 = 1200 𝑘𝑔. 𝑚−3 𝜌𝐴 = 600 𝑘𝑔. 𝑚−3 𝜌𝐴 . 𝑉𝐴 = 𝜌𝐵 . 𝑉𝐵 600.1 = 1200. 𝑉𝐵 1 𝑉𝐵 = 2 Jawaban : C



= 800 ·



ρ2



=



V2 V1



=



¾V ½v



=



25. wA − wzA = FA FA = ρ . V . g



3



V=



2



21. FA = ρ v g = 1000 . (0,4 . 10-4) . 10 = 0,4 N Jawaban : D



4



800 . 5V V



Volume yg muncul : Vmuncul= Vb - Vtercelup = Vb - 0,75 Vb = 0,25 Vb 𝑉𝑡𝑒𝑟𝑐𝑒𝑙𝑢𝑝 0,75𝑉𝑏 = 0,25𝑉 = 3 𝑉



= 64 kg. m−3 Jawaban : B



𝑚𝑢𝑛𝑐𝑢𝑙



23. ρb ∶ Vb = ρc ∶ Vc ρc



=



Vc Vb



=



0,3V 0,5v



=



𝑏



Jawaban : B 3 5



Jawaban : A



Kunci Jawaban:an Fisika XIA



= 2 . 10−3



26. Benda terapung : b . Vb = f . Vtercelup (0,9).Vb = (1,2) Vtercelup 0,9 Vtercelup = 1,2 Vb = 0,75 Vb



22. Vb . ρb = Vza . ρza



ρb



50−30 103 . 10



wu 50 g 10 ρ= = V 2. 10−3 ρ = 2500 kg. m−3 Jawaban : C



Jawaban : B



ρp =



· 10



= 0,2 N Fkawat = 2 – 0,2 = 1,8 N Jawaban : C



20. ρ1 ∶ V1 = ρ2 ∶ V2 ρ1



0,2 7900



27. Karena tegangan permukaan zat cair cenderung memperkecil permukaan Jawaban : C 23



Sagufindo Kinarya



2. P



28. 0,4



= Po + ρ g h = 1 . 105 + 1030 . 10 . 80 = 9,24 . 105 Pa



W



3. W = ρ .v.g = 1000 . 0,4 (10 . 0,3) . 10 = 12 KN Jawaban : C



F1 F  2 A1 A 2 F1 6000  20 300 F1 = 1200 N



4. a. FA 29. F = m . g



= ρ .V . g = 1000 . A . h . g = 1000 . 40 . 5 . 5 . 10 = 1 . 107 Jawaban : D



2 γ cos θ 30. y  ρgr 2 . 0,48 . 0,8 = 13620 .10 . 12 . 23 .10 3



b. FA 16 V



= ρ .V . g = 800 . V . 10 = 2 . 10-3 m3



c. ρ



=



m V



=



5 2 .10 3



= 2.500 kg/m3



y = 0,0169 m = 1,69 cm Jawaban : B



5. P1 h1 = P2 h2 1000 . 40 . 10-2 = ρ 2 . 31 . 10-2 ρ 2 = 1290,3 kg/m3



URAIAN 1. a. P = gh = 1000.10.15 = 15 . 104 Pa b.



6. Wu = 0,3825 N Wa = 0,3622 N FA = Wu – Wa = 0,3825 – 0,3622 = 0,0203 N



P= ρ.g.h = 1000 . 10 . 4 = 4 . 104 Pa



FA



Kunci Jawaban:an Fisika XIA



= (50 - 34) = 16 N



24



= ρ .V . g



Sagufindo Kinarya



V



=



0,0203  1,052 .107 m 3 19300 .10



ρ b = 18



Vη + 4 r2 g



= 18 7. -







ρc



0,15 . 10







4 1 . 10 2 2 . 10



mesin pengangkat mobil alat pengepres alat pengukur tensi sistem pembuangan material pada mobil



+ 1000



= 6750 +1000 = 7.750 kg/m3 10. h 



8. V = (10.10-2)3 =1000.10-6 = 1 . 10-3 m3



2 γ cos θ ρgr o = 2 . 6,4 . cos1203



2000 .10 . 2 .10



= 0,16 m ρc=1200 kg/m3



ρ k = 800 kg/m3



a.



V



h



ρk Vk  ρc Vc ρ k Vc  ρ c Vk



V 800  c 1200 1.10 3



Vc h h 9.



V



= 6,67 . 10-4 m3 Vc 6,67.10-4



ρ b  ρ c  Vb g 6π ηr



ρ b  ρ c   6 π V η r



Vb  g 6π V ηr = 3 4 3 πr g



= 18 4 Kunci Jawaban:an Fisika XIA



Vη r2 g 25



Sagufindo Kinarya



7. A1 V1 = A2 V2 10 . 2 = 5 . V2 V2 = 4 𝑚⁄𝑠



UJI KOMPETENSI FLUIDA BERGERAK



P1 + ½ ρ V1 2 + 0 = P2 + ½ρV1 2 + ρgh2



SOAL PILIHAN



40000 + ½ 1000 .4 = P2 + ½ . 103 .16 + 103 . 10 . 0,6



v = √25 = 5,0 m/s Jawaban : C



1. Kontinuitas : vA . AA = vc . Ac vA . 6 = v . 4 2 vA = 3 v Jawaban : D



8. ½𝜌V12 + 𝜌gh1 = ½𝜌V22 + 𝜌gh2 ½ . 1000 . 36 + 1000 . 10 . 0,7 = ½1000V22 + 1000 . 10 . 1,25



2. Kontinuitas : vc . Ac = vA . AA Vc . 3 = vA . 8 8 Vc = 3 v Jawaban : C



18000 + 7000 = 500V22 + 12500 25000 – 12500 = 500V22 25 = V22 V22 = 5 𝑚⁄𝑠



Jawaban : C



8



3. v2 = 3 v Jawaban : D 2



9. V1 . A1 = V2 . A2 V2 =



5. v = √2 g ∆h = √2.10.1.8 = 6 m/s Jawaban : A



2



P1 – P2



2𝑃



2 .9 . 106 1000 .10 .90



3 Q = 20 𝑚 ⁄𝑠 Jawaban : C Kunci Jawaban:an Fisika XIA



=







v



2 2 1 2 1 2 1 . 1000(92-12) 2



ρ v







105 – P2 = P2 = 100000 – 40000 = 60000 N/m2 Jawaban : E



= 𝜌𝑔ℎ



Q=



=4



10. A1 V1 = A2 V2 𝑅 15 V2 = (𝑅1 )2. V1 =( 5 )2 . 1 = 9 m/s



6. 50% mgh = P . t 1 𝜌𝑉𝑔ℎ =P.t 2 𝑡



¼ . π . 62 2



p + ½ ρ v + ρ .g .h = C p + ½. 103 . 12 + 103 . 10.0 = 52,5+½. 103 . 42 + 103 . 10.0,2 P = 80 kPa Jawaban : C



2



4. v1 = 8 v2 = 8 . 2 = 0,5 Jawaban : A



𝑉



1 . ¼ . π . 122



26



Sagufindo Kinarya



16. v = 2 g h  2  10  0,8 = 4 m/s Jawaban : C



11. V  A . v t



100 . 10-6 =



25 .v 100



v = 0,4 mm/s Jawaban : A



17. v = √2 g ∆h = √2 . 10 . 0,2 = 2 m/s



12. Sama dengan soal no. 9 p + ½ ρ v2 + ρ . g . h = C 1 9,1 . 105 + 2 . 103 + 103 . 10 . 5 = 5



1



3



2



2.ℎ



R = v . t = 2 .1 = 2 m Jawaban : A



3



2 . 10 + 2 . 10 . 𝑣2 + 10 . 10 . 1 𝑣2 = 40 𝑚/𝑠 Jawaban : D m.g.h



18. t = √



2.ℎ 𝑔



=√



2.0,5 10



=



1 √10



s



jarak pancar air : x = v . t 1 1 = v. 10



ρ.V.g.h



13. P = t = t P = 103 . 10.20 = 2. 105 Pguna = 55%. P = 1100 kw Jawaban : B



√



v = √10 m/s Jawaban : B 19. h



14. h1 = 2 m h2 = 20 cm = 0,2 m θ = 60o g = 10 m/s v = 2 . g . h1  h 2 



= 0,8 m



A = 5 cm2 v = 2 g h  2  10  0,8 = 4 m/s Q = A . v = 5 . 10-4 . 4 = 20 . 10-4 m3/s Q =V



= 2 .10 . 2  0,2 = 6 m/s Jawaban : B



2.



10-3



=



t V 60



. V = 0,12 m3 = 120 liter Jawaban : E



15. v = 6 m/s h = ......? v2 62 h=  2 g 2 .10 = 180 cm Jawaban : D



Kunci Jawaban:an Fisika XIA



2.5



𝑡 = √ 𝑔 = √ 10 = 1 s



20. kecepatan kebocoran air : v = √2𝑔ℎ𝑘𝑒𝑑𝑎𝑙𝑎𝑚𝑎𝑛 = √2.10.1 = √20 m/s Jarak pancar air : 27



Sagufindo Kinarya



X = vo . t 2 t = 20 = √



26. v1 = 1 √5



1



= 5 √5 s



 A1   A2



Jawaban : B = 21. v



2 g h  2 10 1,8



=



27. v



=



22. Jawaban : C =



23. Jawaban : C 1



2



m



s



Q1 = Q2 A1 v1= A2 v2



π r  . v = π r  . v 2



2



1



2



2 v1 = v 2 d 2 2



1 2 d 2 



v1 = 4 v2 v1 = 4 . 3 = 12 m/s 2 2 P1 – P2 = 21 ρ v 2  v1







5



P1 = 2,675 . 10 N. m



2 g h ρ' ρ



2 .10 .13600 . 2 , 6 .102 32



29. Jawaban : B







30. Jawaban : B



P1 − 2. 105 = ½. 103 (32 − 122 )



URAIAN



−2



1. v1 = 0,5 m/s d = 4 cm Q, v2, Pabsolut = ......? a. Q = A . v = (π r2) . v



Jawaban : C 25. Jawaban : C



Kunci Jawaban:an Fisika XIA



= 4 m/s



28. A1 v1 = A2 v2 1 1 π D12 . v1 = 4π D22 . v2 4 42 . 0,8 = 122 . v2 Q = A2 v2 1 = π . 122 . 10-4 . 0,089 4 = 0,001 m3/s Jawaban : E



P1 = ......?



2 1



9 25 1 16



2 .10 . 45 .10 2  5 .10 4    4   4 .10 



= 47 m/s Catatan : h seharusnya 2,6 cm Jawaban : C



d2



P = 2 . 10-5 N/m2 v2 = 3



2



   1 







Jawaban : B



= 6 m/s Jawaban : C



24. d1 =



2gh



28



Sagufindo Kinarya



= π (2 . 10-2) 2 . 0,5 = 6,28 . 10-4 m3/s 1 dm3 1 kg b. A1 v1 = A2 v2 d12 v1 = d22 v2 v2 =  d  v1



P1 + 1



2



P1  1 = 1







2



d   1



 4 .102    . 0,5 2   0,6 . 10 



=







5. h raksa



ρudara ρraksa v=



c. P1  1 2 ρ v12  ρ g h 1  C P2 = 4,18.105 – 0,92945.105 = 3,25.105 Pa



Kunci Jawaban:an Fisika XIA



1 . 1,3 602 - 152  2



= 0,8 cm = 1,36 kg/m3 = 13,6 . 103 kg



vudara = ......?



= 9,375 π . 10-4 m3/s



(1 . 10-2)2 . 15 v2 10-4 . 15 v2



2



= 2193,75 N/m2 1 cmHg = 1333,2 h = 1,645 cm 4. Qaorta = Qkapiler π . 12 . 30= N .π .(4.10-4)2 5.10-4 30 = N . 16 . 10-8 . 5.10-4 N = 3,75 . 1011



1 π . 22 . 10-4 . 9,375 4



3. A = π r2 A r2 A1 v1 r12 v1







2



=



= 22,2 m/s 2 2 c. P1 – P2 = 21 ρ v 2  v1 P1 – 1.10-5= 2 2 1 2 1000 22,2  0,5  P1 = 3,47 . 105 Pa 2. a. A1 v1 = A2 v2 1 π D12 . v1 = 1 π D22 . v2 4 4 2 2,5 . 6 = 22 . v2 v2 = 9,375 m/s b. Q = A2 . v2







2



2



ρu v2 - v1



2



=



= 60 m/s 2 ρu v1 = P2 + 1 ρ v 2 2 2



=



2 ρr g h ρudara 2 .13,6 .103 .10 . 0,8 .102 1,36



= 40 m/s



= A2 v2 = r22 v2 1



= ( 2 . 10-2)2 . = 0,25 . 10-24 . v2 = 15 0,25



29



Sagufindo Kinarya



T0 = 1100 – 600 = 500 C Jawaban : C



UJI KOMPETENSI SUHU DAN KALOR



8. Jawaban : C 9.



1. 313 K = …. R 313−273 100



=



𝑡°𝑅 80



t⁰R = 32



Jawaban : A



10. 𝑉𝑡𝑢𝑚𝑝𝑎ℎ = 𝑉𝑡 𝑎𝑖𝑟 − 𝑉𝑡 𝑔𝑒𝑙𝑎𝑠 𝑉𝑡 𝑎𝑖𝑟 = 𝑉0 (1 + 𝛾. ∆𝑇) 𝑉𝑡 𝑎𝑖𝑟 = 1000(1 + 10−4 . 70) = 1007 𝑐𝑚2 𝑉𝑡 𝑔𝑒𝑙𝑎𝑠 = 𝑉0 (1 + 3. 𝛼. ∆𝑇) 𝑉𝑡 𝑔𝑒𝑙𝑎𝑠 = 1000(1 + 3.9. 10−6 . 70) = 1001,89 𝑐𝑚3 𝑉𝑡𝑢𝑚𝑝𝑎ℎ = 1007 − 1001,89 = 5,11𝑐𝑚3 Jawaban : C



2. Soal kurang lengkap, tidak bisa dikerjakan. 3.



∆l1 l01 .∆t1



=



∆l2 l02 .∆t2 2



∆l2 = 100.100 6.120 = 0,144 mm Jawaban : tidak ada di optional 4. ∆A = A0 . 2α. ∆t =15 . 2 . 1,8 . 10-5 . 80 = 4,32 . 10-1 cm2 Jawaban : B



11. 𝑉𝑡𝑢𝑚𝑝𝑎ℎ = 𝑉𝑡 𝑎𝑎𝑠𝑒𝑡𝑜𝑛 − 𝑉𝑡 𝑔𝑒𝑙𝑎𝑠 𝑉𝑡 𝑎𝑠𝑒𝑡𝑜𝑛 = 𝑉0 (1 + 𝛾. ∆𝑇) 𝑉𝑡 𝑎𝑠𝑒𝑡𝑜𝑛 = 6(1 + 1,5. 10−3 . 40) = 6,36 𝐿 𝑉𝑡 𝑔𝑒𝑙𝑎𝑠 = 𝑉0 (1 + 3. 𝛼. ∆𝑇) 𝑉𝑡 𝑔𝑒𝑙𝑎𝑠 = 6(1 + 3. 10−5 . 40) = 6,0072 𝐿 𝑉𝑡𝑢𝑚𝑝𝑎ℎ = 6,36 − 6,0072 = 0,3528 Jawaban : A



5. ∆l = l0 . α. ∆t = 1 . 10−5 .40 = −4 l0 . 10 .4 lt = l0 + ∆l 50,05 = 1,0004.l0 L0 = 50 cm Jawaban : A 6. Jawaabn : E 7. ∆l = l0 . α. ∆t 0,0288 = 0,12.0,004 . Δt Δt = 600 C Kunci Jawaban:an Fisika XIA



Qserap = Qlepas 75(40-20) = 50 (T-40) 1500 = 50 (T-40) 30 = T – 40 T = 70℃ Jawaban : A



12.



30



Qlepas = Qserap 250(100 - TA) = 400 (TA – 35) 2500 – 25TA = 40TA – 1400 Sagufindo Kinarya



3 . 1400 (80 − 20) = 10 . 4200 (20 − 𝑇𝑎𝑖𝑟 )



3900 = 65TA TA = 60℃ Jawaban : B



𝑇𝑎𝑖𝑟 = 14℃ Jawaban : C 19.



Qlepas = Qserap



13.



𝑄𝑠𝑒𝑟𝑎𝑝 = 𝑄𝑙𝑒𝑝𝑎𝑠



50 . Clogam (41,8 – 37) = 50 . 1 (37 – 29,8) 𝑚𝑎𝑖𝑟 . 𝐶𝑎𝑖𝑟 . ∆𝑇 = 𝑚𝐴𝐿 . 𝐶𝐴𝐿 . ∆𝑇 400 . 1 (40 − 25) = 500 . 0,2 (𝑇𝐴𝐿 − 40)



Clogam . 4,8 = 7,2 Clogam = 1,5 𝐾𝑎𝑙⁄℃ Jawaban : C 14.



𝑇𝐴𝐿 = 100℃ Jawaban : C



𝑄𝑙𝑒𝑝𝑎𝑠 = 𝑄𝑡𝑒𝑟𝑖𝑚𝑎 𝑚1 . 𝐶1 . ∆𝑇1 = 𝑚2 . 𝐶2 . ∆𝑇2 100 . 1 (90 − 𝑥 ) = 200 . 1 (𝑥 − 30) 𝑥 = 50𝑜 𝐶 Jawaban : C



20.



𝑚𝐴 . 𝐶𝑎𝑖𝑟 . ∆𝑇 = 𝑚𝐵 . 𝐶𝑎𝑖𝑟 . ∆𝑇 𝑚𝐴 . 1 (20 − 0) = 60 . 1 (50 − 20)



𝑚𝐴 = 40 𝑔𝑟𝑎𝑚 Jawaban : A



𝑄𝑠𝑒𝑟𝑎𝑝 = 𝑄𝑙𝑒𝑝𝑎𝑠



15.



21. Soal tidak lengkap. Tidak bisa dikerjakan.



𝑚𝑒𝑠 . 𝐿+ 𝑚𝑒𝑠 . 𝐶𝑎𝑖𝑟 . ∆𝑇 = 𝑚𝑎𝑖𝑟 . 𝐶𝑎𝑖𝑟 . ∆𝑇



𝑚 .80 + 𝑚 .1(5 − 0) = 2 . 1 (𝑥 − 30)



85 𝑚 = 5100 m = 60 gram



22. Q = m . u 20 . 60 . 100 = 10-2 . u 1,2 .105 u= 10−2 = 1,2 . 10-7 J/Kg Jawaban : A



Jawaban : A 16.



𝑄𝑠𝑒𝑟𝑎𝑝 = 𝑄𝑙𝑒𝑝𝑎𝑠 (satuan m = kg) 𝑚𝑐𝑘 . 𝐶𝑐𝑘 . ∆𝑇 = 𝑚𝑡 . 𝐶𝑡 . ∆𝑇



190 . 𝐶𝑐𝑘 . (36 − 20) = 20 . 8𝐶𝑐𝑘 (𝑇𝑡 − 36)



𝑇𝑡 = 55𝑜 𝐶



23. Jawaban : D



Jawaban : B 17.



24. Soal tidak sesuai dengan babnya.



𝑄𝑠𝑒𝑟𝑎𝑝 = 𝑄𝑙𝑒𝑝𝑎𝑠



25. Jawaban : E



40 (𝑇𝐴 − 25) = 60 (90 − 𝑇𝐴 )



𝑇𝐴 = 64℃ Jawaban : D 18.



𝑄𝑠𝑒𝑟𝑎𝑝 = 𝑄𝑙𝑒𝑝𝑎𝑠



26.



𝑚𝑡𝑏 . 𝐶𝑡𝑏 . ∆𝑇 = 𝑚𝑎𝑖𝑟 . 𝐶𝑎𝑖𝑟 . ∆𝑇 𝑚𝑡𝑏 . 0,1 (100 − 36) = 128 . 1 (36 − 30)



𝑄𝑙𝑒𝑝𝑎𝑠 = 𝑄𝑠𝑒𝑟𝑎𝑝 𝑚𝑡 . 𝐶𝑡 . ∆𝑇 = 𝑚𝑎𝑖𝑟 . 𝐶𝑎𝑖𝑟 . ∆𝑇



Kunci Jawaban:an Fisika XIA



𝑄𝑙𝑒𝑝𝑎𝑠 = 𝑄𝑠𝑒𝑟𝑎𝑝



31



𝑚𝑡𝑏 = 120 𝑔𝑟𝑎𝑚 Jawaban : B Sagufindo Kinarya



𝑄𝑠𝑒𝑟𝑎𝑝 = 𝑄𝑙𝑒𝑝𝑎𝑠



27.



𝑚𝑎𝑖𝑟 . 𝐶𝑎𝑖𝑟 . ∆𝑇 + 𝑚𝐴𝐿 . 𝐶𝐴𝐿 . ∆𝑇 = 𝑚𝑏 . 𝐶𝑏 . ∆𝑇 440 . 1 (𝑡𝐴 − 20) + 500 . 0,22 (𝑡𝐴 − 20) = 200 . 0,11



𝑡𝐴 = 22,12 ℃ Jawaban : A 28. Q = m C ∆T 420= 0,1 . C . 10 C = 420 J/Kg.K Jawaban : B



Kunci Jawaban:an Fisika XIA



32



Sagufindo Kinarya



7. Dengan cara seperti no.6 Jawaban A



UJI KOMPETENSI RAMBATAN KALOR



8. Jawaban A



1. Jawaban D 2.



9. Dengan cara seperti no.6 Jawaban D



HR = HS 𝐾 . 𝐴. = 𝑆 𝑙



𝐾𝑅 . 𝐴 . ∆𝑇 𝑙𝑅



𝐾𝑅 . (𝑇−0)



∆𝑇



10.



𝑆



3𝐾 . (100−𝑇)



𝐻𝐴 𝐾𝐴 𝐻𝐴



= 𝑅 60 T = 80℃ Jawaban C 80



𝐾𝐵



= =



𝐻𝐵 𝐾𝐵 𝐾𝐴 𝐾𝐵



= 1⁄4



3. Jawaban E



𝐾𝐵 𝐾𝐵



= 1⁄4



Jawaban A



4. Jawaban A



11. KP = 2KQ



5. Jawaban E



HP = HQ 𝐾 . 𝐴. = 𝑄 𝑙



𝐾𝑃 . 𝐴 . ∆𝑇



6. HR = HS 4K (TX1 – 20) = 2K (TX2 – TX1) 2TX1 – 40 = TX2 – TX1 3TX1 – TX2 = 40… (1)



𝑙



2KQ (60 - TX) = KQ (TX – 30) TX = 50℃ Jawaban D 12. K1 = 4K2



HQ = HR 2K (TX2 – TX1) = K (160 – TX2) 2TX2 – 2TX1 = 160 – TX2 3TX2 – 2TX1 = 160… (2)



H1 = H2 𝐾 . 𝐴. = 2 𝑙



𝐾1 . 𝐴 . ∆𝑇 𝑙



∆𝑇



4K2 (50 - TX) = K2 (TX – 0) 200 – 4TX = TX 200 = 5TX TX = 40℃ Jawaban B



3TX1 – TX2 = 40 |x 3| -2TX1 + 3TX2 = 160 |x 1| 9TX1 – 3TX2 = 120 -2TX1 + 3TX2 = 160+ 7TX1 = 280 TX1 = 40℃ Jawaban E Kunci Jawaban:an Fisika XIA



∆𝑇



33



Sagufindo Kinarya



13. KP = 4KQ HP = HQ KP (TX – 25) = KQ (200 – TX) 4KQ (TX – 25) = KQ (200 – TX) 4TX – 100 = 200 – TX 5TX = 300 TX = 60℃ Jawaban E 14. Jawaban D 15.



H =



𝐾



. 𝐴 . ∆𝑇



𝑙 0,8 . 4 . 5



=



3,2 . 10−3



= 3750 Jawaban D



𝐽⁄ 𝑠



16. Jawaban A 17. KP = 4KQ ∆TP = ∆TQ LP = 4L LQ = 3L AP = 2A AQ = A 𝐻𝑃 𝐻𝑄



𝐾𝑃 . 𝐴𝑃 . ∆𝑇𝑃 𝑙𝑃 𝐾𝑄 . 𝐴𝑄 . ∆𝑇𝑄 𝑙𝑄



=



𝐴 . 𝑙



= 𝐴𝑃 . 𝑙 𝑄 𝑄



=



𝑃



2𝐴 3𝐿 𝐴



3



. 4𝐿 = 2



Jawaban C Kunci Jawaban:an Fisika XIA



34



Sagufindo Kinarya



UJI KOMPETENSI



3 V . T P1 = 2 P2 V . 2T P1 3 = P2 4



TEORI KINETIK GAS A.



Jawaban : D



SOAL PILIHAN 1. Jawaban : B



11. Soal sama dengan no.10 (perubahan V tidak ada) Sehingga tidak bisa dikerjakan



2. Jawaban : C 3. Jawaban : A



12. T = 300oK , V = 3 . 10-3 m3



4. Jawaban : D



Pers. Gas ideal P.V = n.R.T



5. Jawaban : C 5



10 .3.10-3 = n.8,31.300 6. P1 v1 = P2 v2 1 v2 = 2 v1 P1 v1 = P2 P2 = 2 P1 Jawaban : B



n = 0,12 mol N = n.NA = 0,12 . 6,02 . 1023 1 2 v1



= 0,72 . 1023partikel Jawaban : B 13. Pers. Gas ideal P.V = n.R.T



7. Jawaban : B



P . 1 = 5 . 8,31 . 350



8. P1 v1 = P2 v2 1 v2 = 2 v1



P = 1,4 . 104 N/m2



P1 v1 = P2 P2 = 2 P1 Jawaban : D



Jawaban : D 1 2 v1



14.



P1 P2



=



Kunci Jawaban:an Fisika XIA



T1



=



P2 . V 2



P. V 2P. V = 3 T T 2 3 V2 = V 4 Jawaban : A



9. Jawaban : A



10.



P1 . V 1



V1 . T1



T2



V2 . T2 35



Sagufindo Kinarya



15.



P1 T1



=



𝑃



P2 T2



= 120.000



4𝑃



=



300



= 100.000 + 1000 . 10 . 2



P1 V1 P2 V2  T1 T2



𝑇2



T2 = 1200oK = 927oK Jawaban : D



1,2 .105 . V1 105 . 1,25 V1  T1 300



16. Tidak bisa dikerjakan. Perubahan volume tidak ada



T1 = 288 K = 15 oC



17. Pers. Gas P.V = N.k.T



Jawaban : D



30. 1,38 = N.1,38.10-23.300 N = 1022 partikel



21.



Jawaban : C



Jawaban : D



18.



P. V T



22. P1 . V1 = P2 . V2



3P . 2V



=



T1



P1 . V1 = 2P1 . V2



T1 = 6 T



V2 = ½V1



Jawaban : E



Jawaban ∶ C



19. Soal kurang data suhu dinaikkan mejadi 127oC dan tekanan dinaikkan dua kali. P1 V 1 T1 1.300 300



= =



23.



P2 V2 T2



V1  V2 V



2𝑉2



V2



400



V2 = 200 cm3







T1 T2 300 450



V2 = 1,5 V Jawaban : C



Jawaban : D 20. P2 = Po = 100 kPa P1 = Po + Kunci Jawaban:an Fisika XIA



ρ gh 36



Sagufindo Kinarya



24.



2P1



T2 V2  V1 T1 V2  V1



T2



31. Jawaban : C 32. P =



m2



500 4  v2 32



m =



P0 .V0 .4T0 2.V0 .T0



Jawaban : B T1 P.V T



=



5 T 4



3



T2



=



P1



36. U =



T1



Kunci Jawaban:an Fisika XIA



4P T2



Jawaban : D



Jawaban : D P2



300



=



T2 = 1200 K = 927 0C



P2 = P



29.



P



35.



T2 3 P2 4V



5



3ρ v v2



34. Jawaban : B



P2 V2



=



3ρ . v m



3 .105 .1,5 .103 = 7502 = 0,00089 kg = 0,8 gram Jawaban : D



26. Jawaban : B



P1 V1



3ρ  ρ



33. v =



v2 = 1414 m/s Jawaban : C



28.



2 NEk 3 v



P Ek Jawaban : C



m1



27. P2 =



300



30. Jawaban : A



V2 = 2 V1 Jawaban : C



25.



P1



T2 = 600 K = 327 ℃ Jawaban : D



2T1 T1



v1  v2



=



37



5 Nkt 2 Sagufindo Kinarya



=2,5.6,02.1023.1,38.10-23.300



=



50 2  75 2  100 2  125 4



= 6,23.103 J Jawaban : D 37. Soal tidak sesuai dengan



=



babnya



= 3437,5 = 91,85 m/s



B. SOAL URAIAN



4. a. Erata-rata



P1 V1 P V  2 2 T1 T2



1.



1,5. 0,75 P2 . 3  300 500



2. Vef



P2



= 1,6 Pa



=



3P ρ



= 9,936.10-20 Joule b. U = n NA Ek = 5 .6,02.1023 . 9,936.10-20 = 2,99 . 105 J



n NA 2 EK 3 v 3 PV n = 2 Ek  NA



5. P 



3. 3.10 5 0,9 3 = 10 m/s



3 2 .105. 4 .103 = 2 3,6.1022 . 6,02 .1023



v1  v 2  v 3  v 4 N



= 5,573



= 50  75  100  125 4 350 = = 87,5 m/s 4 b. Vef



6. Ek



2



2



v1  v 2  v 3  v 4 N Kunci Jawaban:an Fisika XIA



=



5 KT 2



=



5 1, 38 .1023. 600 2



= 2,07.10-20 J



=



2



1



= 12 ( 2 K T) =6KT = 6 .1,38.10-23



.1200



=



3. a. vrata-rata=



33750 4



2



38



Sagufindo Kinarya



7.



Vef . N 2  Vef . H 2



MH 2



Vef . N 2  Vef . H 2



2 28



Vef . N 2  Vef . H 2



1 14



V2



MN 2



10. U =



=



Vef . H 2  493 14



5 n NA K T 2



5 .2.6,02.1023.1,38.10-23.300 2 = 1,246 . 104 J



= 1844,6 m/s 8. a. P.V



= 1.100 liter



=NKT



N



=



PV KT



=



1.105. 4 3 π .103.106 1, 38 .1023. 293



= 3,29 π . 1022 molekul b. Ek = =



5 KT 2



5 1, 38 .1023. 293 2



= 1,01 . 10-20 J



3R t M 3 . 8,314 . 293 = 4



c. Vefektif =



= 9.



P1 V1



1827 =42,74 m/s



= P2 V2



440 . 50 = 20 V2 Kunci Jawaban:an Fisika XIA



39



Sagufindo Kinarya