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contoh soal grafik fungsi linear beserta jawabanya

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Electric Fireplaces - International Builders' Show

Electric Fireplaces BUI LT- I N | HD L I N E A R Just plug in one of our electric fireplaces for instant ambience in any room. All of our Built-in and HD Linear models are CSA tested for safety, fit any room style, and are easy to use. They are cool to the touch but will add warmth to any space. Columbia Corner Surround with Arched Rectangle built-in fireplace front shown in Cherry Stain Finish. A division of the Outdoor GreatRoom Company Electric LINEAR Cool to the touch and warm to the eye This hot fireplace feels cool to the touch, yet, puts out heat when you want it. It hangs like a piece of art – easy to install and sets up in minutes. Place it on any interior wall with an electric outlet. Just plug it in and have a beautiful fire instantly. Model GE-58 Select your ideal fire. Fireplace Features Choose from a large range of options – from fire intensity, to backlighting color, to heat range and more! • Ambient backlighting comes standard on all models. • High efficiency LED on fire and backlighting— Uses only 15 watts. • Operating costs as low as a penny per day with flame and backlighting, and 9-18 cents per hour with the heater on.

LM1949 Injector Drive Controller (Rev. C) - Texas Instruments

APPLICATIONS The LM1949 linear integrated circuit serves as an excellent control of fuel injector drive circuitry in modern automotive systems. The IC is designed to control an external power NPN Darlington transistor that drives the high current injector solenoid. The current required to open a solenoid is several times greater than the current necessary to merely hold it open; therefore, the LM1949, by directly sensing the actual solenoid current, initially saturates the driver until the “peak” injector current is four times that of the idle or “holding” current (Figure 19–Figure 22). This guarantees opening of the injector. The current is then automatically reduced to the sufficient holding level for the duration of the input pulse. In this way, the total power consumed by the system is dramatically reduced. Also, a higher degree of correlation of fuel to the input voltage pulse (or duty cycle) is achieved, since opening and closing delays of the solenoid will be reduced.

LM35 Precision Centigrade Temperature Sensors (Rev. D)

Product Folder Sample & Buy Technical Documents Support & Community Tools & Software LM35 www.ti.com SNIS159D – AUGUST 1999 – REVISED OCTOBER 2013 LM35 Precision Centigrade Temperature Sensors FEATURES DESCRIPTION The LM35 series are precision integrated-circuit temperature sensors, with an output voltage linearly proportional to the Centigrade temperature. Thus the LM35 has an advantage over linear temperature sensors calibrated in ° Kelvin, as the user is not required to subtract a large constant voltage from the output to obtain convenient Centigrade scaling. The LM35 does not require any external calibration or trimming to provide typical accuracies of ±¼°C at room temperature and ±¾°C over a full −55°C to +150°C temperature range. Low cost is assured by trimming and calibration at the wafer level. The low output impedance, linear output, and precise inherent calibration of the LM35 make interfacing to readout or control circuitry especially easy. The device is used with single power supplies, or with plus and minus supplies. As the LM35 draws only 60 μA from the supply, it has very low self-heating of less than 0.1°C in still air. The LM35 is rated to operate over a −55°C to +150°C temperature range, while the LM35C is rated for a −40°C to +110°C range (−10° with improved accuracy). The LM35 series is available packaged in hermetic TO transistor packages, while the LM35C, LM35CA, and LM35D are also available in the plastic TO-92 transistor package. The LM35D is also available in an 8-lead surface-mount smalloutline package and a plastic TO-220 package.

BAHASA INGGRIS IPA PAKET B Kunci Jawaban Soal B 1. Jawaban ...

BAHASA INGGRIS IPA PAKET B Kunci Jawaban Soal B 1. Jawaban : B Dari judul menunjukan bahwa petunjuk diperuntukan bagi mereka yang akan menikah. Regulations for Wedding/Receptions 2. Jawaban : A Informasinya ada di text kalimat no 2. The cost for a wedding or reception is £8.00 per person. If the reception includes catering, the cost will be an additional £20.00 per person. 3. Jawaban : A Informasinya ada di text kalimat no 5. This deposit will be refunded provided the grounds are left free of damage and litter. 4. Jawaban : B Reception artinya resepsi. Prompt: cepat, Welcome: selamat datang, Release: melepas, Party:pesta, Invitation: undangan. Jadi jawabannya adalah Welcome. 5. Jawaban : A Rehearsal artinya latihan. Preparation: persiapan, Implementation: pelaksanaan, Orientation: orientasi, Correlation: korelasi, Duration: lamanya waktu. Jadi jawabannya adalah preparation 6. Jawaban : C Terdapat pada paragrap pertama, yaitu “Although speech is the most advanced form of communication, there are many ways of communicating without using speech. Signals, signs, symbols, and gestures may be found in every known culture.” Bahwa Signals, signs, symbols, and gestures adalah bagian dari communication.

PEMBAHASAN DAN KUNCI JAWABAN GEOGRAFI KELAS XII ...

PEMBAHASAN DAN KUNCI JAWABAN GEOGRAFI KELAS XII PAKET B 1. Berdasarkan soal nomor 1 a. Konsep aglomerasi adalah merupakan gabungan, kumpulan, 2 atau lebih pusat kegiatan dalam 1 lokasi/kawasan terterntu seperti kawasan industri, pemukiman, perdagangan, dsb. b. Konsep morfologi menjelaskan kenampakan bentuk-bentuk muka bumi, seperti dataran rendah, lereng, bukit/dataran tinggi. c. Konsep pola menitik beratkan pada pola keruangan baik fisik maupun sosialnya seperti pola permukiman penduduk, pola aliran sungai, dsb. d. Konsep lokasi mengkaji letak suatu objek dipermukaan bumi. Pada konsep ini utamanya dalam menjawab pertanyaan dimana (where). e. Konsep ketergantungan adalah konsep yang menunjukkan keterkaitan keruangan antar wilayah akibat adanya perbedaan potensi antar wilayah. Seperti keterkaitan antara desa dengan kota. Kunci jawaban D 2. Prinsip-prinsip geografi ada 4 a. Prinsip deskripsi, merupakan penjelasan lebih jauh mengenai gejala-gejala yang diselidiki/dipelajari. Deskripsi disajikan dalam bentuk tulisan, diagram tabel/gambar/peta. b. Prinsip korologi, merupakan gejala, fakta/masalah geografi disuatu tempat yang ditinjau dari sebaran, interelasi, interaksi, dan integrasinya dalam ruang. c. Prinsip persebaran, merupakan suatu gejala dan fakta yang tersebar tidak merata dipermukaan bumi. d. Prinsip interelasi, merupakan suatu hubungan yang saling terkait dalam ruang antara gejala yang 1 dengan gejala lain. e. Prinsip distribusi, merupakan suatu gejala dan fakta yang tidak merata dipermukaan bumi.

Kunci Jawaban Soal Essai Paket A.pdf

Kunci jawaban Babak FINAL Jenis soal : ESSAY 1. Kinerja bensin diukur berdasarkan nilai oktan (octane number) yaitu keberadaan senyawa 2,2,4 - trimetil pentane (isooktana) dengan nilai oktan 100, sedangkan nheptana nilai oktannya adalah nol. a. Gambarkan struktur 2,2,4 – trimetil pentane dan n – heptana (20 Point) Penyelesaian : b. Gambarkan semua isomer struktur n-heptana dan namai secara IUPAC (30 Point) Penyelesaian : c. Gambarkan struktur dan nama IUPAC alkena paling sederhana yang mempunyai isomer cis dan trans. (30 Point) Penyelesaian : d. Jelaskan pengertian bensin dengan angka oktan 75 % (20 Point) Penyelesaian : Angka oktan pada bensin ditentukan dengan adanya senyawa trimetil pentane dan nheptana dimana apabila pada bensin memiliki angka oktan 100 % maka pada besin tersebut terkandung senyawa trimetil pentane banding senyawa n-heptana yaitu 100 : 0, sehingga apabila bensin dengan angka oktan 75 % maka dalam bensin tersebut terkandung 75 % senyawa trimetil pentane dan 25 % senyawa n-heptana. 2. Reaksi : 2NOBr (g)  2NO (g) + Br2 (g) H = +16,1 kJ Diketahui : Tekanan awal NOBr = 0,65 atm. : NOBr telah terurai sebanyak 28% (Saat Kstb) (a) Tuliskan bentuk tetapan kesetimbangan, Kp. (10 poin) Penyelesaian : Kp  [p NO ] 2 [p Br2 ] [p NOBr ] 2 (b) Tentukan tekanan parsial gas NOBr, NO, dan Br2 setelah tercapai keadaan kesetimbangan. (30 poin) Penyelesaian : 100  28 p NOBr   0,65 atm  0,468 atm 100 28 p NO   0,65 atm  0,182 atm 100 p Br2  28 2 100  0,65 atm  0,091 atm (c) Tentukan tekanan total sesudai tercapai kesetimbangan (20 poin) Penyelesaian : (100  28 )  (28  14 ) 114 p tot  [ ]  0,65 atm   0,65 atm  0,741 atm 100 100 (d) Hitung nilai tetapan kesetimbangan, Kp pada temperatur tersebut. (20 poin) Penyelesaian :

KUNCI JAWABAN ERLANGGA FOKUS UN SMA/MA IPS 2014 - Yimg

KUNCI JAWABAN ERLANGGA FOKUS UN SMA/MA IPS 2014 UNIT 2 : SOAL UJIAN NASIONAL UN 2011/2012 1. A 2. D 3. D 4. E 5. D 6. A 7. E 8. E 9. A 10. D 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. B D A D E A C A E E 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. E A D C D A B D D C 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. C D A E D C C E A D 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. B D E C D D D B B A UN 2012/2013 1. E 2. E 3. A 4. A 5. D 6. C 7. A 8. D 9. E 10. E 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. C C D C C B E B A E 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. E B E A D A B A C B 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. D B C B A C A E A C 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. A C D C C B D B A E 11. 12. 13. 14. 15. D D B C A 16. 17. 18. 19. 20. E A A A D 21. 22. 23. 24. 25. B C C A E UNIT 3: PREDIKSI SOAL UN 2014 1. 2. 3. 4. 5. A B E D E 6. 7. 8. 9. 10. B D A E A 26. 27. 28. 29. 30. E D A E D 31. 32. 33. 34. 35. B A D A B 36. 37. 38. 39. 40. D E A D E 41. 42. 43. 44. 45. E C C A B 46. 47. 48. 49. 50. C D A D C 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. A D E A B C C E B E 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. A C A E D C D C B E 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. C E A D A E D D B B 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. A E A C C E A D B C 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. B E D C B D E A A D 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. A A E A B D C B C D 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. A A D C B E C A E D 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. E A B C B D E A D C 31. 32.

KUNCI JAWABAN SOAL FISIKA - Uhamka
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KUNCI JAWABAN SOAL FISIKA NO 1 2 3 4 5 6 7 8 9 10 JAWABAN D C D B B E B C A D NO 11 12 13 14 15 16 17 18 19 20 JAWABAN D D D C B C C C C D NO 21 22 23 24 25 26 27 28 29 30 JAWABAN A B A A C E D A C D NO 31 32 33 34 35 36 37 38 39 40 JAWABAN B E C E C E E E E A NO 41 42 43 44 45 46 47 48 49 50 JAWABAN C A A D B B D C A E KUNCI JAWABAN SOAL BIOLOGI NO JAWABAN NO JAWABAN NO JAWABAN NO JAWABAN NO JAWABAN 1 D 11 A 21 D 31 E 41 A 2 B 12 C 22 B 32 E 42 B 3 A 13 D 23 E 33 B 43 D 4 C 14 C 24 A 34 C 44 B 5 E 15 C 25 A 35 D 45 A 6 C 16 C 26 E 36 A 46 C 7 D 17 C 27 C 37 A 47 E 8 D 18 B 28 B 38 B 48 B 9 E 19 D 29 A 39 B 49 A 10 E 20 E 30 C 40 C 50 C

KUNCI JAWABAN TUKPD DKI TAHAP 2 MATA PELAJARAN : IPA ...

KUNCI JAWABAN TUKPD DKI TAHAP 2 MATA PELAJARAN : IPA TAHUN PELAJARAN : 2012-2013 Pilihlah jawaban yang benar! 1. Dua orang atlet mulai berlari dengan waktu bersamaan, dan saat mencapai garis finish dicatat dengan hasil seperti gambar. Selisih waktu antara Atlet 1 dengan Atlet 2 adalah .... A. 45s B. 28s C. 17s D. 11s Pembahasan : Waktu tempuh Atlet 1 adalah 17sekon dan Atlet 2 adalah 28sekon. Selisih waktu kedua Atlet = 28s – 22s = 6s Kunci : # (Anulir) 2. 4. 5. Pembahasan : 3. Termometer dinding yang menggunakan skala suhu Celsius menunjukkan suhu ruang kelas 300C. Suhu ruang tersebut sama dengan .... A. 560F B. 540F C. 620F D. 860F Pembahasan: Persamaan antara skala Celsius ke skala Fahrenheit adalah : 9 9 t oC  t  32 o F  30  32 5 5  54  32  86o F Kunci : D Grafik berikut melukiskan pemanasan 3kg air. Jika kalor jenis air 4200J/kg0C, kalor yang diperlukan pada proses R ke S adalah .... A. 12600J B. 126000J C. 252000J D. 378000J Pembahasan: Massa (m) = 3kg, C= 4200J/kg0C dan perubahan suhu (t) = 30 – 10 = 20oC Kalor yang diperlukan : Q = m.C. t = 3kg . 4200J/kg0C . 20oC Q = 126000J Kunci : C

Integrated Circuit FM Receiver using Bipolar Linear ... - Working Group

Even if FM radio broadcasting is a fundamental means of communication, it can’t escape one of the big trends in electronics today: miniaturization. As mobile communication devices are gaining more and more functionalities, people still want to have the option of listening to FM radio on their smart phones or MP3 players. This can only be achieved through the miniaturization of the FM reception process by implementing all its functions into a microchip. This miniaturization process is widely used in the industry, but has not been explored in depth by undergraduate students. In order to appreciate the complexity of such a task, we decided to design an FM receiver in a microchip. To make the project more interesting, we chose to create a user interface facilitating its use.

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