{"id":278835,"date":"2020-01-15T16:12:31","date_gmt":"2020-01-15T09:12:31","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=278835"},"modified":"2022-03-01T19:23:06","modified_gmt":"2022-03-01T12:23:06","slug":"utbk-fisika-2018","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-fisika-2018\/","title":{"rendered":"Soal UTBK SBMPTN Fisika 2018"},"content":{"rendered":"

\"\"<\/p>\n

Hai Quipperian, sudah siapkah kamu menyambut seleksi akbar, SBMPTN? Atau justru kamu masih bingung mencari soal-soal SBMPTN? Bagi kamu yang saat ini masih bingung mencari soal-soal SBMPTN, tampaknya kamu berada di artikel yang tepat, nih.\u00a0<\/span><\/p>\n

Kali ini, Quipper Blog akan membahas latihan soal SBMPTN Fisika TKD SAINTEK 2018. Kabar baiknya, latihan soal yang disediakan Quipper Blog, lengkap dengan jawaban beserta pembahasannya, <\/span>lho<\/span><\/em>. Semakin penasaran? Yuk, segera kerjakan soalnya!<\/span><\/p>\n

Latihan Soal 1<\/strong><\/h2>\n

Perhatikan gambar sistem mekanik berikut.<\/span><\/p>\n

\"\"<\/p>\n

Sudut kemiringan bidang\u00a0\u03b8=30o<\/sup><\/span>\u00a0dan bidang miring licin. Sistem berada dalam keadaan setimbang serta massa katrol dan massa pegas diabaikan. Jika setiap massa dijadikan dua kali semula, salah satu cara yang dapat dilakukan agar sistem tetap setimbang adalah\u2026<\/span><\/p>\n

    \n
  1. Konstanta pegas tetao dan pertambahan panjang pegas menjadi 2 kali semula.<\/span><\/li>\n
  2. Konstanta pegas menjadi 0,5 kali semula dan pertambahan panjang pegas menjadi 2 kali semula.<\/span><\/li>\n
  3. Konstanta pegas tetap dan pertambahan panjang pegas menjadi 0,5 kali semula.<\/span><\/li>\n
  4. Konstanta pegas tetap dan pertambahan panjang pegas menjadi 4 kali semula.<\/span><\/li>\n
  5. Konstanta pegas menjadi dua kali semula dan pertambahan panjang pegas tetap.<\/span><\/li>\n<\/ol>\n

    Jawaban: E<\/strong><\/p>\n

    Pembahasan:<\/strong><\/p>\n

    Diketahui:<\/span><\/p>\n

    \"\"<\/p>\n

    Ditanya: Salah satu cara agar sistem tetap setimbang =\u2026?<\/span><\/p>\n

    Penyelesaian:<\/span><\/p>\n

    Pada katrol tetap, berlaku:<\/span><\/p>\n

    \"\"<\/p>\n

    Pada benda 2:<\/span><\/p>\n

    \"\"<\/p>\n

    Substitusikan nilai <\/span>T<\/span><\/em>2<\/span><\/em> pada persamaan (2) ke persamaan (1).<\/span><\/p>\n

    \"\"<\/p>\n

    Pada katrol bergerak, berlaku:<\/span><\/p>\n

    \"\"<\/p>\n

    Substitusikan nilai <\/span>T<\/span><\/em>1<\/span><\/em> pada persamaan (3) ke persamaan (4).<\/span><\/p>\n

    \"\"<\/p>\n

    Lalu, ganti <\/span>m<\/span><\/em>1<\/span><\/em><\/sub> dan <\/span>m<\/span><\/em>2<\/span><\/em><\/sub> dengan <\/span>m<\/span><\/em>1<\/span><\/em><\/sub>\u2019 dan <\/span>m<\/span><\/em>2<\/span><\/em><\/sub>\u2019.<\/span><\/p>\n

    \"\"<\/p>\n

    Jadi, salah satu cara yang dapat dilakukan agar sistem tetap setimbang adalah dengan cara mengubah konstanta pegas menjadi dua kali semula dan pertambahan panjang pegas dijaga tetap.<\/span><\/p>\n

    Latihan Soal 2<\/strong><\/h2>\n

    Perhatikan gambar berikut.<\/span><\/p>\n

    \"\"<\/p>\n

      \n
    1. 600 gram<\/span><\/li>\n
    2. 700 gram<\/span><\/li>\n
    3. 800 gram<\/span><\/li>\n
    4. 900 gram<\/span><\/li>\n
    5. 500 gram<\/span><\/li>\n<\/ol>\n

      Jawaban: B<\/strong><\/p>\n

      Pembahasan:<\/strong><\/p>\n

      Diketahui:<\/span><\/p>\n

      \"\"<\/p>\n

      Ditanya: <\/span>m<\/span><\/em> =\u2026?<\/span><\/p>\n

      Penyelesaian:<\/span><\/p>\n

      Terjadinya pemuaian panjang akibat adanya penggantungan beban\u00a0m\u00a0menyebabkan tali menjadi bertambah panjang sebesar:<\/span><\/p>\n

      \"\"<\/p>\n

      Berlaku Hukum I Newton.<\/span><\/p>\n

      \"\"<\/p>\n

      Jadi, nilai <\/span>m<\/span><\/em> adalah 700 gram.<\/span><\/p>\n

      Latihan Soal 3<\/strong><\/h2>\n

      Sebuah balok kayu bermassa\u00a07,5 kg\u00a0dan bervolume\u00a00,01 m<\/span>3<\/span><\/sup>\u00a0dimasukkan ke dalam air\u00a0(\u03c1<\/span>air<\/span><\/sub> = 1,0 \u00d7 10<\/span>3<\/span><\/sup>\u00a0kg\/m<\/span>3<\/span><\/sup>). Besar gaya untuk menahan balok agar terbenam seluruhnya di dalam air adalah\u2026<\/span><\/p>\n

        \n
      1. 50 N<\/span><\/li>\n
      2. 100 N<\/span><\/li>\n
      3. 12,5 N<\/span><\/li>\n
      4. 25 N<\/span><\/li>\n
      5. 10 N<\/span><\/li>\n<\/ol>\n

        Jawaban: D<\/strong><\/p>\n

        Pembahasan:<\/strong><\/p>\n

        Diketahui:<\/span><\/p>\n

        \"\"<\/p>\n

        Ditanya: <\/span>\u2206<\/span>F<\/span><\/em> =\u2026?<\/span><\/p>\n

        Penyelesaian:<\/span><\/p>\n

        Pertama, Quipperian harus mencari besarnya gaya ke atas.<\/span><\/p>\n

        \"\"<\/p>\n

        Selanjutnya, Quipperian bisa menentukan besarnya gaya untuk menahan balok agar tidak terbenam seluruhnya di dalam air.<\/span><\/p>\n

        \"\"<\/p>\n

        Jadi, besarnya gaya untuk menahan balok agar terbenam seluruhnya di dalam air adalah\u00a025\u00a0N.<\/span><\/p>\n

        Latihan Soal 4<\/strong><\/h2>\n

        Mobil pemadam kebakaran sedang bergerak dengan laju\u00a020\u00a0m\/s\u00a0sambil membunyikan sirene pada frekuensi\u00a0400\u00a0Hz\u00a0(cepat rambat bunyi\u00a0300\u00a0m\/s). Jika mobil pemadam kebakaran bergerak menjauhi seseorang yang sedang berdiri di tepi jalan, manakah di antara pernyataan berikut yang benar?<\/span><\/p>\n

        (1) Panjang gelombang bunyi menurut pendengar adalah\u00a090\u00a0cm.<\/span><\/p>\n

        (2) Orang tersebut akan mendengar sirine pada frekuensi\u00a0370\u00a0Hz.<\/span><\/p>\n

        (3) Orang tersebut akan mendengar sirine dengan frekuensi\u00a0400\u00a0Hz.<\/span><\/p>\n

        (4) Orang tersebut akan mendengar frekuensi sirine lebih rendah daripada sirine mobil pemadam kebakaran.<\/span><\/p>\n

          \n
        1. jika (1), (2), dan (3) yang benar<\/span><\/li>\n
        2. jika (1) dan (3) yang benar<\/span><\/li>\n
        3. jika (2) dan (4) yang benar<\/span><\/li>\n
        4. jika hanya (4) saja yang benar<\/span><\/li>\n
        5. jika semua jawaban benar<\/span><\/li>\n<\/ol>\n

          Jawaban: D<\/strong><\/p>\n

          Pembahasan:<\/strong><\/p>\n

          Diketahui:<\/span><\/p>\n

          v<\/span><\/em>s<\/span><\/em><\/sub> = + 20 m\/s<\/span><\/p>\n

          f<\/span><\/em>s<\/span><\/sub> = 400 Hz<\/span><\/p>\n

          v<\/span><\/em> = 300 m\/s<\/span><\/p>\n

          v<\/span><\/em>p<\/span><\/em><\/sub> = 0<\/span><\/p>\n

          Ditanya: Pernyataan yang benar =\u2026?<\/span><\/p>\n

          Penyelesaian:<\/span><\/p>\n

          Pertama, Quipperian harus mencari <\/span>f<\/span><\/em>p<\/span><\/sub>.<\/span><\/p>\n

          \"\"<\/p>\n

          Selanjutnya, Quipperian harus mencari \u03bbp<\/sub><\/span>.<\/span><\/p>\n

          \"\"<\/p>\n

          Orang tersebut akan mendengar sirine dengan frekuensi 375 Hz, sedangkan besar frekuensi sirinenya 400 Hz (<\/span>f<\/span><\/em>p<\/span><\/sub> < <\/span>f<\/span><\/em>s<\/span><\/sub>).<\/span><\/p>\n

          Jadi, hanya pernyataan (4) saja yang benar.<\/span><\/p>\n

          Latihan Soal 5<\/strong><\/h2>\n

          Indeks bias kaca lebih besar daripada indeks bias air.<\/span><\/p>\n

          SEBAB<\/span><\/p>\n

          Cepat rambat cahaya dalam kaca lebih besar daripada cepat rambat cahaya yang melewati air.<\/span><\/p>\n

            \n
          1. Jika pernyataan benar, alasan benar, dan keduanya menunjukkan hubungan sebab akibat.<\/span><\/li>\n
          2. Jika pernyataan benar, alasan benar, dan keduanya tidak menunjukkan hubungan sebab akibat.<\/span><\/li>\n
          3. Jika pernyataan benar dan alasan salah.<\/span><\/li>\n
          4. Jika pernyataan salah dan alasan benar.<\/span><\/li>\n
          5. Jika pernyataan dan alasan keduanya salah.<\/span><\/li>\n<\/ol>\n

            Jawaban: C<\/strong><\/p>\n

            Pembahasan:<\/strong><\/p>\n

            Diketahui:<\/span><\/p>\n

            n<\/span><\/em>kaca<\/span><\/em><\/sub> > <\/span>n<\/span><\/em>air<\/span><\/em>\u00a0<\/span><\/sub><\/p>\n

            Ditanya: Hubungan antara pernyataan dan alasan =\u2026?<\/span><\/p>\n

            Penyelesaian:<\/span><\/p>\n

            Berikut ini hubungan antara indeks bias dan cepat rambat cahaya.<\/span><\/p>\n

            \"\"<\/p>\n

            Dari persamaan di atas, besarnya indeks bias suatu medium berbanding terbalik dengan besarnya cepat rambat cahaya dalam medium tersebut. Ketika indeks bias kaca lebih besar daripada indeks bias air, maka kecepatan cahaya di dalam kaca akan lebih kecil dibandingkan dengan kecepatan cahaya di dalam air.<\/span><\/p>\n

            Jadi, pernyataan benar tetapi alasan salah.<\/span><\/p>\n

            Bagaimana Quipperian, apakah kamu sudah mulai paham mengerjakan latihan soal SBMPTN Fisika TKD SAINTEK 2018 di atas? Agar pemahamanmu semakin terasah, sering-seringlah mengerjakan latihan soal. Ingat, selangkah lagi kamu akan memasuki PTN impian.\u00a0<\/span><\/p>\n

            Untuk mendukung kesuksesanmu masuk PTN, Quipper Video hadir dengan Paket Intensif UTBK SBMPTN 2020. Semua kebutuhan sudah tersedia di dalamnya, mulai dari materi persiapan, materi pemantapan, latihan soal, bank soal, hingga pembahasan lengkapnya. Jadi, tunggu apalagi. Yuk, gabung Quipper Video sekarang!<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

            Hai Quipperian, sudah siapkah kamu menyambut seleksi akbar, SBMPTN? Atau justru kamu masih bingung mencari soal-soal SBMPTN? Bagi kamu yang saat ini masih bingung…<\/p>\n","protected":false},"author":99400369,"featured_media":278858,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_crdt_document":"","_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[679385273,679385209],"tags":[679385465],"ppma_author":[679386827],"class_list":["post-278835","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbmptn","category-soal-sbmptn","tag-soal-utbk-saintek-2018"],"acf":[],"yoast_head":"\nSoal UTBK SBMPTN Fisika 2018 - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Buat kamu yang lagi cari-cari kumpulan soal Fisika SBMPTN 2018, langsung klik di sini, deh! 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