{"id":279199,"date":"2020-01-23T13:11:50","date_gmt":"2020-01-23T06:11:50","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=279199"},"modified":"2021-11-03T18:32:08","modified_gmt":"2021-11-03T11:32:08","slug":"utbk-fisika-2017","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-fisika-2017\/","title":{"rendered":"Soal UTBK SBMPTN Fisika 2017"},"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, <\/span>nih<\/span><\/em>.\u00a0<\/span><\/p>\n

Kali ini, Quipper Blog akan membahas latihan soal SBMPTN Fisika TKD SAINTEK 2017. 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

Sebuah kontainer besi dengan massa 10 kg terletak pada sebuah bidang datar. Dari keadaan diam, kontainer didorong dengan gaya tetap 40 Newton sejajar bidang datar selama 5 detik dan kemudian dilepaskan. Koefisien gesek kinetis dan statis antara kontainer dan bidang datar berturut-turut adalah 0,2 dan 0,3. Jika percepatan gravitasi sebesar\u00a010\u00a0m\/s<\/span>2<\/span><\/sup>, yang terjadi pada kontainer dalam waktu 5 detik berikutnya adalah\u2026<\/span><\/p>\n

    \n
  1. bergerak kemudian berhenti sebelum detik ke-5 setelah dilepaskan<\/span><\/li>\n
  2. bergerak dengan kecepatan tetap selama 5 detik setelah dilepaskan<\/span><\/li>\n
  3. bergerak dengan kecepatan tidak tetap selama lebih dari 5 detik<\/span><\/li>\n
  4. bergerak kemudian berhenti lebih dari 5 detik setelah dilepaskan<\/span><\/li>\n
  5. bergerak kemudian akhirnya diam tepat pada detik ke-5 setelah dilepaskan<\/span><\/li>\n<\/ol>\n

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

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

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

    \"\"<\/p>\n

    Ditanya: <\/span>v<\/span><\/em>t2<\/span><\/em><\/sub> =\u2026?<\/span><\/p>\n

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

    Mari kita tinjau gerak kontainer pada 5 detik pertama dan 5 detik kedua.<\/span><\/p>\n

    Gerak kontainer pada 5 detik pertama\u00a0(<\/span>t<\/span><\/em> = 0\u00a0s sampai\u00a0<\/span>t<\/span><\/em> = 5\u00a0s).<\/span><\/p>\n

    Dalam selang waktu\u00a0<\/span>t<\/span><\/em> = 0\u00a0s sampai\u00a0<\/span>t<\/span><\/em> = 5\u00a0s, kontainer mengalami percepatan karena mendapat gaya 40 N. Selain itu, gerak kontainer juga mendapat perlawanan gaya gesek. Oleh karena kontainer bergerak, maka yang digunakan adalah koefisien gesek kinetis. Dengan demikian, besar gaya gesek yang bekerja pada kontainer adalah sebagai berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Percepatan kontainer tersebut dapat ditentukan dengan hukum II Newton berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Dengan demikian, kecepatan kontainer pada detik ke-5 (<\/span>v<\/span><\/em>t1<\/span><\/em><\/sub>) adalah sebagai berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Gerak kontainer pada 5 detik kedua\u00a0(<\/span>t<\/span><\/em> = 5\u00a0s sampai\u00a0<\/span>t<\/span><\/em> = 10\u00a0s).<\/span><\/p>\n

    Pada 5 detik kedua, gaya yang bekerja hanya gaya gesek, sehingga kontainer mengalami perlambatan. Besar perlambatan yang dialami kontainer adalah sebagai berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Dengan demikian, kecepatan kontainer pada detik ke-10 (5 detik setelah dilepaskan) adalah sebagai berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Ini berarti, kontainer berhenti tepat pada detik ke-10.<\/span><\/p>\n

    Jadi, kontainer tetap bergerak kemudian akhirnya diam tepat pada detik ke-5 setelah dilepaskan.<\/span><\/p>\n

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

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

    \"\"<\/span><\/p>\n

    Jika udara\u00a0(\u03c1<\/span>udara <\/span><\/sub>= 1,36\u00a0kg\/m<\/span>3<\/span><\/sup>)\u00a0dialirkan ke dalam tabung pitot dan perbedaan tinggi air raksa\u00a0(\u03c1<\/span>raksa <\/span><\/sub>= 13.600\u00a0kg\/m<\/span>3<\/span><\/sup>)\u00a0pada manometer adalah 2 cm, kecepatan aliran udara tersebut adalah … m\/s.<\/span><\/p>\n

      \n
    1. \n
        \n
      1. 20\u221a<\/span>2<\/span><\/li>\n
      2. 200<\/span><\/li>\n
      3. 20<\/span><\/strong><\/li>\n
      4. 400<\/span><\/li>\n
      5. 20\u221a<\/span>10<\/span><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

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

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

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

        \u03c1<\/span>udara <\/span><\/sub>= \u03c1 = 1,36\u00a0kg\/m<\/span>3<\/span><\/sup><\/p>\n

        \u03c1<\/span>raksa <\/span><\/sub>= \u03c1\u2019 = 13.600\u00a0kg\/m<\/span>3<\/span><\/sup><\/p>\n

        h<\/span><\/em> = 2 cm = 0,02 m<\/span><\/p>\n

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

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

        Kecepatan aliran udara dapat ditentukan dengan penerapan persamaan Bernoulli berikut.<\/span><\/p>\n

        \"\"<\/p>\n

        Jadi, kecepatan aliran udara tersebut adalah <\/span>20\u221a<\/span>10<\/span> m\/s.<\/span><\/p>\n

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

        Tiga kawat panjang dibentuk seperti gambar berikut.<\/span><\/p>\n

        \"\"<\/p>\n

        \"\"<\/p>\n

        \"\"<\/p>\n

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

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

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

        \"\"\u00a0<\/span><\/p>\n

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

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

        \"\"<\/p>\n

        Segmen kawat AC dan CB melewati titik C, sehingga kedua kawat tersebut tidak menimbulkan medan magnet. Kawat yang menimbulkan medan magnet hanyalah segmen kawat AB. Untuk mengetahui medan magnetnya, kita harus menentukan jarak kawat AB ke titik C, yaitu \u03b1.<\/span><\/p>\n

        Mula-mula, tentukan panjang BC. Dengan teorema Pythagoras, diperoleh:<\/span><\/p>\n

        \"\"<\/p>\n

        Panjang\u00a0\u03b1\u00a0dapat diketahui dengan menganggap ABC sebagai segitiga, sehingga luasnya dapat dihitung sebagai berikut.<\/span><\/p>\n

        \"\"<\/span><\/p>\n

        Dengan demikian, medan magnet di titik C dapat ditentukan sebagai berikut.<\/span><\/p>\n

        \"\"<\/p>\n

        \"\"<\/p>\n

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

        Kelereng bermassa\u00a0<\/span>m<\/span><\/em>1<\/span><\/sub>\u00a0yang bergerak sepanjang sumbu-X dengan kecepatan dari arah kiri menumbuk kelereng kedua bermassa\u00a0<\/span>m<\/span><\/em>2<\/span><\/sub>\u00a0<\/sub>yang diam pada titik pusat (0, 0). Jika kecepatan kelereng pertama dan kedua setelah tumbukan adalah\u00a0<\/span>v<\/span><\/em>1<\/span><\/sub>\u2032\u00a0dan\u00a0<\/span>v<\/span><\/em>2<\/span><\/sub>\u2032, dengan arah\u00a030o<\/sup><\/span>\u00a0dan\u00a0300o<\/sup><\/span>\u00a0terhadap sumbu-X, perbandingan antara\u00a0<\/span>v<\/span><\/em>1<\/span><\/sub>\u00a0dan\u00a0<\/span>v<\/span><\/em>1<\/span><\/sub>\u2032\u00a0adalah\u2026<\/span><\/p>\n

        \"\"<\/p>\n

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

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

        Diketahui: Kelereng bermassa\u00a0<\/span>m<\/span><\/em>1<\/span><\/sub>\u00a0yang bergerak sepanjang sumbu-X dengan kecepatan dari arah kiri menumbuk kelereng kedua bermassa\u00a0<\/span>m<\/span><\/em>2<\/span>\u00a0yang diam pada titik pusat (0, 0). Jika kecepatan kelereng pertama dan kedua setelah tumbukan adalah\u00a0<\/span>v<\/span><\/em>1<\/span><\/sub>\u2032\u00a0dan\u00a0<\/span>v<\/span><\/em>2<\/span><\/sub>\u2032.<\/span><\/p>\n

        Ditanya: <\/span>v<\/span><\/em>1<\/span><\/sub>\u00a0:\u00a0<\/span>v<\/span><\/em>1<\/span><\/sub>\u2032\u00a0=\u2026?<\/span><\/p>\n

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

        Permasalahan pada soal dapat digambarkan sebagai berikut.<\/span><\/p>\n

        \"\"<\/p>\n

        Saat terjadi tumbukan, berlaku hukum kekekalan momentum.<\/span><\/p>\n

        Pada arah vertikal (sumbu-Y):<\/span><\/p>\n

        \"\"<\/p>\n

        Pada arah horizontal (sumbu-X):<\/span><\/p>\n

        \"\"<\/p>\n

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

        Dua buah pesawat terbang yang cukup canggih menempuh jalur terbang yang sama berupa sebuah garis lurus. Pesawat pertama memiliki kecepatan\u00a00,8<\/span>c<\/span><\/em>\u00a0terhadap bumi, sedangkan pesawat yang kedua (berada di belakang pesawat pertama) memiliki kecepatan\u00a00,2<\/span>c<\/span><\/em>\u00a0terhadap bumi. Sebuah benda bergerak lurus sejajar dengan kedua pesawat dengan kecepatan\u00a00,5<\/span>c<\/span><\/em>\u00a0terhadap pesawat pertama. Jika diukur dari pesawat kedua, kecepatan benda tersebut adalah…<\/span><\/p>\n

          \n
        1. 0,94<\/span>c<\/span><\/em><\/li>\n
        2. 0,98<\/span>c<\/span><\/em><\/li>\n
        3. 0,89<\/span>c<\/span><\/em><\/li>\n
        4. 0,84<\/span>c<\/span><\/em><\/li>\n
        5. 0,80<\/span>c<\/span><\/em><\/li>\n<\/ol>\n

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

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

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

          \"\"<\/p>\n

          Ditanya: kecepatan benda terhadap pesawat 2 (<\/span>v<\/span><\/em>A.2<\/span><\/sub>) =\u2026?<\/span><\/p>\n

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

          Permasalahan pada soal dapat digambarkan sebagai berikut.<\/span><\/p>\n

          \"\"<\/p>\n

          Pesawat dan benda bergerak mendekati kecepatan cahaya, sehingga berlaku kecepatan relativistik berikut.<\/span><\/p>\n

          \"\"<\/p>\n

          Kecepatan pesawat 1 terhadap pesawat 2\u00a0(<\/span>v<\/span><\/em>1.2<\/span>)\u00a0atau sebaliknya belum diketahui, sehingga perlu dicari terlebih dahulu.<\/span><\/p>\n

          \"\"<\/p>\n

          Dengan demikian, diperoleh:<\/span><\/p>\n

          \"\"<\/p>\n

          Jadi, kecepatan benda tersebut jika diukur dari pesawat kedua adalah\u00a00,89<\/span>c<\/span><\/em>.<\/span><\/p>\n

          Latihan Soal 6<\/strong><\/h2>\n

          Radiasi termal dipancarkan oleh suatu benda bersuhu\u00a0<\/span>T<\/span><\/em>\u00a0dengan panjang gelombang yang bervariasi. Radiasi dengan panjang gelombang 600 mikrometer memiliki intensitas maksimum. Jika suhu benda dinaikkan menjadi\u00a03<\/span>T<\/span><\/em>,\u00a0panjang gelombang radiasi dengan intensitas maksimum berubah menjadi…<\/span><\/p>\n

            \n
          1. 260 <\/span>\u03bc<\/span>m<\/span><\/li>\n
          2. 250 <\/span>\u03bc<\/span>m<\/span><\/li>\n
          3. 280 <\/span>\u03bc<\/span>m<\/span><\/li>\n
          4. 200 <\/span>\u03bc<\/span>m<\/span><\/li>\n
          5. 300 <\/span>\u03bc<\/span>m<\/span><\/li>\n<\/ol>\n

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

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

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

            \"\"<\/p>\n

            Ditanya: <\/span>\u03bbm2<\/sub><\/span><\/p>\n

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

            Berdasarkan hukum pergeseran Wien, diperoleh:<\/span><\/p>\n

            \"\"<\/p>\n

            Jadi, panjang gelombang radiasi dengan intensitas maksimum berubah menjadi 200 \u00b5m.<\/span><\/p>\n

            Latihan Soal 7<\/strong><\/h2>\n

            Pada sebuah mesin panas, kalor yang diserap dari reservoir temperatur\u00a0T1\u00a0sebesar 3 kali kalor yang digunakan untuk melakukan kerja. Jika kalor yang diserap dalam satu siklusnya sebesar 1.650 J dan waktu yang dibutuhkan untuk melakukan dua siklus adalah 3 detik, daya yang dihasilkan mesin tersebut sebesar…<\/span><\/p>\n

              \n
            1. 767 watt<\/span><\/li>\n
            2. 667 watt<\/span><\/li>\n
            3. 367 watt<\/span><\/li>\n
            4. 567 watt<\/span><\/li>\n
            5. 467 watt<\/span><\/li>\n<\/ol>\n

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

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

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

              Q<\/span><\/em>1 <\/span><\/sub>= 3<\/span>W<\/span><\/em> = 1.650 J<\/span><\/p>\n

              t<\/span><\/em> = 3 detik untuk 2 siklus<\/span><\/p>\n

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

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

              \"\"<\/p>\n

              Kemudian, tentukan daya yang dihasilkan dengan rumus berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Jadi, daya yang dihasilkan mesin tersebut adalah 367 watt.<\/span><\/p>\n

              Bagaimana Quipperian, apakah kamu sudah mulai paham mengerjakan latihan soal SBMPTN Fisika TKD SAINTEK 2017 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":279235,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_post_was_ever_published":false},"categories":[679385273,679385209],"tags":[],"ppma_author":[679386827],"class_list":["post-279199","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbmptn","category-soal-sbmptn"],"acf":[],"yoast_head":"\nSoal UTBK SBMPTN Fisika 2017 - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Ayo, buat kamu yang sedang mencari-cari soal Fisika SBMPTN 2017, langsung mampir ke sini ya! 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