{"id":279746,"date":"2020-02-24T22:27:29","date_gmt":"2020-02-24T15:27:29","guid":{"rendered":"https:\/\/quipperhome.wpcomstaging.com\/?p=279746"},"modified":"2020-02-24T22:27:29","modified_gmt":"2020-02-24T15:27:29","slug":"utbk-2020-fisika","status":"publish","type":"post","link":"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-2020-fisika\/","title":{"rendered":"Prediksi Soal UTBK Fisika 2020"},"content":{"rendered":"

\"\"<\/p>\n

SBMPTN sebentar lagi, soal-soal harus segera dikuasai. Itulah sebabnya, Quipperian harus <\/span>gercep<\/span><\/em> alias gerak cepat. Nah, Bagi kamu yang saat ini masih bingung mencari soal-soal SBMPTN, tampaknya kamu berada di <\/span>website<\/span><\/em> yang tepat.\u00a0<\/span><\/p>\n

Kali ini, Quipper Blog akan membahas Prediksi UTBK SBMPTN Fisika SAINTEK 2020. Kabar baiknya, latihan soal yang disediakan Quipper Blog, lengkap dengan jawaban beserta pembahasannya, <\/span>lho<\/span><\/em>. Semakin penasaran? <\/span>Kuy<\/span><\/em>, segera kerjakan soalnya!<\/span><\/p>\n

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

Sebuah bola bermassa\u00a0<\/span>m<\/span><\/em>\u00a0digantungkan pada seutas tali sepanjang\u00a0\u2113\u00a0dan digerakkan melingkar secara horizontal dengan sudut\u00a0\u03b8\u00a0seperti pada gambar.<\/span><\/p>\n

\"\"<\/p>\n

Bola bergerak melingkar dengan kecepatan\u00a0<\/span>v<\/span><\/em>\u00a0dalam radius\u00a0<\/span>R<\/span><\/em>\u00a0sehingga tali mengalami gaya tegangan tali yang besarnya\u00a0<\/span>T<\/span><\/em>.\u00a0Besarnya\u00a0<\/span>v<\/span><\/em>\u00a0adalah\u2026<\/span><\/p>\n

\"\"<\/p>\n

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

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

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

    \n
  • Massa bola = <\/span>m<\/span><\/em><\/li>\n
  • Panjang tali = \u2113<\/span><\/li>\n
  • Sudut yang terbentuk = \u03b8<\/span><\/li>\n
  • Tegangan tali = <\/span>T<\/span><\/em><\/li>\n
  • Radius = <\/span>R<\/span><\/em><\/li>\n
  • Kecepatan = <\/span>v<\/span><\/em><\/li>\n<\/ul>\n

    Ditanya: Besar kecepatan\u00a0(<\/span>v<\/span><\/em>) =…?<\/span><\/p>\n

    Penyelesaian:<\/span>
    \n<\/span>Besar sinus\u00a0\u03b8\u00a0dapat ditentukan dari gambar segitiga oleh tali berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Ini berarti, besarnya sinus\u00a0\u03b8\u00a0berdasarkan hubungan antara\u00a0\u2113\u00a0dan\u00a0<\/span>R<\/span><\/em>:<\/span><\/p>\n

    \"\"<\/p>\n

    Mencari besarnya <\/span>v<\/span><\/em>:<\/span><\/p>\n

    Sebelumnya, uraikan dahulu gayanya seperti berikut.<\/span><\/p>\n

    \"\"<\/p>\n

    Ini berarti:<\/span><\/p>\n

    \"\"<\/p>\n

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

    Sebuah mobil derek digunakan untuk menderek sebuah mobil\u00a0<\/span>pick up<\/span><\/em>\u00a0yang mogok di tepi jalan raya sehingga percepatannya sebesar\u00a02 m\/s<\/span>2<\/span><\/sup>.\u00a0Kabel horizontal mobil derek yang dipakai untuk mendereknya akan putus jika tegangan di dalamnya melebihi\u00a0<\/span>T<\/span><\/em>.\u00a0Saat mobil derek digunakan untuk menarik sebuah sepeda motor yang mogok di tepi jalan raya, menyebabkan percepatan sebesar\u00a018 m\/s<\/span>2<\/span><\/sup>.\u00a0Jika mobil derek digunakan untuk menderek mobil\u00a0<\/span>pick up<\/span><\/em>\u00a0dan sepeda motor bersama-sama (sepeda motor dinaikkan ke atas\u00a0<\/span>pick up<\/span><\/em>) akan menyebabkan percepatan sebesar\u2026<\/span><\/p>\n

      \n
    1. 18 m\/s<\/span>2<\/span><\/sup><\/li>\n
    2. 20\u00a0<\/span>m\/s<\/span>2<\/span><\/sup><\/li>\n
    3. 1,8\u00a0<\/span>m\/s<\/span>2<\/span><\/sup><\/li>\n
    4. 0,2\u00a0<\/span>m\/s<\/span>2<\/span><\/sup><\/li>\n
    5. 2\u00a0<\/span>m\/s<\/span>2<\/span><\/sup><\/li>\n<\/ol>\n

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

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

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

      \"\"<\/p>\n

      Ditanya: <\/span>a<\/span><\/em>ps<\/span><\/em><\/sub> =\u2026?<\/span><\/p>\n

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

      Mula-mula, Quipperian harus mencari <\/span>m<\/span><\/em>p<\/span><\/em><\/sub>:<\/span><\/p>\n

      \"\"<\/p>\n

      Lalu, kamu bisa mencari <\/span>m<\/span><\/em>s<\/span><\/em><\/sub>:<\/span><\/p>\n

      \"\"<\/p>\n

      Terakhir, mencari <\/span>a<\/span><\/em>ps<\/span><\/em><\/sub>:<\/span><\/p>\n

      \"\"<\/p>\n

      Quipperian juga bisa menggunakan SUPER \u201cSolusi Quipper\u201d.<\/span><\/p>\n

      \"\"<\/p>\n

      Jadi, percepatan saat mobil derek digunakan untuk menderek mobil\u00a0<\/span>pick up<\/span><\/em>\u00a0dan sepeda motor bersama-sama adalah\u00a01,8\u00a0<\/span>m\/s<\/span>2<\/span><\/sup>.<\/span><\/p>\n

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

      Dua buah pegas dengan konstanta pegas\u00a0<\/span>k<\/span><\/em>1<\/span><\/em><\/sub>\u00a0dan\u00a0<\/span>k<\/span><\/em>2<\/span><\/em><\/sub>\u00a0disusun secara paralel. Kemudian, sebuah kawat XY sepanjang\u00a0\u2113\u2113,\u00a0masing-masing ujungnya digantungkan ke susunan pegas paralel tersebut sehingga ujung kawat X diikat ke ujung bebas pegas dengan kontanta pegas\u00a0<\/span>k<\/span><\/em>1<\/span><\/em>\u00a0dan ujung kawat Y diikat ke ujung bebas pegas dengan konstanta\u00a0<\/span>k<\/span><\/em>2<\/span><\/em>.\u00a0Lalu, kawat XY dialiri arus listrik sebesar\u00a0<\/span>i<\/span><\/em>\u00a0berarah ke utara dan sebuah magnet U diletakkan sehingga kawat XY berada di antara kedua kutub magnet U tersebut. Hal ini mengakibatkan kawat XY berada dalam medan magnet sebesar\u00a0<\/span>B<\/span><\/em>\u00a0dari magnet U tersebut sedemikian hingga arah medan magnet\u00a0<\/span>B<\/span><\/em>\u00a0tersebut berarah ke timur. Jika massa kawat XY diabaikan, maka pada saat posisi setimbang susunan pegas akan bertambah panjang sebesar\u2026<\/span><\/p>\n

      \"\"<\/p>\n

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

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

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

      \"\"<\/p>\n

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

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

      Berdasarkan aturan arah tangan kanan, arus listrik ke arah utara dan medan magnet ke arah timur, maka gaya lorentz berarah ke bawah. Sementara arah gaya pemulih pegas berarah ke atas. Ini berarti:<\/span><\/p>\n

      \"\"<\/p>\n

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

      Sebuah proton diikat pada ujung bebas sebuah pegas dengan konstanta pegas\u00a0400\u00a0N\/m\u00a0yang digantung. Kemudian susunan pegas dan proton tersebut diletakkan pada suatu medan listrik berarah ke bawah sehingga saat setimbang, pegas bertambah panjang\u00a04\u00a0mm.\u00a0Jika massa proton diabaikan, maka besarnya medan listrik tersebut adalah\u2026<\/span><\/p>\n

        \n
      1. 10<\/span>-19<\/span><\/sup> N\/C<\/span><\/li>\n
      2. 0<\/span><\/li>\n
      3. 256 \u00d7 10<\/span>21<\/span><\/sup> N\/C<\/span><\/li>\n
      4. 10<\/span>19<\/span><\/sup> N\/C<\/span><\/li>\n
      5. 256 \u00d7 10<\/span>-21<\/span><\/sup> N\/C<\/span><\/li>\n<\/ol>\n

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

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

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

        \"\"<\/p>\n

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

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

        Oleh karena jenis muatan proton adalah positif, maka arah gaya listrik yang timbul akan searah dengan arah medan listriknya, yaitu ke bawah. Untuk itu, saat setimbang:<\/span><\/p>\n

        \"\"<\/p>\n

        Jadi, besarnya medan listrik tersebut adalah 10<\/span>19<\/span><\/sup> N\/C.<\/span><\/p>\n

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

        Sebuah kabel listrik memiliki hambatan\u00a013 \u03a9.\u00a0Oleh seorang anak kecil, kabel tersebut ditarik secara beraturan sehingga panjangnya meregang menjadi\u00a04\u00a0kali semula. Jika volume kabel tidak berubah, maka besarnya hambatan listrik sekarang adalah\u2026<\/span><\/p>\n

          \n
        1. 221 \u03a9<\/span><\/li>\n
        2. 212 \u03a9<\/span><\/li>\n
        3. 13 \u03a9<\/span><\/li>\n
        4. 208 \u03a9<\/span><\/li>\n
        5. 52 \u03a9<\/span><\/li>\n<\/ol>\n

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

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

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

          \"\"<\/p>\n

          Ditanya: <\/span>R<\/span><\/em>2<\/span><\/em><\/sub> =\u2026?<\/span><\/p>\n

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

          Dengan menganggap bentuk kabel adalah tabung dengan tinggi tabung berupa panjang kabel, maka perbandingan luas penampang kabel:<\/span><\/p>\n

          \"\"<\/p>\n

          \"\"<\/p>\n

          Jadi, besarnya hambatan listrik sekarang adalah 208 \u03a9.<\/span><\/p>\n

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

          Sebuah prisma segienam saat dicelupkan ke dalam air, hanya separuh dari volumenya yang tercelup. Kemudian, prisma segienam tersebut dimasukkan ke dalam campuran air dan minyak, sehingga\u00a040%\u00a0volumenya berada di dalam air. Volume prisma segienam yang berada di udara sebesar\u2026<\/span><\/p>\n

            \n
          1. 12,5%<\/span><\/li>\n
          2. 50%<\/span><\/li>\n
          3. 47,5%<\/span><\/li>\n
          4. 40%<\/span><\/li>\n
          5. 60%<\/span><\/li>\n<\/ol>\n

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

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

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

            \"\"<\/p>\n

            Saat dicelupkan ke dalam campuran air dan minyak (air berada di bawah karena memiliki massa jenis yang lebih besar daripada minyak):<\/span><\/p>\n

            \"\"<\/p>\n

            Ditanya: <\/span>V<\/span><\/em>udara<\/span><\/em><\/sub> =\u2026?<\/span><\/p>\n

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

            Saat dicelupkan ke dalam air, berlaku persamaan berikut.<\/span><\/p>\n

            \"\"<\/p>\n

            Saat dicelupkan ke dalam campuran air dan minyak (air berada di bawah karena memiliki massa jenis yang lebih besar daripada minyak):<\/span><\/p>\n

            \"\"<\/p>\n

            Substitusikan nilai \u03c1<\/span>benda <\/span><\/sub>. <\/span>V<\/span><\/em>benda<\/span><\/sub> dari persamaan (1) ke persamaan (2).<\/span><\/p>\n

            \"\"<\/p>\n

            Mencari volume prisma segienam yang berada di udara:<\/span><\/p>\n

            \"\"<\/p>\n

            Jadi, volume prisma segienam yang berada di udara adalah sebesar 47,5%.<\/span><\/p>\n

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

            Pada percobaan tabung resonansi menggunakan kecepatan bunyi\u00a0320 m\/s,\u00a0terjadi resonansi keempat jika tinggi air\u00a020 cm\u00a0dan terjadi resonansi kelima jika tinggi air\u00a015 cm.\u00a0Frekuensi bunyi yang digunakan pada percobaan tersebut adalah\u2026<\/span><\/p>\n

              \n
            1. 3.200 Hz<\/span><\/li>\n
            2. 28 Hz<\/span><\/li>\n
            3. 4.800 Hz<\/span><\/li>\n
            4. 2.800 Hz<\/span><\/li>\n
            5. 32 Hz<\/span><\/li>\n<\/ol>\n

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

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

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

              \"\"<\/p>\n

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

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

              Mula-mula, Quipperian harus mencari <\/span>L<\/span><\/em>4<\/span><\/em><\/sub>.<\/span><\/p>\n

              \"\"<\/p>\n

              Oleh karena panjang tabung yang digunakan pada resonansi keempat dan kelima sama, maka:<\/span><\/p>\n

              \"\"<\/p>\n

              Frekuensi bunyi yang digunakan adalah sebagai berikut.<\/span><\/p>\n

              \"\"<\/p>\n

              Jadi, frekuensi bunyi yang digunakan pada percobaan tersebut adalah\u00a03.200\u00a0Hz.<\/span><\/p>\n

              Latihan Soal 8<\/strong><\/h2>\n

              Tiga mol gas ideal mengalami proses isobarik pada volume\u00a0V\u00a0sehingga suhunya menjadi tiga kali semula. Jika\u00a0R\u00a0adalah konstanta gas molar, maka besarnya usaha yang dikerjakan oleh gas selama proses tersebut adalah\u2026<\/span><\/p>\n

                \n
              1. RT<\/span><\/em><\/li>\n
              2. 3<\/span>RT<\/span><\/em><\/li>\n
              3. 6<\/span>RT<\/span><\/em><\/li>\n
              4. 2<\/span>RT<\/span><\/em><\/li>\n
              5. 9<\/span>RT<\/span><\/em><\/li>\n<\/ol>\n

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

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

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

                \"\"<\/p>\n

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

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

                Mencari <\/span>V<\/span><\/em>2<\/span><\/em><\/sub> pada proses isobarik.<\/span><\/p>\n

                \"\"<\/p>\n

                Mencari <\/span>W<\/span><\/em>.<\/span><\/p>\n

                \"\"<\/p>\n

                Berdasarkan persamaan umum gas ideal, diperoleh:<\/span><\/p>\n

                \"\"<\/p>\n

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

                \"\"<\/p>\n

                Jadi, besarnya usaha yang dikerjakan oleh gas selama proses tersebut adalah 6<\/span>RT<\/span><\/em>.<\/span><\/p>\n

                Latihan Soal 9<\/strong><\/h2>\n

                Setelah\u00a030 menit,\u00a0suatu unsur radioaktif tinggal\u00a080 gram.\u00a0Tiga puluh menit kemudian, unsur tersebut tinggal\u00a010 gram.\u00a0Massa mula-mula unsur tersebut adalah\u2026<\/span><\/p>\n

                  \n
                1. 80 gram<\/span><\/li>\n
                2. 640 gram<\/span><\/li>\n
                3. 320 gram<\/span><\/li>\n
                4. 40 gram<\/span><\/li>\n
                5. 10 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>0<\/span><\/em><\/sub> =\u2026?<\/span><\/p>\n

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

                  Mencari <\/span>m<\/span><\/em>1<\/span><\/em><\/sub>:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Mencari <\/span>m<\/span><\/em>2<\/span><\/em><\/sub>:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Substitusikan nilai <\/span>m<\/span><\/em>0<\/span><\/em><\/sub> dari persamaan\u00a0(1)\u00a0ke persamaan\u00a0(2),\u00a0sehingga:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Mencari <\/span>m<\/span><\/em>0<\/span><\/em><\/sub>:<\/span><\/p>\n

                  \"\"<\/p>\n

                  Jadi, massa mula-mula unsur tersebut adalah 640 gram.<\/span><\/p>\n

                  Latihan Soal 10<\/strong><\/h2>\n

                  Sebuah kubus bergerak dengan kecepatan\u00a00,6<\/span>c<\/span><\/em>.\u00a0Apabila dilihat oleh pengamat yang diam, maka volume kubus akan menyusut sebesar\u2026<\/span><\/p>\n

                    \n
                  1. 20%<\/span><\/li>\n
                  2. 80%<\/span><\/li>\n
                  3. 60%<\/span><\/li>\n
                  4. 100%<\/span><\/li>\n
                  5. 40%<\/span><\/li>\n<\/ol>\n

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

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

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

                    \"\"<\/p>\n

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

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

                    Pertama, kamu harus mencari volume kubus mula-mula.<\/span><\/p>\n

                    \"\"<\/p>\n

                    Panjang rusuk kubus yang searah dengan arah perpindahannya adalah sebagai berikut.<\/span><\/p>\n

                    \"\"<\/p>\n

                    Sementara kedua rusuk kubus yang lain tidak mengalami penyusutan karena tidak searah dengan arah perpindahannya. Ini berarti, volume kubus akhirnya:<\/span><\/p>\n

                    \"\"<\/p>\n

                    Besarnya penyusutan volume kubus dirumuskan sebagai berikut.<\/span><\/p>\n

                    \"\"<\/p>\n

                    Jadi, volume kubus akan menyusut sebesar 20%.<\/span><\/p>\n

                    Bagaimana Quipperian, apakah kamu sudah mulai paham mengerjakan Prediksi UTBK SBMPTN Fisika SAINTEK 2020 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. <\/span>Kuy<\/span><\/em>, gabung Quipper Video sekarang!<\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

                    SBMPTN sebentar lagi, soal-soal harus segera dikuasai. Itulah sebabnya, Quipperian harus gercep alias gerak cepat. Nah, Bagi kamu yang saat ini masih bingung mencari…<\/p>\n","protected":false},"author":99400369,"featured_media":279749,"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-279746","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-sbmptn","category-soal-sbmptn"],"acf":[],"yoast_head":"\nPrediksi Soal UTBK Fisika 2020 - Quipper Blog<\/title>\n<meta name=\"description\" content=\"Seperti apa prediksi soal UTBK 2020 Fisika? Yuk, coba kerjakan di sini! Selain soal, ada pula jawaban dan pembahasannya, Quipperian.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/quipperhome.wpcomstaging.com\/masuk-ptn\/sbmptn\/soal-sbmptn\/utbk-2020-fisika\/\" \/>\n<meta property=\"og:locale\" content=\"id_ID\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Prediksi Soal UTBK Fisika 2020 - Quipper Blog\" \/>\n<meta property=\"og:description\" content=\"Seperti apa prediksi soal UTBK 2020 Fisika? Yuk, coba kerjakan di sini! 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