Kuis Interaktif Ukm Ke-1 Kelas Xi IPA

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| By Achmad Yunus Arb
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Achmad Yunus Arb
Community Contributor
Quizzes Created: 1 | Total Attempts: 166
Questions: 5 | Attempts: 166

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Kuis Interaktif Ukm Ke-1 Kelas Xi IPA - Quiz

SOAL INI UNTUK REMIDI UKM KE - 1 KELAS XI IPA 2009/2010


Questions and Answers
  • 1. 

    1. Koordinat suatu benda sebagai fungsi waktu adalah x = 2t3 – 18t, x dalam meter dan t dalam sekon. Percepatan pada saat t = 2 sekon adalah … .

    • A.

      6

    • B.

      12

    • C.

      16

    • D.

      24

    • E.

      32

    Correct Answer
    D. 24
    Explanation
    The given equation represents the position of an object as a function of time. To find the acceleration at t = 2 seconds, we need to find the second derivative of the position equation with respect to time. Taking the derivative of 2t^3 - 18t, we get 6t^2 - 18. Plugging in t = 2, we get 6(2)^2 - 18 = 24. Therefore, the acceleration at t = 2 seconds is 24 m/s^2.

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  • 2. 

    3. Benda dengan massa 10 kg dalam keadaan diam berada di bidang datar kasar ( s = 0,40 ; k = 0,35), g = 10 m/s2. Bila diberi gaya horisontal yang tetap sebesar 30 N, besarnya gaya gesekan yang bekerja pada benda tersebut adalah … .

    • A.

      20

    • B.

      25

    • C.

      30

    • D.

      35

    • E.

      40

    Correct Answer
    C. 30
    Explanation
    The question states that a 10 kg object is at rest on a rough surface with a coefficient of friction of 0.40. When a horizontal force of 30 N is applied, we need to determine the magnitude of the frictional force acting on the object. The frictional force can be calculated using the equation Ff = μN, where Ff is the frictional force, μ is the coefficient of friction, and N is the normal force. The normal force can be calculated as the weight of the object, which is mass times gravity, N = mg. Substituting the given values, we get N = 10 kg * 10 m/s^2 = 100 N. Now substituting the values into the frictional force equation, Ff = 0.40 * 100 N = 40 N. Therefore, the correct answer is 40.

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  • 3. 

    Besar gaya gravitasi antara dua benda yang berinteraksi adalah …..

    • A.

      Sebanding dengan massa masing-masing benda.

    • B.

      Berbanding terbalik dengan jarak kedua benda.

    • C.

      Sebanding dengan jarak kedua benda.

    • D.

      Sebanding dengan kuadrat jarak kedua benda.

    • E.

      Berbanding terbalik dengan massa masing-masing benda.

    Correct Answer
    A. Sebanding dengan massa masing-masing benda.
    Explanation
    The gravitational force between two objects is directly proportional to the mass of each object. This means that as the mass of either object increases, the gravitational force between them also increases. Conversely, if the mass of either object decreases, the gravitational force decreases as well. Therefore, the correct answer is that the gravitational force is proportional to the mass of each object.

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  • 4. 

    6. Planet A mempunyai periode TA, planet B mempunyai periode TB. Jika perbandingan jarak planet A terhadap jarak planet B ke matahari adalah 4 : 9, berapakah nilai perbandingan periode TA : TB ?

    • A.

      2 : 3

    • B.

      4 : 9

    • C.

      8 : 27

    • D.

      16 : 81

    • E.

      64 : 729

    Correct Answer
    C. 8 : 27
    Explanation
    The correct answer can be obtained by using Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is directly proportional to the cube of its average distance from the Sun. In this case, since the ratio of the distances is 4:9, the ratio of the orbital periods will be the square root of 4: the square root of 9, which simplifies to 2:3. Therefore, the ratio of the orbital periods TA:TB will be 2:3.

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  • 5. 

    Pada benda yang mengalami getaran harmonik, maka jumlah energi kinetik dan energi potensialnya adalah …..

    • A.

      Maksimum pada simpangan maksimum

    • B.

      Maksimum pada simpangan nol.

    • C.

      Tetap besarnya pada simpangan berapapun.

    • D.

      Berbanding lurus dengan simpangannya.

    • E.

      Berbanding terbalik pada simpangannya.

    Correct Answer
    C. Tetap besarnya pada simpangan berapapun.
    Explanation
    In a harmonic vibration, the total mechanical energy (kinetic energy + potential energy) remains constant regardless of the amplitude. This means that the kinetic energy and potential energy are always equal and opposite at any given displacement. Therefore, the correct answer is "tetap besarnya pada simpangan berapapun" which means "remains constant at any displacement".

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Apr 23, 2010
    Quiz Created by
    Achmad Yunus Arb
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