Soal Semester Kelas 11
Questions and Answers
- 1.
1. Bila posisi sebuah benda dinyatakan dengan persamaan x = 5 t3 + 2t 2 -3t, maka percepatan benda tersebut dinyatakan oleh ….
- A.
A.30 t
- B.
B. 30 t + 4
- C.
C.15 t 2 + 4t
- D.
D. t 6+ t 4-3t
- E.
E. t 4 +t 3 -3
Correct Answer
B. B. 30 t + 4Explanation
The given equation represents the position of an object as a function of time. To find the acceleration of the object, we need to take the second derivative of the position equation with respect to time. Taking the derivative of x = 5t^3 + 2t^2 - 3t gives us the velocity equation v = 15t^2 + 4t - 3. Taking the derivative of the velocity equation gives us the acceleration equation a = 30t + 4. Therefore, the correct answer is B. 30t + 4.Rate this question:
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- 2.
2.Percepatan sebuah partikel pada saat t adalah 6 t i – 4 j. Mula-mula partikel bergerak dengan kecepatan 2i. Vector kecepatan partikel pada saat t adalah …
- A.
A. (2 +3 t)i – 4t j
- B.
B. (2 – 3 t)i + 4 t j
- C.
C. (2 -3 t2) i + 4 t j
- D.
D. (2 + 3 t2) i - 4 t j
- E.
E. (3 + 2t t2) i – 4 t j
Correct Answer
D. D. (2 + 3 t2) i - 4 t jExplanation
The given question provides information about the acceleration of a particle at time t, which is 6t i - 4j. It also states that the initial velocity of the particle is 2i. To find the velocity of the particle at time t, we need to integrate the acceleration with respect to time. Integrating 6t i - 4j with respect to t gives us (3t^2)i - 4tj. Adding the initial velocity of 2i to this, we get (2 + 3t^2)i - 4tj. Therefore, the correct answer is D. (2 + 3t^2) i - 4t j.Rate this question:
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- 3.
3.Sebuah bola ditendang dengan kecepatan awal 20 m s–1 dan sudut elevasi 30o dan waktu yang diperlukan 2 sekon. Jika g = 10 m s–2 , jarak mendatar yang dicapai bola adalah …
- A.
A. 20√3 m
- B.
B. 20 m
- C.
C. 10√3 m
- D.
D. 10 m
- E.
E. 5 m
Correct Answer
A. A. 20√3 mExplanation
The horizontal distance covered by the ball can be calculated using the formula: distance = initial velocity * time. In this case, the initial velocity is 20 m/s and the time is 2 seconds. Therefore, the horizontal distance covered by the ball is 20 m/s * 2 s = 40 m. However, since the ball was kicked at an angle of 30 degrees, only the horizontal component of the initial velocity contributes to the horizontal distance. The horizontal component can be calculated using the formula: horizontal component = initial velocity * cos(angle). In this case, the horizontal component is 20 m/s * cos(30) = 20 * (√3/2) = 10√3 m. Therefore, the correct answer is A. 20√3 m.Rate this question:
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- 4.
4.Sebuah peluru ditembakkan dengan kecepatan awal 100 m s–1 dan sudut elevasi 300. Jika gravitasi di tempat itu 10 m s–2, maka waktu yang diperlukan peluru tersebut untuk mencapai titik tertinggi adalah …
- A.
A. 2 sekon
- B.
B. 5 sekon
- C.
C. 6 sekon
- D.
D. 10 sekon
- E.
E. 15 sekon
Correct Answer
B. B. 5 sekonExplanation
The time taken for the bullet to reach its highest point can be calculated using the equation of motion for vertical motion. The initial velocity of the bullet is 100 m/s and the angle of elevation is 30 degrees. The vertical component of the initial velocity can be found by multiplying the initial velocity by the sine of the angle of elevation. The time taken to reach the highest point can be found by dividing the vertical component of the initial velocity by the acceleration due to gravity. Using these calculations, the time is found to be 5 seconds.Rate this question:
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- 5.
5.Suatu benda berotasi mengitari sebuah poros dengan posisi sudutnya, θ = 2t 2 -9 t + 4; θ dalam rad dan t dalam sekon. Kecepatan sudut suatu partikel pada benda pada t = 1,0 sekon, adalah ….
- A.
A. – 6,0 rad/s
- B.
B. – 5,0 rad/s
- C.
C. – 4,0 rad/s
- D.
D. – 3,0 rad/s
- E.
E. – 2 rad/s
Correct Answer
B. B. – 5,0 rad/sExplanation
The given equation represents the angular displacement (θ) of an object rotating about an axis as a function of time (t). To find the angular velocity (ω), which represents the rate of change of angular displacement with respect to time, we need to differentiate the equation with respect to time.
Differentiating the equation, we get ω = d(θ)/dt = 4t - 9.
Substituting t = 1.0 into the equation, we get ω = 4(1.0) - 9 = -5.0 rad/s.
Therefore, the correct answer is B. – 5.0 rad/s.Rate this question:
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- 6.
6.Suatu benda berotasi mengitari sebuah poros dengan kecepatan sudutnya, ω dapat dinyatakan sebagai ω = t 2 -5,0; ω dalam rad/s dan t dalam sekon. Percepatan sudut partikel pada benda t =1,0 sekon adalah ….
- A.
A. 2,0 rad/s
- B.
B. 2,5 rad/s
- C.
C. 3,0 rad/s
- D.
D. 3,5 rad/s
- E.
E. 4,0 rad/s
Correct Answer
A. A. 2,0 rad/sExplanation
The given equation ω = t^2 - 5.0 represents the angular velocity of the rotating object as a function of time. To find the angular acceleration at t = 1.0 second, we need to take the derivative of the equation with respect to time. Taking the derivative of ω = t^2 - 5.0 gives us the angular acceleration, which is a constant value of 2.0 rad/s. Therefore, the correct answer is A. 2.0 rad/s.Rate this question:
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- 7.
Perhatikan pernyataan berikut: (1) Berbanding lurus dengan percepatan sudut (2) Berbanding terbalik dengan jari-jari (3) Berbanding lurus dengan jari-jari (4) Berbanding lurus dengan pangkat dua kecepatan luncur Yang berlaku untuk percepatan tangensial pada gerakan lengkung adalah …
- A.
A. (1) dan (2)
- B.
B. (1) dan (3)
- C.
C. (2) dan (4)
- D.
D. (3) dan (4)
- E.
E. (4) saja
Correct Answer
B. B. (1) dan (3)Explanation
Percepatan tangensial pada gerakan lengkung berbanding lurus dengan percepatan sudut (pernyataan 1) dan berbanding lurus dengan jari-jari (pernyataan 3).Rate this question:
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- 8.
8.Pada sebuah benda yang bergerak beraturan dengan lintasan melingkar, kecepatan liniernya bergantung pada …
- A.
A. massa dan jari-jari lingkaran
- B.
B. massa dan periode
- C.
C. massa dan frekuensi
- D.
D. periode dan jari-jari lintasan
- E.
E. kecepatan sudut dan jari-jari lingkaran
Correct Answer
E. E. kecepatan sudut dan jari-jari lingkaranExplanation
The correct answer is E because the linear velocity of an object moving in a circular path depends on the angular velocity (kecepatan sudut) and the radius of the circle (jari-jari lingkaran). The linear velocity is the distance traveled per unit of time, and in a circular motion, it is equal to the product of the angular velocity and the radius of the circle. Therefore, both the angular velocity and the radius of the circle determine the linear velocity of the object.Rate this question:
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- 9.
9.Partikel melakukan gerak rotasi dengan persaman posisi sudut θ = t 3 – t 2 + 5, θ dalam radian dan t dalam sekon. Percepatan sudut partikel tersebut saat t = 2 s adalah ... .
- A.
A. 4 rad/s2
- B.
B. 6 rad/s2
- C.
C. 8 rad/s2
- D.
D. 10 rad/s2
- E.
E. 15 rad/s2
Correct Answer
D. D. 10 rad/s2Explanation
The given equation for the position angle of the particle is θ = t^3 - t^2 + 5, where θ is in radians and t is in seconds. To find the angular acceleration of the particle at t = 2 s, we need to differentiate the equation twice with respect to time. The first derivative gives us the angular velocity, and the second derivative gives us the angular acceleration. Differentiating the equation twice, we get θ' = 3t^2 - 2t and θ'' = 6t - 2. Substituting t = 2 s into the second derivative, we get θ'' = 6(2) - 2 = 12 - 2 = 10 rad/s^2. Therefore, the correct answer is D. 10 rad/s^2.Rate this question:
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- 10.
10.Percepatan gravitasi rata-rata di permukaan bumi sama dengan a. Untuk tempat di ketinggian R (R = jari-jari bumi) dari permukaan bumi, memiliki percepatan gravitasi sebesar …
- A.
A. 0,125 a
- B.
B. 4,000 a
- C.
C. 0,500 a
- D.
D. 1,000 a
- E.
E. 0,250 a
Correct Answer
E. E. 0,250 aExplanation
The average gravitational acceleration at the surface of the Earth is equal to 1g, where g is the acceleration due to gravity. The acceleration due to gravity decreases as we move away from the surface of the Earth. Since the question states that the height R is equal to the radius of the Earth, the acceleration due to gravity at that height would be half of the average gravitational acceleration at the surface. Therefore, the acceleration due to gravity at height R would be 0.5g, which is equivalent to 0.250 a.Rate this question:
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Mar 21, 2023Quiz Edited by
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Dec 14, 2011Quiz Created by
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