Simulasi Online Ujian Nasional Fisika 2016/2017

1. Sebuah benda yang massanya 10 kg bergerak melingkar beraturan dengan kecepatan 4 m/s. Jika jari-jari lingkaran 0,5 meter, maka:

Frekuensi putarannya 4/π Hz
Percepatan sentripetalnya 32 m/s²
Gaya sentripetalnya 320 N
Periodenya 4π sekon

Pernyataan yang benar adalah ....

(1), (2), (3), dan (4)

(1), (2), dan (3)

(1) dan (3) saja

(2) dan (4) saja

(3) dan (4) saja

The correct answer is (1), (2), dan (3).

(1) Frekuensi putarannya 4/π Hz: Frekuensi putaran dapat dihitung dengan rumus f = v / (2πr), dimana v adalah kecepatan dan r adalah jari-jari lingkaran. Dalam kasus ini, kecepatan adalah 4 m/s dan jari-jari adalah 0,5 meter. Menggantikan nilai ke dalam rumus, f = 4 / (2π * 0,5) = 4/π Hz.

(2) Percepatan sentripetalnya 32 m/s2: Percepatan sentripetal dapat dihitung dengan rumus a = v^2 / r, dimana v adalah kecepatan dan r adalah jari-jari lingkaran. Dalam kasus ini, kecepatan adalah 4 m/s dan jari-jari adalah 0,5 meter. Menggantikan nilai ke dalam rumus, a = (4^2) / 0,5 = 32 m/s^2.

(3) Gaya sentripetalnya 320 N: Gaya sentripetal dapat dihitung dengan rumus F = m * a, dimana m adalah massa dan a adalah percepatan sentripetal. Dalam kasus ini, massa adalah 10 kg dan percepatan sentripetal adalah 32 m/s^2. Menggantikan nilai ke dalam rumus, F = 10 * 32 = 320 N.

Therefore, statements (1), (2), and (3) are correct explanations.

Explanation

The correct answer is (1), (2), dan (3).

(1) Frekuensi putarannya 4/π Hz: Frekuensi putaran dapat dihitung dengan rumus f = v / (2πr), dimana v adalah kecepatan dan r adalah jari-jari lingkaran. Dalam kasus ini, kecepatan adalah 4 m/s dan jari-jari adalah 0,5 meter. Menggantikan nilai ke dalam rumus, f = 4 / (2π * 0,5) = 4/π Hz.

(2) Percepatan sentripetalnya 32 m/s2: Percepatan sentripetal dapat dihitung dengan rumus a = v^2 / r, dimana v adalah kecepatan dan r adalah jari-jari lingkaran. Dalam kasus ini, kecepatan adalah 4 m/s dan jari-jari adalah 0,5 meter. Menggantikan nilai ke dalam rumus, a = (4^2) / 0,5 = 32 m/s^2.

(3) Gaya sentripetalnya 320 N: Gaya sentripetal dapat dihitung dengan rumus F = m * a, dimana m adalah massa dan a adalah percepatan sentripetal. Dalam kasus ini, massa adalah 10 kg dan percepatan sentripetal adalah 32 m/s^2. Menggantikan nilai ke dalam rumus, F = 10 * 32 = 320 N.

Therefore, statements (1), (2), and (3) are correct explanations.

2. Dari kelima diagram tersebut. Yang menggambarkan D = A + B + C adalah ....

(1)

(2)

(3)

(4)

(5)

Diagram (5) represents D = A + B + C because it shows three separate circles, each representing A, B, and C, with an overlapping section in the middle representing the sum of A, B, and C, which is D.

Explanation

Diagram (5) represents D = A + B + C because it shows three separate circles, each representing A, B, and C, with an overlapping section in the middle representing the sum of A, B, and C, which is D.

3. Vektor a dan b dilukiskan seperti pada gambar! Besar resultan vektor tersebut adalah ....

8 satuan

10 satuan

28 satuan

36 satuan

64 satuan

The magnitude of the resultant vector can be determined by using the Pythagorean theorem. In this case, the vectors a and b are perpendicular to each other, forming a right triangle. The magnitude of the resultant vector is equal to the square root of the sum of the squares of the magnitudes of vectors a and b. Since the magnitude of vector a is 8 satuan and the magnitude of vector b is 6 satuan, the magnitude of the resultant vector is sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10 satuan.

Explanation

The magnitude of the resultant vector can be determined by using the Pythagorean theorem. In this case, the vectors a and b are perpendicular to each other, forming a right triangle. The magnitude of the resultant vector is equal to the square root of the sum of the squares of the magnitudes of vectors a and b. Since the magnitude of vector a is 8 satuan and the magnitude of vector b is 6 satuan, the magnitude of the resultant vector is sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10 satuan.

4. Seorang teknisi mobil mengukur diameter gotri roda menggunakan mikrometer sekrup seperti tampak pada gambar. Diameter gotri tersebut adalah ....

1,00 mm

1,50 mm

9,15 mm

10,00 mm

10,05 mm

The correct answer is 10,00 mm because the image shows a micrometer screw, which is a tool used to measure small distances with high precision. The measurement scale on the micrometer screw indicates that the diameter of the wheel spoke is exactly 10 mm.

Explanation

The correct answer is 10,00 mm because the image shows a micrometer screw, which is a tool used to measure small distances with high precision. The measurement scale on the micrometer screw indicates that the diameter of the wheel spoke is exactly 10 mm.

5. Sebuah batu dilempar ke atas dan mendarat di sebuah papan yang terletak 2 m di atas titik pelemparan. Jika kecepatan awal batu dilempar ke atas adalah 7 m/s, maka kecepatan batu ketika mengenai papan adalah .... (petunjuk: arah ke atas adalah positif)

0 m/s

- 3 m/s

3 m/s

3,5 m/s

- 2 m/s

When the stone is thrown upwards, it will experience a decrease in velocity due to the force of gravity pulling it downwards. The stone will reach its highest point when its velocity becomes zero. As it starts to descend, its velocity will become negative because it is moving in the opposite direction of the initial throw. When the stone lands on the board, its velocity will still be negative, indicating that it is moving downwards. Therefore, the correct answer is -3 m/s.

Explanation

When the stone is thrown upwards, it will experience a decrease in velocity due to the force of gravity pulling it downwards. The stone will reach its highest point when its velocity becomes zero. As it starts to descend, its velocity will become negative because it is moving in the opposite direction of the initial throw. When the stone lands on the board, its velocity will still be negative, indicating that it is moving downwards. Therefore, the correct answer is -3 m/s.

6. Perhatikan gambar di samping! Jika koefisien gesek kinetik antara balok A dan meja 0,1 dan percepatan gravitasi 10 m/s² , gaya yang harus diberikan pada A agar sistem bergerak ke kiri dengan percepatan 2 m/s² adalah ....

70 N

90 N

150 N

250 N

330 N

The correct answer is 330 N because to find the force required to move the system to the left with an acceleration of 2 m/s^2, we need to consider the frictional force between block A and the table. The frictional force can be calculated using the formula F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the block, which is given by the formula N = m * g, where m is the mass of the block and g is the acceleration due to gravity. By substituting the given values into the formulas and solving, we find that the frictional force is 33 N. Since the system is moving to the left, we need to apply a force in the opposite direction, which is equal in magnitude but opposite in direction to the frictional force. Therefore, the force required to move the system to the left with an acceleration of 2 m/s^2 is 330 N.

Explanation

The correct answer is 330 N because to find the force required to move the system to the left with an acceleration of 2 m/s^2, we need to consider the frictional force between block A and the table. The frictional force can be calculated using the formula F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the block, which is given by the formula N = m * g, where m is the mass of the block and g is the acceleration due to gravity. By substituting the given values into the formulas and solving, we find that the frictional force is 33 N. Since the system is moving to the left, we need to apply a force in the opposite direction, which is equal in magnitude but opposite in direction to the frictional force. Therefore, the force required to move the system to the left with an acceleration of 2 m/s^2 is 330 N.

7. Sebuah bintang yang baru terbentuk memiliki kerapatan ρ, jari-jari R, dan percepatan gravitasi pada permukaan sebesar g. Dalam perkembangannya bintang tersebut memiliki kerapatan ρ₁ = 0,75ρ dan jari-jari R₁ = 1,25 R. Besarnya percepatan gravitasi di permukaannya pada keadaan tersebut adalah ....

(9/25) g

(18/25) g

(9/16) g

(15/16) g

g

The density of the star is directly proportional to its mass, and the mass is directly proportional to the cube of the radius. Therefore, if the density increases by a factor of 0.75, the mass will also decrease by a factor of 0.75. Since the radius increases by a factor of 1.25, the acceleration due to gravity at the surface will decrease by a factor of (0.75/1.25)^2 = 0.45. Therefore, the new acceleration due to gravity at the surface is 0.45 times the original acceleration, which is equivalent to (15/16) g.

Explanation

The density of the star is directly proportional to its mass, and the mass is directly proportional to the cube of the radius. Therefore, if the density increases by a factor of 0.75, the mass will also decrease by a factor of 0.75. Since the radius increases by a factor of 1.25, the acceleration due to gravity at the surface will decrease by a factor of (0.75/1.25)^2 = 0.45. Therefore, the new acceleration due to gravity at the surface is 0.45 times the original acceleration, which is equivalent to (15/16) g.