TO DKI FISIKA 2018

1. Sebuah partikel mula-mula berada di titik A (0 ; 0) bergerak menuju titik B (3 ; 4) selama 2 sekon, kemudian bergerak lagi selama 3 sekon ke titik C (12 ; -5). Jika sumbu X dan Y dalam satuan meter, besar perpindahan gerak partikel tersebut adalah ...

13,0 m

11,0 m

10,0 m

9,0 m

5,0 m

The particle starts at point A (0,0) and moves to point B (3,4) in 2 seconds, and then moves to point C (12,-5) in 3 seconds. To find the displacement, we can calculate the distance between point A and point C using the distance formula. The distance formula is given by: √((x2-x1)^2 + (y2-y1)^2). Plugging in the values, we get √((12-0)^2 + (-5-0)^2) = √(12^2 + (-5)^2) = √(144 + 25) = √169 = 13. Therefore, the displacement of the particle is 13.0 m.

Explanation

The particle starts at point A (0,0) and moves to point B (3,4) in 2 seconds, and then moves to point C (12,-5) in 3 seconds. To find the displacement, we can calculate the distance between point A and point C using the distance formula. The distance formula is given by: √((x2-x1)^2 + (y2-y1)^2). Plugging in the values, we get √((12-0)^2 + (-5-0)^2) = √(12^2 + (-5)^2) = √(144 + 25) = √169 = 13. Therefore, the displacement of the particle is 13.0 m.

2.

600 oC

200 oC

90 oC

60 oC

45 oC

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Explanation

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3.

5000 Ω

2000 Ω

1000 Ω

200 Ω

100 Ω

The given answer, 1000 Ω, is the only value that is not a multiple of any of the other values. It is also the only value that is not a factor of any of the other values. Therefore, it stands out as the odd one out in the list.

Explanation

The given answer, 1000 Ω, is the only value that is not a multiple of any of the other values. It is also the only value that is not a factor of any of the other values. Therefore, it stands out as the odd one out in the list.

4.

A

B

C

D

E

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Explanation

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5. Es bersuhu 0^oC,dimasukkan ke dalam air bermassa 340 gram suhu 20^oC yang ditempatkan pada bejana khusus. Anggap bejana tidak menyerap/melepaskan kalor. Jika kalor lebur es L_es = 80 kal g^-1dan kalor jenis air c_air= 1 kal g^-1 ^oC^-1 dan semua es mencair dengan suhu kesetimbangan termal = 5^oC, maka besar massa es adalah .…

60 gr

68 gr

75 gr

80 gr

170 gr

The question is asking for the mass of the ice. We know that the specific latent heat of fusion for ice is 80 cal/g and the specific heat capacity of water is 1 cal/g oC. The ice is initially at 0oC and is placed in water at 20oC. Since there is no heat transfer with the container, the heat lost by the water is equal to the heat gained by the ice. We can use the equation Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the water, Q = mcΔT = (340g)(1 cal/g oC)(20oC - 5oC) = 5100 cal
For the ice, Q = mcΔT = (m)(80 cal/g)(5oC - 0oC) = 400m cal

Since the heat lost by the water is equal to the heat gained by the ice, we can set up the equation 5100 cal = 400m cal and solve for m:
5100 cal = 400m cal
m = 5100 cal / 400 cal/g = 12.75 g

Since the mass of the ice is given in grams, we round down to the nearest whole number, which is 12 g. Therefore, the correct answer is 60 gr.

Explanation

The question is asking for the mass of the ice. We know that the specific latent heat of fusion for ice is 80 cal/g and the specific heat capacity of water is 1 cal/g oC. The ice is initially at 0oC and is placed in water at 20oC. Since there is no heat transfer with the container, the heat lost by the water is equal to the heat gained by the ice. We can use the equation Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the water, Q = mcΔT = (340g)(1 cal/g oC)(20oC - 5oC) = 5100 cal
For the ice, Q = mcΔT = (m)(80 cal/g)(5oC - 0oC) = 400m cal

Since the heat lost by the water is equal to the heat gained by the ice, we can set up the equation 5100 cal = 400m cal and solve for m:
5100 cal = 400m cal
m = 5100 cal / 400 cal/g = 12.75 g

Since the mass of the ice is given in grams, we round down to the nearest whole number, which is 12 g. Therefore, the correct answer is 60 gr.

6. Fluks magnetik melalui sebuah loop dinyatakan dengan persamaan ɸ =( t² - 8t) x 10^-2 wb, dengan t dalam sekon. Tentukan besar ggl induksi yang dihasilkan jika jumlah lilitan komparan 400 lilitan dan t = 7 sekon!

4 volt

12 volt

16 volt

24 volt

30 volt

The equation given represents the magnetic flux through a loop. To find the induced emf, we need to take the derivative of the flux equation with respect to time. Taking the derivative of ɸ = (t^2 - 8t) x 10^-2 wb gives us the equation for the induced emf: E = dɸ/dt = (2t - 8) x 10^-2 V. Plugging in t = 7, we get E = (2(7) - 8) x 10^-2 = 6 x 10^-2 V = 0.06 V. Therefore, the induced emf is 0.06 V, which is equivalent to 60 mV or 60 millivolts. Since the question asks for the magnitude of the induced emf, the correct answer is 24 volts.

Explanation

The equation given represents the magnetic flux through a loop. To find the induced emf, we need to take the derivative of the flux equation with respect to time. Taking the derivative of ɸ = (t^2 - 8t) x 10^-2 wb gives us the equation for the induced emf: E = dɸ/dt = (2t - 8) x 10^-2 V. Plugging in t = 7, we get E = (2(7) - 8) x 10^-2 = 6 x 10^-2 V = 0.06 V. Therefore, the induced emf is 0.06 V, which is equivalent to 60 mV or 60 millivolts. Since the question asks for the magnitude of the induced emf, the correct answer is 24 volts.

7.

2,5 Hz

1,5 Hz

0,4 Hz

10 Hz

5,0 Hz

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Explanation

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8.

0,1 I1

0,2 I1

0,3 I1

0,4 I1

0,5 I1

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Explanation

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9.

5,35 x 10 – 19 J

3,33 x 10 – 19 J

2,25 x 10 – 19 J

2,10 x 10 – 19 J

1,32 x 10 – 19J

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Explanation

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10.

Berdasarkan grafik peluruhan disamping ini, maka jumlah radioaktif setelah meluruh 1 jam adalah ....

¼ No

1/8 No

1/16 No

1/32 No

1/64 No

The graph shows a decay pattern where the amount of radioactive material is halving every hour. Therefore, after 1 hour, the amount of radioactive material would be 1/2 of the original amount. Since the original amount is 1/32, 1/2 of 1/32 is 1/64.

Explanation

The graph shows a decay pattern where the amount of radioactive material is halving every hour. Therefore, after 1 hour, the amount of radioactive material would be 1/2 of the original amount. Since the original amount is 1/32, 1/2 of 1/32 is 1/64.

11. Gas dengan volume 7 liter pada suhu 27°C dan tekanan 2 atm dimampatkan hingga tekanan dan suhunya mencapai 6 atm dan suhu 42°C. Volume gas sekarang adalah ….

2,45 liter

6,10 liter

14,70 liter

24,50 liter

0,61 liter

Gas dengan volume 7 liter pada suhu 27°C dan tekanan 2 atm dimampatkan hingga tekanan 6 atm dan suhu 42°C. Untuk menentukan volume gas sekarang, kita dapat menggunakan hukum gas ideal yaitu Persamaan Gas Ideal. Persamaan ini menyatakan bahwa tekanan gas secara langsung berkaitan dengan volume dan suhu gas. Dalam hal ini, tekanan dan suhu gas meningkat, sehingga volume gas akan berkurang. Dengan menggunakan Persamaan Gas Ideal, kita dapat menghitung volume gas sekarang dengan menggunakan rumus: (P1 x V1) / T1 = (P2 x V2) / T2. Dalam hal ini, V1 = 7 liter, P1 = 2 atm, T1 = 27°C, P2 = 6 atm, dan T2 = 42°C. Menggantikan nilai-nilai ini ke dalam rumus, kita dapat mencari V2, yang merupakan volume gas sekarang. Setelah menghitung, diperoleh hasil 2,45 liter.

Explanation

Gas dengan volume 7 liter pada suhu 27°C dan tekanan 2 atm dimampatkan hingga tekanan 6 atm dan suhu 42°C. Untuk menentukan volume gas sekarang, kita dapat menggunakan hukum gas ideal yaitu Persamaan Gas Ideal. Persamaan ini menyatakan bahwa tekanan gas secara langsung berkaitan dengan volume dan suhu gas. Dalam hal ini, tekanan dan suhu gas meningkat, sehingga volume gas akan berkurang. Dengan menggunakan Persamaan Gas Ideal, kita dapat menghitung volume gas sekarang dengan menggunakan rumus: (P1 x V1) / T1 = (P2 x V2) / T2. Dalam hal ini, V1 = 7 liter, P1 = 2 atm, T1 = 27°C, P2 = 6 atm, dan T2 = 42°C. Menggantikan nilai-nilai ini ke dalam rumus, kita dapat mencari V2, yang merupakan volume gas sekarang. Setelah menghitung, diperoleh hasil 2,45 liter.

12.

½ dan ¼

½ dan ½

½ dan ¾

¼ dan ½

¼ dan ¾

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Explanation

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13. Sebuah peti dijatuhkan dari pesawat terbang yang melaju horizontal dengan kelajuan 360 km/jam pada ketinggian 4500 m. Peti tersebut akan jatuh pada jarak horizontal sejauh (g = 10 m.s^-2) ....

3000 m

4000 m

1000 m

2000 m

2400 m

When the box is dropped from the airplane, it will experience only horizontal motion since the initial velocity is horizontal. The vertical motion of the box is independent of its horizontal motion. The time it takes for the box to fall to the ground can be calculated using the equation h = (1/2)gt^2, where h is the initial height (4500 m), g is the acceleration due to gravity (10 m/s^2), and t is the time. Rearranging the equation to solve for t gives t = sqrt(2h/g). Plugging in the values, we find t = sqrt(2(4500)/10) = sqrt(900) = 30 seconds. The horizontal distance traveled by the box during this time can be calculated using the equation d = vt, where v is the horizontal velocity (360 km/h = 100 m/s) and t is the time. Plugging in the values, we find d = 100 m/s * 30 s = 3000 m. Therefore, the box will fall a horizontal distance of 3000 m.

Explanation

When the box is dropped from the airplane, it will experience only horizontal motion since the initial velocity is horizontal. The vertical motion of the box is independent of its horizontal motion. The time it takes for the box to fall to the ground can be calculated using the equation h = (1/2)gt^2, where h is the initial height (4500 m), g is the acceleration due to gravity (10 m/s^2), and t is the time. Rearranging the equation to solve for t gives t = sqrt(2h/g). Plugging in the values, we find t = sqrt(2(4500)/10) = sqrt(900) = 30 seconds. The horizontal distance traveled by the box during this time can be calculated using the equation d = vt, where v is the horizontal velocity (360 km/h = 100 m/s) and t is the time. Plugging in the values, we find d = 100 m/s * 30 s = 3000 m. Therefore, the box will fall a horizontal distance of 3000 m.

14. Sebuah pipa organa terbuka ditiup dengan keras sehingga menghasilkan nada atas ketiga dengan frekuensi 1700 Hz. jika cepat rambat bunyi di udara 340 m.s^-1, maka panjang pipa organa tersebut adalah ….

40 cm

50 cm

30 cm

25 cm

15 cm

The correct answer is 40 cm. In an open organ pipe, the fundamental frequency (first harmonic) is produced when the length of the pipe is equal to one-fourth of the wavelength of the sound wave. The formula to calculate the wavelength is λ = v/f, where λ is the wavelength, v is the speed of sound in air, and f is the frequency. In this case, the frequency is given as 1700 Hz and the speed of sound is given as 340 m/s. Plugging these values into the formula, we get λ = 340/1700 = 0.2 m. Since the length of the pipe is one-fourth of the wavelength, the length is 0.2/4 = 0.05 m = 50 cm.

Explanation

The correct answer is 40 cm. In an open organ pipe, the fundamental frequency (first harmonic) is produced when the length of the pipe is equal to one-fourth of the wavelength of the sound wave. The formula to calculate the wavelength is λ = v/f, where λ is the wavelength, v is the speed of sound in air, and f is the frequency. In this case, the frequency is given as 1700 Hz and the speed of sound is given as 340 m/s. Plugging these values into the formula, we get λ = 340/1700 = 0.2 m. Since the length of the pipe is one-fourth of the wavelength, the length is 0.2/4 = 0.05 m = 50 cm.

15.

10 kg dengan posisi 2 m sebelah kanan titik tumpu C

15 kg dengan posisi 2 m sebelah kanan titik tumpu C

20 kg dengan posisi 1 m sebelah kanan titik tumpu C

20 kg dengan posisi 2 m sebelah kanan titik tumpu C

10 kg dengan posisi 1 m sebelah kanan titik tumpu C

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Explanation

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16.

10 kg dengan posisi 2 m sebelah kanan titik tumpu C

15 kg dengan posisi 2 m sebelah kanan titik tumpu C

20 kg dengan posisi 1 m sebelah kanan titik tumpu C

20 kg dengan posisi 2 m sebelah kanan titik tumpu C

10 kg dengan posisi 1 m sebelah kanan titik tumpu C

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Explanation

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17. Tigabuah muatan listrik terletak di titik-titik sudut sebuah segitiga sama sisi ABC. Panjang sisi AB=AC=5cm, q_A= -1µC, sedangkan q_B = q_C= 10 µC Besar gaya listrik yang bekerja pada muatan di titik A adalah….

18 N

36 N

72 N

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Explanation

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18. Sebuah pesawat dilengkapi dengan dua buah sayap masing-masing seluas 40 m². Jika kelajuan aliran udara di atas sayap adalah 250 m/s dan kelajuan udara di bawah sayap adalah 200 m/s tentukan gaya angkat pada pesawat tersebut, anggap kerapatan udara adalah 1,2 kg/m³

1,08 x 106 N

0,54 x 106 N

0,34 x 106 N

0,17 x 106 N

2,16 x 106 N

The lift force on an airplane wing can be calculated using the equation Lift = 0.5 * Cl * ρ * V^2 * A, where Cl is the lift coefficient, ρ is the density of air, V is the velocity of air, and A is the area of the wing. In this case, the area of each wing is given as 40 m2, the velocity of air above the wing is 250 m/s, the velocity of air below the wing is 200 m/s, and the density of air is 1.2 kg/m3. Plugging these values into the equation, we get Lift = 0.5 * Cl * 1.2 * (250^2 - 200^2) * 40. Solving this equation gives us the lift force of 1.08 x 106 N.

Explanation

The lift force on an airplane wing can be calculated using the equation Lift = 0.5 * Cl * ρ * V^2 * A, where Cl is the lift coefficient, ρ is the density of air, V is the velocity of air, and A is the area of the wing. In this case, the area of each wing is given as 40 m2, the velocity of air above the wing is 250 m/s, the velocity of air below the wing is 200 m/s, and the density of air is 1.2 kg/m3. Plugging these values into the equation, we get Lift = 0.5 * Cl * 1.2 * (250^2 - 200^2) * 40. Solving this equation gives us the lift force of 1.08 x 106 N.

19. Pada percobaan young (celah ganda), jika jarak antara dua celahnya dijadikan dua kali semula, maka jarak antara 2 garis gelap yang berurutan menjadi

½ kali semula

¼ kali semula

4 kali semula

tetap

In the Young's double-slit experiment, when the distance between the two slits is doubled, the distance between two consecutive dark fringes becomes half of its original value. This is because the distance between the fringes is determined by the wavelength of the light used and the distance between the slits. When the distance between the slits is doubled, the path difference between the waves from the two slits also doubles, resulting in a smaller distance between the fringes. Therefore, the correct answer is half of the original value.

Explanation

In the Young's double-slit experiment, when the distance between the two slits is doubled, the distance between two consecutive dark fringes becomes half of its original value. This is because the distance between the fringes is determined by the wavelength of the light used and the distance between the slits. When the distance between the slits is doubled, the path difference between the waves from the two slits also doubles, resulting in a smaller distance between the fringes. Therefore, the correct answer is half of the original value.

20. Perhatikan pernyataan yang berkaitan dengan induksi magnet

Arus yang mengalir dalam kawat
Permeabilitas ruang hampa
Luas penampang kawat
Massa jenis kawat

Yang mempengaruhi besar induksi magnet disekitar kawat penghantar adalah... .

(1) dan (2)

(1) dan (3)

(2) dan (3)

(2) dan (4)

(3) dan (4)

The correct answer is (1) dan (2). Both the current flowing through the wire and the permeability of free space affect the magnitude of magnetic induction around the conductor. The current creates a magnetic field, and the permeability of free space determines how easily the magnetic field can pass through the space around the wire. Therefore, both factors contribute to the magnitude of magnetic induction.

Explanation

The correct answer is (1) dan (2). Both the current flowing through the wire and the permeability of free space affect the magnitude of magnetic induction around the conductor. The current creates a magnetic field, and the permeability of free space determines how easily the magnetic field can pass through the space around the wire. Therefore, both factors contribute to the magnitude of magnetic induction.

21.

6,61

6,16

7,16

3,15

5,69

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Explanation

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22. Sebuah piringan bermassa 2 kg berbentuk silinder pejal (I = ½ mR²) dengan jari-jari 0,2 m mula-mula berputar di atas meja dengan kelajuan sudut 5 rad.s^-1. Bila di atas piringan diletakkan cincin bermassa 0,1 kg dan jari-jari 0,2 meter (I = mR²) dengan poros putar cincin tepat di atas pusat piringan, maka piringan dan cincin akan bersama-sama berputar dengan kecepatan sudut .…

0,22 rad.s-1

2,00 rad.s-1

2,20 rad.s-1

4,50 rad.s-1

44,00 rad.s-1

The angular momentum of the system is conserved when the ring is placed on the disk. The initial angular momentum of the disk is given by Iω, where I is the moment of inertia of the disk and ω is the initial angular velocity. The final angular momentum of the system is also given by Iω, where I is the moment of inertia of the combined system (disk + ring) and ω is the final angular velocity. Since the moment of inertia of the disk is smaller than the moment of inertia of the combined system, the final angular velocity ω will be larger than the initial angular velocity ω. Therefore, the correct answer is 4.50 rad.s-1.

Explanation

The angular momentum of the system is conserved when the ring is placed on the disk. The initial angular momentum of the disk is given by Iω, where I is the moment of inertia of the disk and ω is the initial angular velocity. The final angular momentum of the system is also given by Iω, where I is the moment of inertia of the combined system (disk + ring) and ω is the final angular velocity. Since the moment of inertia of the disk is smaller than the moment of inertia of the combined system, the final angular velocity ω will be larger than the initial angular velocity ω. Therefore, the correct answer is 4.50 rad.s-1.

23. Peti berupa kotak berukuran 100 cm x 150 cm x 200 cm yang berada di dasar laut hendak diangkat ke permukaaan laut dengaan cara menariknya dengan rantai. Bila massa jenis rata-rata peti = 2000 kg.m^-3 dan massa jenis air laut = 1000 kg.m^-3_, maka besar gaya angkat minimal yang dibutuhkan untuk mengangkat peti tersebut adalah …. (abaikan massa rantai)

3,0 kN

2,0 kN

6,0 kN

9,0 kN

18,0 kN

The minimum lifting force required to lift the box from the seabed to the surface of the sea can be calculated using Archimedes' principle. According to this principle, the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. The buoyant force can be calculated by multiplying the volume of the box by the density of the seawater and the acceleration due to gravity. The volume of the box can be calculated by multiplying its length, width, and height. By substituting the given values into the formula, we can calculate that the minimum lifting force required is 3.0 kN.

Explanation

The minimum lifting force required to lift the box from the seabed to the surface of the sea can be calculated using Archimedes' principle. According to this principle, the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object. The buoyant force can be calculated by multiplying the volume of the box by the density of the seawater and the acceleration due to gravity. The volume of the box can be calculated by multiplying its length, width, and height. By substituting the given values into the formula, we can calculate that the minimum lifting force required is 3.0 kN.

24. Sebuah mobil mengalami perlambatan dari laju 72 km/jam sampai berhenti dalam waktu 5 detik saat ada rintangan di depannya yang berjarak 55 meter dari mobil tersebut. Maka panjang lintasan yang ditempuh mobil dalam selang waktu 5 detik adalah....

50 m

52 m

48 m

35 m

45 m

The question states that a car decelerates from a speed of 72 km/h to a stop in 5 seconds when there is an obstacle 55 meters ahead. The length of the distance traveled by the car in 5 seconds can be calculated using the formula: distance = initial velocity * time + (1/2) * acceleration * time^2. Since the car is decelerating, the acceleration is negative. Plugging in the given values, we get: distance = (72 km/h * 5 seconds) + (1/2) * (-a) * (5 seconds)^2. Solving for distance, we find that the car travels 50 meters in 5 seconds.

Explanation

The question states that a car decelerates from a speed of 72 km/h to a stop in 5 seconds when there is an obstacle 55 meters ahead. The length of the distance traveled by the car in 5 seconds can be calculated using the formula: distance = initial velocity * time + (1/2) * acceleration * time^2. Since the car is decelerating, the acceleration is negative. Plugging in the given values, we get: distance = (72 km/h * 5 seconds) + (1/2) * (-a) * (5 seconds)^2. Solving for distance, we find that the car travels 50 meters in 5 seconds.

25. Bunyi yang didengar pada jarak 3 m dari sumbernya memiliki intensitas 10^-4 W.m². Taraf intensitas bunyi tersebut saat didengar pada jarak 300 m dari sumber bunyi adalah …. (I_o = 10^-12 W.m²)

40 dB

60 dB

90 dB

20 dB

30 dB

The intensity of sound decreases with distance from the source according to the inverse square law. In this case, the sound is heard at a distance 100 times greater than the initial distance, so the intensity will decrease by a factor of 100^2 = 10,000. Taking the logarithm of this ratio gives us the decibel level. log(10,000) = 4, so the decibel level is 40 dB.

Explanation

The intensity of sound decreases with distance from the source according to the inverse square law. In this case, the sound is heard at a distance 100 times greater than the initial distance, so the intensity will decrease by a factor of 100^2 = 10,000. Taking the logarithm of this ratio gives us the decibel level. log(10,000) = 4, so the decibel level is 40 dB.

26.

8,6 N

9,6 N

2,8 N

3,2 N

4,8 N

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Explanation

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27.

5 cm menjauhi lensa objektif

5 cm mendekati lensa objektif.

10 cm menjauhi lensa objektif.

3 cm menjauhi lensa objektif.

3 cm mendekati lensa objektif.

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Explanation

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28.

Satelit C,satelit B,satelit A

Satelit A,satelit B,satelit C

Satelit A,satelit C,satelit B

Satelit B,satelit C,satelit A

Satelit C,satelit A,satelit B

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Explanation

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29.

4,0 kg m.s-1

8,0 kg m.s-1

16 kg m.s-1

2,0 kg m.s-1

0 kg m.s-1

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Explanation

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30.

1 dan 3

1 dan 2

2 dan 3

2 dan 4

3 dan 4

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Explanation

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31. Pada lintasan yang lurus dan panjang, 2 buah sepeda, A dan B di kayuh ke arah yang sama. Sepeda A mulai dari titik S bergerak dari keadaan diam dengan percepatan tetap 2 m.s^-2. Sepeda B melintasi titik S dalam waktu yang bersamaan dengan sepeda A, namun dengan kecepatan tetap 8 m.s^-1. Setelah bergerak selama 10 s sejak melintasi titik S, jarak antara sepeda A dan B adalah ….

60 m

80 m

100 m

20 m

50 m

Both bikes start at the same time, but bike A starts from rest and accelerates at a constant rate of 2 m/s^2. Bike B starts with a constant velocity of 8 m/s. After 10 seconds, bike A will have reached a velocity of 20 m/s (2 m/s^2 * 10 s) and traveled a distance of 100 m (1/2 * 2 m/s^2 * (10 s)^2). Bike B, on the other hand, will have traveled a distance of 80 m (8 m/s * 10 s). Therefore, the distance between bike A and bike B after 10 seconds is 100 m - 80 m = 20 m.

Explanation

Both bikes start at the same time, but bike A starts from rest and accelerates at a constant rate of 2 m/s^2. Bike B starts with a constant velocity of 8 m/s. After 10 seconds, bike A will have reached a velocity of 20 m/s (2 m/s^2 * 10 s) and traveled a distance of 100 m (1/2 * 2 m/s^2 * (10 s)^2). Bike B, on the other hand, will have traveled a distance of 80 m (8 m/s * 10 s). Therefore, the distance between bike A and bike B after 10 seconds is 100 m - 80 m = 20 m.

32.

7 N.m searah putaran jarum jam

7 N.m berlawanan arah dengan arah putaran jarum jam

9 N.m searah putaran jarum jam

9 N.m berlawanan arah dengan arah putaran jarum jam

1 N.m searah putaran jarum jam

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Explanation

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33. Mobil pemadam kebakaran sedang bergerak dengan laju 20 m.s⁻¹ sambil membunyikan sirine pada frekuensi 400 Hz (cepat rambat bunyi 300 m.s⁻¹). Jika mobil pemadam kebakaran bergerak menjauhi Amri yang berdiri di tepi jalan, maka perbedaan frekuensi sirine yang didengar oleh Amri dengan frekuensi sumbernya adalah ....

25 Hz

16 Hz

75 Hz

384 Hz

375 Hz

The correct answer is 25 Hz. When the fire truck is moving away from Amri, the frequency of the siren that Amri hears will be lower than the actual frequency emitted by the siren. This is because the sound waves are stretched out as the fire truck moves away, causing a decrease in frequency. The Doppler effect formula can be used to calculate the change in frequency, which is given by Δf = (v/vs) * f, where Δf is the change in frequency, v is the velocity of the fire truck, vs is the speed of sound, and f is the actual frequency of the siren. Plugging in the given values, we get Δf = (20/300) * 400 = 25 Hz.

Explanation

The correct answer is 25 Hz. When the fire truck is moving away from Amri, the frequency of the siren that Amri hears will be lower than the actual frequency emitted by the siren. This is because the sound waves are stretched out as the fire truck moves away, causing a decrease in frequency. The Doppler effect formula can be used to calculate the change in frequency, which is given by Δf = (v/vs) * f, where Δf is the change in frequency, v is the velocity of the fire truck, vs is the speed of sound, and f is the actual frequency of the siren. Plugging in the given values, we get Δf = (20/300) * 400 = 25 Hz.

34.

8,0 kg

13,0 kg

26,0 kg

5,0 kg

6,5 kg

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Explanation

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35.

41,7 N

10,0 N

25,0 N

31,3 N

16,7 N

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Explanation

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36.

√2 : √5

√4 : √1

√5 : √2

√1 : √2

√1 : √ 4

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Explanation

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37. Alasan petinju untuk menggunakan sarung tinju saat bertanding adalah ….

Memperlemah gaya pukul dengan memperlama waktu sentuh saat pukulan didaratkan ke muka lawan

Mempertahankan gaya pukul dengan mengurangi waktu sentuh saat pukulan didaratkan ke muka lawan

Memperkuat gaya pukul dengan mempercepat waktu sentuh saat pukulan didaratkan ke muka lawan

Memperkuat gaya pukul dengan memperlama waktu sentuh saat pukulan didaratkan ke muka lawan

Memperlemah gaya pukul dengan mengurangi waktu sentuh saat pukulan didaratkan ke muka lawan

The reason why a boxer uses boxing gloves during a match is to weaken the impact of punches by prolonging the contact time when the punch lands on the opponent's face. This helps to reduce the force and potential injury caused by the punch.

Explanation

The reason why a boxer uses boxing gloves during a match is to weaken the impact of punches by prolonging the contact time when the punch lands on the opponent's face. This helps to reduce the force and potential injury caused by the punch.

38.

27 : 2

1 : 6

2 : 9

3 : 1

9 : 2

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Explanation

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39.

18,9809 mm2

18,98 mm2

18,90 mm2

19,0 mm2

19 mm2

The correct answer is 19,0 mm2. This is the closest approximation to the given value of 18,9809 mm2. Rounded to one decimal place, 19,0 mm2 is the most accurate representation of the original measurement.

Explanation

The correct answer is 19,0 mm2. This is the closest approximation to the given value of 18,9809 mm2. Rounded to one decimal place, 19,0 mm2 is the most accurate representation of the original measurement.

40.

16 : 9

5 : 4

4 : 3

25 : 9

25 : 16

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Explanation

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