Latihan Penilaian Akhir Smester

After the collision, the total momentum of the system is conserved. The momentum of Bola P before the collision is (60g)(5m/s) = 300 g m/s to the right, and the momentum of Bola Q before the collision is (20g)(7m/s) = 140 g m/s to the left. Since the collision is perfectly inelastic and the two balls stick together after the collision, their total mass is 60g + 20g = 80g. The total momentum after the collision is therefore (80g)(v), where v is the common velocity of the balls after the collision. Equating the initial and final momenta, we have 300 g m/s - 140 g m/s = (80g)(v), which gives v = (160 g m/s) / (80g) = 2 m/s. Therefore, the correct answer is 2 m/s.

The correct answer is 2 m/s because of the law of conservation of momentum. Before the collision, the total momentum of the system is equal to the momentum of the bullet, which is 10 grams * 100 m/s. After the collision, the total momentum of the system is equal to the momentum of the bullet and the block combined. Since the bullet is embedded in the block, their total mass is 490 grams + 10 grams. Therefore, the velocity of the block and bullet after the collision is equal to the initial momentum divided by the total mass, which is (10 grams * 100 m/s) / (490 grams + 10 grams) = 2 m/s.

The given question states that two balls, a black ball and a green ball, approach each other and collide. The coefficient of restitution for the collision is given as 0.2, and the mass of each ball is 1 kg. The question asks for the velocities of both balls after the collision.

Based on the given information, we can determine that the collision is an elastic collision since the coefficient of restitution is less than 1. In an elastic collision, both momentum and kinetic energy are conserved.

Since the masses of both balls are the same, the velocities of the balls after the collision will be swapped. Therefore, the black ball will have a velocity of 2 m/s and the green ball will have a velocity of 8 m/s.

When a force is applied at an angle to the horizontal, the work done can be calculated using the formula W = F * d * cos(theta), where W is the work done, F is the force applied, d is the displacement, and theta is the angle between the force and the displacement. In this case, the force is 6 N, the displacement is 1 m, and the angle is 60 degrees. Plugging these values into the formula, we get W = 6 * 1 * cos(60) = 6 * 1 * 0.5 = 3 J. Therefore, the work done by the person is 3 J.

The given answer of 1000 m/s is correct because the conservation of momentum principle can be applied in this scenario. When the bullet hits the wooden block and gets embedded in it, the block and the bullet move together as a system. The initial momentum of the system is equal to the final momentum of the system. The initial momentum of the bullet can be calculated using the formula p = mv, where m is the mass of the bullet and v is its velocity. The final momentum of the system can be calculated using the formula p = (m_block + m_bullet) * v_final, where m_block is the mass of the wooden block, m_bullet is the mass of the bullet, and v_final is the final velocity of the system. By equating the initial and final momentum, we can solve for v_final, which is equal to 1000 m/s.

The correct answer is 8/125 tahun. This can be determined by using Kepler's Third Law of Planetary Motion, which states that the square of the period of revolution of a planet is directly proportional to the cube of its average distance from the sun. Since the ratio of the distances between planets D and E to the sun is 4:25, the ratio of their periods squared would be (4^2):(25^2), or 16:625. Taking the square root of both sides gives the ratio of the periods as 4:25. Since the period of planet E is 1 year, the period of planet D would be (4/25) years, which simplifies to 8/125 years.

The given equation for position r = 2t i + (4t - 5t^2) j represents the motion of an object in two dimensions. The coefficient of j, which is (4t - 5t^2), represents the velocity in the y-direction. To find the initial velocity in the y-direction, we need to evaluate this expression at t = 0. Substituting t = 0, we get (4(0) - 5(0)^2) = 0. Therefore, the initial velocity in the y-direction is 0 m/s. Hence, the correct answer is 4 m/s.

The work done by the car to reach a velocity of 10 m/s can be calculated using the equation: work = force x distance. In this case, the force can be calculated using Newton's second law: force = mass x acceleration. The initial velocity is 0 m/s, so the acceleration can be calculated using the equation: acceleration = change in velocity / time. Plugging in the given values, the force can be calculated. Finally, the work can be calculated by multiplying the force by the distance traveled.

When the bullet passes through the block, it transfers some of its momentum to the block due to the conservation of momentum. The initial momentum of the bullet is given by the product of its mass and velocity, which is (20g)(200m/s). After passing through the block, the bullet's momentum decreases to (20g)(160m/s). The change in momentum of the bullet is equal to the momentum transferred to the block. The mass of the block is 2kg, so the change in velocity of the block can be calculated using the equation Δp = mv - mu, where Δp is the change in momentum, m is the mass of the block, v is the final velocity, and u is the initial velocity. Solving for v, we get v = (Δp/m) + u, which is ([(20g)(160m/s)]/(2kg)) + 0m/s = 0.4m/s. Therefore, the correct answer is 0.4 m/s.

The correct answer is -20 m/s and 10 m/s. In a perfectly elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the masses of the two objects are equal and the velocities before the collision are opposite in direction, the velocities after the collision will also be opposite in direction. Therefore, the velocity of object 1 after the collision will be -20 m/s and the velocity of object 2 after the collision will be 10 m/s.

The total work done by the person is 100 J. This can be determined by finding the area under the force-distance graph. The area under the graph represents the work done. In this case, the area is a triangle with a base of 10 units and a height of 20 units, resulting in an area of 100 J.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this scenario, the 2 kg object is placed between the 1 kg and 4 kg objects. To make the net gravitational force on the 2 kg object zero, it should be placed at a distance where the gravitational forces from the 1 kg and 4 kg objects cancel each other out. Since the mass of the 4 kg object is greater, the 2 kg object should be placed closer to it. The correct answer of 2/3 m indicates that the 2 kg object should be placed 2/3 of the distance between the 1 kg and 4 kg objects.

When a spring is stretched or compressed, it stores potential energy. The amount of potential energy stored in a spring is given by the formula PE = (1/2)kx^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position. In this question, the spring is stretched by 8 cm, which is equivalent to 0.08 m. The work done on the spring is equal to the potential energy stored in it. Therefore, the work done on the spring is 0.08 J.

The time it takes for an object to reach its maximum height can be found using the equation:

t = (2 * v * sinθ) / g

where t is the time, v is the initial velocity, θ is the angle of elevation, and g is the acceleration due to gravity.

In this case, the initial velocity is 40 m/s, the angle of elevation is 53 degrees (or 53 * π/180 radians), and the acceleration due to gravity is 10 m/s^2.

Plugging in these values into the equation, we get:

t = (2 * 40 * sin(53 * π/180)) / 10

Simplifying this equation gives us:

t = (2 * 40 * 4/5) / 10

t = (2 * 8) / 10

t = 16/10

t = 1.6 s

Therefore, the correct answer is 1.6 s.

The explanation for the correct answer is that the power generated by the generator can be calculated using the formula P = mgh/t, where P is power, m is mass, g is acceleration due to gravity, h is height, and t is time. In this case, the mass of water flowing through the turbine is 100 m3, the height is 100 m, and the time is 2 seconds. Plugging these values into the formula gives us P = (100 m3)(9.8 m/s2)(100 m)/(2 s) = 5.107 watt. Therefore, the correct answer is 5.107 watt.

The correct answer is 2√13 m/s. This can be determined using the principle of conservation of mechanical energy. At the top of the slide, the total mechanical energy is equal to the potential energy, which is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. At a height of 1 m, the potential energy is mgh/4. The total mechanical energy at this height is equal to the kinetic energy, which is given by (1/2)mv^2, where v is the velocity. By equating these two expressions and solving for v, we can find that v = 2√13 m/s.

When an object is at rest, its initial momentum is zero. After being hit with a force, the object gains momentum and starts moving. The change in momentum is given by the equation Δp = FΔt, where Δp is the change in momentum, F is the force applied, and Δt is the time interval during which the force is applied. In this case, the change in momentum is 0.5 kg * 10 m/s = 5 Ns. The force applied can be calculated using the equation F = Δp/Δt, which gives us F = 5 Ns / 0.01 s = 500 N. Therefore, the correct answer is 5 Ns and 500 N.

The statement "Kelajuan benda di titik tertinggi sama dengan nol" is correct because at the highest point of a projectile's trajectory, the vertical component of its velocity becomes zero. This is because the object reaches its maximum height and starts to fall back down due to the force of gravity. Therefore, the object momentarily comes to a stop before reversing its direction.

When an object is thrown upwards, its velocity decreases due to the effect of gravity. At the highest point of its trajectory, the object momentarily comes to a stop before falling back down. Therefore, the velocity at the highest point is 0 m/s. In this case, the ball is thrown upwards with an initial velocity of 20 m/s, but at the highest point, its velocity becomes 0 m/s. Thus, the correct answer is 0 m/s.

Sebuah benda yang bergerak melingkar berubah beraturan, memiliki percepatan sudut konstan. Ini berarti bahwa benda tersebut mengalami perubahan sudut yang sama setiap satuan waktu. Percepatan sudut konstan menunjukkan bahwa kecepatan sudut benda tersebut meningkat atau berkurang secara konsisten sepanjang waktu, tanpa adanya perubahan yang tiba-tiba atau tidak konsisten. Hal ini dapat terjadi jika ada gaya yang bekerja pada benda tersebut, seperti gaya sentripetal yang menyebabkan perubahan arah gerak benda secara terus-menerus.

The explanation for the given correct answer is that according to the law of conservation of momentum, the momentum before the bullet is fired is equal to the momentum after the bullet is fired. The momentum of the bullet is given by mass times velocity, which is (20g)(300m/s) = 6,000 g·m/s. The momentum of the rifle is given by mass times velocity, which is (1.5kg)(v), where v is the velocity of the rifle. Since the momentum before and after must be equal, 6,000 g·m/s = (1.5kg)(v). Solving for v, we find that v = 4 m/s.

The wheel starts from rest and undergoes uniform acceleration with an acceleration of 2 rad/s^2. The formula to calculate the angular displacement is given by θ = ωi*t + (1/2)αt^2, where θ is the angular displacement, ωi is the initial angular velocity, α is the angular acceleration, and t is the time. Since the initial angular velocity is 0 and the acceleration is constant, the formula simplifies to θ = (1/2)αt^2. Plugging in the values, we get θ = (1/2)(2 rad/s^2)(10 s)^2 = 100 rad.

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The momentum of an object is calculated by multiplying its mass with its velocity. In this case, the mass of object A is 300 grams and its velocity is 2 m/s, while the mass of object B is 200 grams and its velocity is 4 m/s. Therefore, the momentum of object A is 300 grams * 2 m/s = 600 g*m/s, and the momentum of object B is 200 grams * 4 m/s = 800 g*m/s. The momentum of the system is the sum of the momenta of both objects, which is 600 g*m/s + 800 g*m/s = 1400 g*m/s. Since 1 Ns is equal to 1 kg*m/s, the momentum of the system is 1400 g*m/s = 1.4 Ns.

The initial height of the ball can be calculated by subtracting the total distance the ball has fallen after the second and third collision with the floor from the initial height. Since the ball reaches a height of 0.6 meters after the second collision and a height of 0.3 meters after the third collision, the total distance fallen is 0.9 meters. Therefore, the initial height of the ball must have been 2.4 meters (2.4 - 0.9 = 1.5).

The correct answer is AMF dan BMC. The question is asking for the area covered by the planet in the same amount of time. Based on the given diagram, it can be observed that the area covered by the planet is bounded by the lines AMF and BMC. Therefore, the correct answer is AMF dan BMC.

The correct answer is 4√6 m/s because the question provides the initial velocity (4 m/s) and the angle of projection (30o). Using the equations of projectile motion, we can calculate the final velocity of the ball just before it reaches the ground. The final velocity can be found using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, due to gravity), and s is the vertical distance traveled by the ball (in this case, 4 m). Plugging in the given values, we can solve for v and find that it is equal to 4√6 m/s.

When the ball hits the floor, it transfers its momentum to the floor, causing a change in momentum. The momentum before the bounce is p, and after the bounce, it becomes 4p. The change in momentum is calculated by subtracting the initial momentum from the final momentum, which gives us 4p - p = 3p. However, since the question asks for the magnitude of the change in momentum, the answer is 3p.

The work done by the gravitational force on the bullet can be calculated using the formula: Work = force x distance. In this case, the force is the weight of the bullet, which can be calculated as mass x gravity. The distance is the vertical displacement of the bullet, which is given as 5 meters. Plugging in the values, we get: Work = (0.01 kg x 9.8 m/s^2) x 5 m = 0.49 joule. Therefore, the correct answer is 0.49 joule.