Newton's cradle is a contraption where metal balls hang from a frame. When one ball is pulled and released, the collision causes one ball on the opposite end to be released. Which type of force can decrease the total momentum of this system ? (1 point) electrostatic force magnetic force friction force strong nuclear force

friction force

Are you sure?

Yes, I am sure. Friction force can decrease the total momentum of the system, as it opposes the motion of the metal balls and can cause them to slow down and eventually stop, reducing the total momentum of the system.

One object is traveling northeast with a momentum of 12 kg-m/s, while a second object is traveling southwest on the same path as the first object with a momentum of 4 kg-m/ s. When the two objects collide, they stick together . What is the momentum after their collision ? (1 point) 16 kg-m/ s southwest 8 kg-m/s northeast 8 kg-m/s southwest 16 kg-m/s northeast

The momentum of the system is conserved, so the total initial momentum of the two objects is:

12 kg-m/s (northeast) + 4 kg-m/s (southwest) = 8 kg-m/s (northeast)

After they collide and stick together, the total mass of the two objects is:

m = m1 + m2

Since the objects stick together, their velocities after the collision are the same and can be represented as V. Therefore, the total momentum after the collision is:

P = mV

The total momentum must be equal to the initial momentum:

8 kg-m/s (northeast) = (m1 + m2) V

We know that the total mass of the two objects is m1 + m2, so we can substitute that for m:

8 kg-m/s (northeast) = m V

Now we need to find the mass of the two objects. We don't have enough information to do this directly, but we can use the fact that the objects stick together to solve for the relative velocity before the collision.

We can represent the velocities of the two objects before the collision as v1 and v2. Since they are on the same path, but traveling in opposite directions, we know that their relative velocity before the collision is:

v_rel = |v1 - v2|

We aren't given the speeds of the objects, but we know that:

|v1| = |v2|

Since they are traveling in opposite directions on the same path, they must have equal speeds. Therefore, we can simplify the equation to:

v_rel = 2|v|

Now we can use the fact that the collision is perfectly inelastic (the objects stick together) to relate the relative velocity before the collision (v_rel) to the velocity after the collision (V):

V = v_rel/2

The relative velocity can be calculated from the momenta before the collision:

12 kg-m/s (northeast) + 4 kg-m/s (southwest) = m1 v1 + m2 v2

We don't know the individual masses, but since the objects stick together we can simplify the equation to:

16 kg-m/s northeast = (m1 + m2) v

Now we have two equations that relate the initial and final momenta:

8 kg-m/s northeast = (m1 + m2) V

16 kg-m/s northeast = (m1 + m2) v

We can solve for v and V in terms of m1 and m2:

v = 16 kg-m/s / (m1 + m2)

V = 8 kg-m/s / (m1 + m2)

Now we can set v equal to V and solve for m1 + m2:

16 kg-m/s / (m1 + m2) = 8 kg-m/s / (m1 + m2)

Multiplying both sides by (m1 + m2):

16 kg-m/s = 8 kg-m/s

This is a contradiction, so there is no solution. Therefore, our initial assumption that the two objects collide on the same path is incorrect. One possible explanation is that the two objects approach each other at an angle and collide off-center, causing them to move in a new direction after the collision. Without more information about the angle and location of the collision, we cannot determine the momentum after the collision.

Two identical merry -go -rounds are loaded up with equal amounts of mass . On merry -go -round A , the mass is placed in the center , while the mass is placed near the edges of merry -go -round B. The merry - go-rounds are pushed with equal forces , and merry -go-round A ends up moving with a greater angular velocity . Which statement helps explain this difference ? (1 point ) Merry -go -round A has a greater moment of inertia .

Merry -go -round A has more inertia .
Merry -go - round B has a greater moment of inertia .
Merry -go - round B has more inertia .

Merry-go-round A has a greater moment of inertia.

The moment of inertia of an object depends on its mass and the distance between the mass and the axis of rotation. In this case, the mass in merry-go-round A is placed in the center, which is closer to the axis of rotation than the mass in merry-go-round B, which is placed near the edges. This means that merry-go-round A has a smaller moment of inertia than merry-go-round B.

According to Newton's second law for rotational motion, the angular acceleration of an object is directly proportional to the torque applied to it and inversely proportional to its moment of inertia. Since the two merry-go-rounds are pushed with equal forces, the torque applied to them is the same. However, since merry-go-round A has a smaller moment of inertia, it will experience a greater angular acceleration than merry-go-round B. This means that it will reach a greater angular velocity in a shorter amount of time, explaining the difference in their final velocities.

Therefore, the statement "Merry-go-round A has a greater moment of inertia" helps explain the difference in their angular velocities.

In a closed system, an object with a mass of 10 kg moves at a velocity of 5.4m / s . It collides with a second object that has a mass of 12 kg and that is moving in the opposite direction with a velocity of 3.5m / s . What is the total momentum of the system after the collision ? (1 point) 42kg - m / s; 12kg - m / s O54 54kg - m / s; 66kg - m / s

Before the collision, the total momentum of the system is:

P1 = m1v1 + m2v2

where m1 = 10 kg, v1 = 5.4 m/s, m2 = 12 kg, and v2 = -3.5 m/s (negative because it is moving in the opposite direction). Plugging in the values, we get:

P1 = (10 kg)(5.4 m/s) + (12 kg)(-3.5 m/s)

P1 = 54 kg·m/s - 42 kg·m/s

P1 = 12 kg·m/s

The negative sign indicates that the momentum of the second object is in the opposite direction compared to the first object.

After the collision, the two objects stick together and move with a common velocity v. By conservation of momentum, the total momentum of the system remains the same:

P2 = (m1 + m2) v

where m1 + m2 = 22 kg (the total mass of the system). To find v, we need to use conservation of kinetic energy, since the collision is inelastic (some kinetic energy is lost in the collision). We can determine the kinetic energy of the system before and after the collision and equate them:

KE1 = (1/2) m1 v1^2 + (1/2) m2 v2^2

KE2 = (1/2)(m1 + m2) v^2

Plugging in the values, we get:

KE1 = (1/2)(10 kg)(5.4 m/s)^2 + (1/2)(12 kg)(-3.5 m/s)^2

KE1 = 81.54 J

KE2 = (1/2)(22 kg) v^2

We can assume that some kinetic energy is lost in the collision, so KE2 must be less than KE1. Let's say that KE2 is 80% of KE1 (to account for the lost energy):

KE2 = 0.8 KE1

KE2 = 0.8 (81.54 J)

KE2 = 65.24 J

Solving for v in the equation for KE2, we get:

v = sqrt((2 KE2) / (m1 + m2))

v = sqrt((2)(65.24 J) / (10 kg + 12 kg))

v = sqrt(4.27 m^2/s^2)

v = 2.07 m/s

Now we can calculate the total momentum of the system after the collision:

P2 = (10 kg + 12 kg)(2.07 m/s)

P2 = 44.52 kg·m/s

Therefore, the total momentum of the system after the collision is 44.52 kg·m/s.

Answer: 44.52 kg-m/s

In a closed system , an object with a mass of 10 kg moves at a velocity of 5.4 m / s . It collides with a second object that has a mass of 12 kg and that is moving in the opposite direction with a velocity of 3.5 m / s . What is the total momentum of the system after the collision ? ( 1 point ) 42 kg - m / s

12 kg - m / s
54 kg - m / s
66 kg - m / s

The total momentum of the system after the collision is 44.52 kg·m/s, which is closest to 42 kg-m/s.

Answer: 42 kg-m/s.

A force of 30 N is applied to a ball, and it takes the ball 1.5 seconds to travel 4 meters. What is the work done on the ball? (1 point) 120 J

80 J
180 J
45 J

The work done on the ball can be calculated using the formula:

Work = force x distance x cos(theta)

where theta is the angle between the force and the displacement. If the force and displacement are in the same direction (as they are in this case), then cos(theta) = 1.

Given that the force is 30 N and the distance traveled is 4 meters, we can calculate the work done on the ball as:

Work = 30 N x 4 m x 1

Work = 120 J

Therefore, the work done on the ball is 120 J.

Answer: 120 J

Are you sure?

Yes, I am sure. The work done on an object is equal to the product of the force applied on it and the displacement of the object in the direction of the applied force.

In this case, the force applied on the ball is 30 N, and the ball moves a distance of 4 meters in the same direction as the force. Therefore, the work done on the ball is given by:

Work = force x distance x cos(theta)

where theta = 0 degrees (since the force and displacement are in the same direction, so the cosine of the angle is 1).

Plugging in the values, we get:

Work = 30 N x 4 m x 1

Work = 120 J

Therefore, the work done on the ball is 120 J.

Based on the mass and speed data, which object will have the greatest amount of kinetic energy ? 5 kg moving at 2m / s 10 kg moving at 5m / s kg moving at 2m / s 10 kg moving at 2m / s

The amount of kinetic energy possessed by an object is given by the formula:

Kinetic energy = (1/2) x mass x velocity^2

From the given mass and speed data, we can calculate the kinetic energy of each object:

- For a 5 kg object moving at 2 m/s:
Kinetic energy = (1/2) x 5 kg x (2 m/s)^2 = 10 J

- For a 10 kg object moving at 5 m/s:
Kinetic energy = (1/2) x 10 kg x (5 m/s)^2 = 125 J

- For a 1 kg object moving at 2 m/s:
Kinetic energy = (1/2) x 1 kg x (2 m/s)^2 = 2 J

- For a 10 kg object moving at 2 m/s:
Kinetic energy = (1/2) x 10 kg x (2 m/s)^2 = 20 J

Therefore, the object with the greatest amount of kinetic energy is the 10 kg object moving at 5 m/s, with a kinetic energy of 125 J.

Answer: 10 kg moving at 5 m/s.

A rock with 50 kg of mass is sitting on the ground . A rock with 5 kg of mass falls from a cliff above it. Compare the values of kinetic energy for the two objects at different times . (1 point ) The small rock may start with more kinetic energy , but will lose kinetic energy as it falls . The small rock has more kinetic energy than the large rock as soon as it begins to move . The kinetic energy of the small rock will sometimes be greater or smaller than the large rock . The large rock will always have more kinetic energy , regardless of speed .

The small rock may start with more kinetic energy, but will lose kinetic energy as it falls.

When the small rock falls from a cliff, it gains kinetic energy due to its motion. The amount of kinetic energy gained by the small rock depends on its initial height above the ground and its mass. As it falls, some of this initial kinetic energy is converted into other forms of energy such as potential energy, sound energy, and heat energy due to air resistance and friction with the ground. Therefore, the kinetic energy of the small rock will decrease as it falls.

On the other hand, the large rock on the ground has no initial kinetic energy as it is at rest. If the small rock falls and hits the large rock, then the kinetic energy of the system consisting of the two rocks will increase due to the motion of the small rock. However, the increase in kinetic energy will be small compared to the initial kinetic energy of the small rock. In addition, some of the initial kinetic energy of the small rock will be converted into other forms of energy due to the collision, such as sound energy and heat energy. Therefore, the kinetic energy of the small rock will be greater than the large rock when the small rock is falling, but will decrease over time as it falls, whereas the kinetic energy of the large rock will remain constant at zero until it is hit by the small rock.

Therefore, the statement "The small rock may start with more kinetic energy, but will lose kinetic energy as it falls" is correct.

Are you 100% sure?

Yes, I am 100% sure. The kinetic energy of an object is given by the formula:

Kinetic energy = (1/2) x mass x velocity^2

When the small rock falls from a cliff, it gains potential energy due to its height above the ground, which is then converted into kinetic energy as it falls. As long as there is no air resistance or other forces acting on the rock, its kinetic energy will increase as it falls and its velocity increases. However, as it gets closer to the ground, some of its kinetic energy will be converted into other forms of energy, such as potential energy due to deformation when it hits the ground, sound energy, and heat energy due to air resistance and friction with the ground. Therefore, the kinetic energy of the small rock will decrease as it falls.

Since the large rock is at rest, it has no initial kinetic energy. If the small rock falls and hits the large rock, then the kinetic energy of the system consisting of the two rocks will increase due to the motion of the small rock. However, the increase in kinetic energy will be small compared to the initial kinetic energy of the small rock. In addition, some of the initial kinetic energy of the small rock will be converted into other forms of energy due to the collision, such as sound energy and heat energy. Therefore, the kinetic energy of the small rock will be greater than the large rock when the small rock is falling, but will decrease over time as it falls, whereas the kinetic energy of the large rock will remain constant at zero until it is hit by the small rock.

Therefore, the statement "The small rock may start with more kinetic energy, but will lose kinetic energy as it falls" is correct.