In Newton's third law, the action and reaction forces D.)act on different objects
Explanation:
Newton's third law of motion states that:
<em>"When an object A exerts a force on object B (action force), then action B exerts an equal and opposite force (reaction force) on object A"</em>
It is important to note from the statement above that the action force and the reaction force always act on different objects. Let's take an example: a man pushing a box. We have:
- Action force: the force applied by the man on the box, forward
- Reaction force: the force applied by the box on the man, backward
As we can see from this example, the action force is applied on the box, while the reaction force is applied on the man: this means that the two forces do not act on the same object. This implies that whenever we draw the free-body diagram of the forces acting on an object, the action and reaction forces never appear in the same diagram, since they act on different objects.
Learn more about Newton's third law of motion:
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The correct answer is B, widespread pollution. If you look closely, you can see that the other answers are not problems at all, but benefits! :)
Answer:
<h2>C. <u>
0.55 m/s towards the right</u></h2>
Explanation:
Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Momentum = Mass (M) * Velocity(V)
BEFORE COLLISION
Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s
Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s
AFTER COLLISION
Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s
<u>x is the final velocity of the 0.25kg ball</u>
Momentum of 0.15kg body moving at 0.75m/s(body at rest) =
0.15 * 0.75kgm/s = 0.1125 kgm/s
Using the law of conservation of momentum;
0.25+0 = 0.25x + 0.1125
0.25x = 0.25-0.1125
0.25x = 0.1375
x = 0.1375/0.25
x = 0.55m/s
Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>
<u></u>
