Newton's Third Law of Motion states that for every action there must exist an equal but opposite reaction.
This ultimately implies that, Newton's Third Law of Motion is a law based on action-reaction force pairs.
In this exercise, you're required to identify the action-reaction forces in the attached picture.
Under condition A, a boy is leaning against a wall;
- The force being exerted by the boy on the wall is a force of action.
- The force being exerted by the wall on the boy is a force of reaction.
Under condition B, a boy is jumping off a diving board.
- The force being exerted by the feet of the boy on the diving board is a force of action.
- The force being exerted by the diving board on the feet of the boy is a force of reaction.
Under condition C, a nail is being hammered into the wall.
- The force being exerted by the hammer on the nail is a force of action.
- The force being exerted by the nail on the hammer is a force of reaction.
Under condition D, a man is walking on a ground (floor).
- The force being exerted by the foot of the man on the ground (floor) is a force of action.
- The force being exerted by the ground (floor) on the foot of the man is a force of reaction.
Under condition E, a boy is holding a ball;
- The force being exerted by the hand of the boy on the ball is a force of action.
- The force being exerted by the ball on the hand of the boy is a force of reaction.
Read more: brainly.com/question/15170643
No it does not! As it is not moving?!
Answer:
The last significant figure is inherently uncertain because of the rounding off. The rounding-off uncertainty is usually half of the last figure's decimal-place value if no other uncertainty is expressed. As a rule of thumb results of measurements and calculations have a limited number of significant figures.
Answer:
L = mp*v₀*(ms*D) / (ms + mp)
Explanation:
Given info
ms = mass of the hockey stick
uis = 0 (initial speed of the hockey stick before the collision)
xis = D (initial position of center of mass of the hockey stick before the collision)
mp = mass of the puck
uip = v₀ (initial speed of the puck before the collision)
xip = 0 (initial position of center of mass of the puck before the collision)
If we apply
Ycm = (ms*xis + mp*xip) / (ms + mp)
⇒ Ycm = (ms*D + mp*0) / (ms + mp)
⇒ Ycm = (ms*D) / (ms + mp)
Now, we can apply the equation
L = m*v*R
where m = mp
v = v₀
R = Ycm
then we have
L = mp*v₀*(ms*D) / (ms + mp)