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Fed [463]
4 years ago
8

At what point in its trajectory does a batted baseball have its minimum speed? If air resistance can be neglected, how does this

speed compare with the horizontal component of its velocity at other points?
Physics
1 answer:
lys-0071 [83]4 years ago
8 0

To understand the problem and give a logical explanation we will use the concepts related to the movement of a Projectile.

On this concept the components concerning the speed of an object that is launched in parabolic motion are characterized. There will be two types of speed components, the vertical and the horizontal.

The vertical component of velocity gradually decreases as it is affected by the force of gravity until it becomes zero at the highest point of the trajectory. After this point, it again begins to increase gradually in effect of its movement towards the gravitational attraction of the Earth.

In the case of horizontal force, if the effect of air resistance is neglected, the speed will be the same throughout the trajectory.

Now comparing this with the movement of the baseball, at the highest point, the vertical component will be zero, and the horizontal velocity will remain. Both at the beginning of the trajectory and at the end of it, there will be components of the vertical velocity, which would increase this value. Therefore the minimum speed value will be at the top, when the vertical component is zero, and the horizontal component is maintained

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identify the action reaction forces in each picture the first picture is done for you write your answer on a separate sheet of p
xxMikexx [17]

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

4 0
3 years ago
Which quantity or quantities is/are increasing for the object represented by line B?
torisob [31]

Answer:

C. Velocity and Position

Explanation:

7 0
3 years ago
Can a body have momentum without energy?
mestny [16]
No it does not! As it is not moving?!
7 0
3 years ago
Read 2 more answers
HELP ASAP!
bonufazy [111]

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.

4 0
3 years ago
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the
krek1111 [17]

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)

5 0
3 years ago
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