Answer:
Choice a.
, assuming that the skating rink is level.
Explanation:
<h3>Net force in the horizontal direction</h3>
There are two horizontal forces acting on the boy:
- The pull of his friend, and
- Frictions.
The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.
The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.
.
<h3>Net force in the vertical direction</h3>
The net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.
However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.
<h3>Net force</h3>
Therefore, the (combined) net force on this boy would be:
.
The ball's horizontal and vertical velocities at time
are


but the ball is thrown horizontally, so
. Its horizontal and vertical positions at time
are


The ball travels 22 m horizontally from where it was thrown, so

from which we find the time it takes for the ball to land on the ground is

When it lands,
and


Light speed is the speed at which light can travel in a "vacuum" (space for example).
The speed of light is constant and is exactly: 299,792,458 meters per second
ONLY IN A VACUUM!!
The speed of light changes when it goes through different mediums (such as from space to Earth, light travels slower in Earth).
Good luck!
The curved surface of water is called the meniscus
Answer:

Explanation:
Given data
Mass m=67.0 kg
Final Speed vf=8.00 m/s
Initial Speed vi=2.00 m/s
Distance d=25.0 m
Force F=30.0 N
From work-energy theorem we know that the work done equals the change in kinetic energy
W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²
And

So

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N
So