Green would be the colour of the paper
Answer:
Work done on an object is equal to
FDcos(angle).
So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.
However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.
But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?
Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.
Explanation:
The kinetic energy and gravitational potential energy changes during its movement from ground to the top height.
<h3>What happens to kinetic and potential energy while motion?</h3>
When the ball moves upward, its gravitational potential energy is increases and kinetic energy begins to decrease but when the ball falls towards the earth, its gravitational potential energy is transformed into kinetic energy. When the ball collides with the ground, the kinetic energy is transformed into other forms of energy.
Learn more about kinetic energy here: brainly.com/question/20658056
Are there any answer choices?
Submarines use <span>buoyancy by filling ballast tanks up with water. When they are filled with water, they are more dense than the surrounding water, so they are able to sink. If they want to rise, they fill these tanks up with air so that the density is less than the water it surrounds.
Hope this helps! :)</span>
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
