Answer:
It's constant everywhere in its trajectory.
Explanation:
the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.
The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.
This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.
A truck is moving with less velocity in the direction in which the truck is moving earlier because the truck has more momentum.
<h3 /><h3>In which direction the truck moves?</h3>
A truck is moving with the velocity of 10 m/s in the same direction in which the truck is moving earlier because the truck has more mass so it has more momentum. Due to collision, the velocity of the truck is slow down but can't be stopped because of high momentum in the truck.
So we can conclude that a truck is moving with less velocity in the direction in which the truck is moving earlier because the truck has more momentum.
Learn more about momentum here: brainly.com/question/7538238
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An object has undergone acceleration if ...
-- it's moving faster than it was before
or
-- it's moving slower than it was before
or
-- it's moving in a different direction that it was before.
Newton has 3 Laws specifically The Three Laws of Motion
Answer:
Ea = 112500[J]
Eb = 87500[J]
Explanation:
To solve this problem we must use the principle of energy conservation which tells us that the energy of a body plus the work done or applied by the body equals the final energy of a body.
This can be easily visualized by the following equation:
Now we must define the energies at points A & B.
<u>For point A</u>
At point A we only have kinetic energy since it moves at 15 [m/s]
So the kinetic energy
![E_{A}=\frac{1}{2}*m*v_{A}^{2} \\E_{A}=\frac{1}{2} *1000*(15)^{2} \\E_{A}=112500[J]](https://tex.z-dn.net/?f=E_%7BA%7D%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7BA%7D%5E%7B2%7D%20%20%5C%5CE_%7BA%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%2A1000%2A%2815%29%5E%7B2%7D%20%5C%5CE_%7BA%7D%3D112500%5BJ%5D)
The final kinetic energy can be calculated as follows:
![112500-25000=E_{B}\\E_{B}=87500[J]](https://tex.z-dn.net/?f=112500-25000%3DE_%7BB%7D%5C%5CE_%7BB%7D%3D87500%5BJ%5D)
