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The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²
The direction of the acceleration of the ball is downwards
The given parameters
initial velocity of the ball, u = 0
height above the ground, h = 2.2 m
time of motion of the ball, t = 96 ms = 0.096 s
The magnitude of the acceleration of the ball while coming to rest is calculated as;
let the downwards direction of the acceleration be positive
The direction of the acceleration of the ball is downwards
Learn more here: brainly.com/question/15407740
Answer: 3 m/s
Explanation:
We can solve the problem by using the law of conservation of momentum: during the collision between the two balls, the total momentum of the system before the collision and after the collision must be conserved:
The total momentum before the collision is given only by the cue ball, since the solid ball is initially at rest, therefore
So, the final total momentum will also be
And the total momentum after the collision is given only by the solid ball, since the cue ball is now at rest, therefore:
from which we find the velocity of the solid ball