In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
<h2>
Answer: Toward the center of the circle.</h2>
This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.
However, in this movement the velocity has a constant magnitude, but its direction varies continuously.
Let's say
is the velocity vector, whose direction is perpendicular to the radius
of the trajectory, therefore
the acceleration
is directed toward the center of the circumference.
Answer:
The system of measurements taking the centimeters ,gram and second as its fundamental units , now replaced by the SI unit System
Evaporation . Because it drys up and turns into gas
Answer:
Acceleration, 
Explanation:
Given that,
Height from a ball falls the ground, h = 17.3 m
It is in contact with the ground for 24.0 ms before stopping.
We need to find the average acceleration the ball during the time it is in contact with the ground.
Firstly, find the velocity when it reached the ground. So,

u = initial velocity=0 m/s
a = acceleration=g

It is in negative direction, u = -18.41 m/s
Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

So, the average acceleration of the ball during the time it is in contact is
.