Answer:
we could use the formula, v=u+at,
65=25+a (10), a=4 , since the motion is declerating we have a=-4 m/s2
Answer:
The ball experiences the greater momentum change
Explanation:
The momentum change of each object is given by:

where
m is the mass of the object
v is the final velocity
u is the initial velocity
Both objects have same mass m and same initial velocity u. So we have:
- For the ball, the final velocity is

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

- For the clay, the final velocity is

since it sticks to the wall. So, the change in momentum is

So we see that the greater momentum change (in magnitude) is experienced by the ball.
String the·o·ry
noun
a cosmological theory based on the existence of cosmic strings.
Answer:

Explanation:
For answer this we will use the law of the conservation of the angular momentum.

so:

where
is the moment of inertia of the merry-go-round,
is the initial angular velocity of the merry-go-round,
is the moment of inertia of the merry-go-round and the child together and
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:



Finally we replace all the data:

Solving for
:

Examples of Newton's three law of motion.
First law of motion: A rocket being launched up in the atmosphere.
Second law of motion:while riding a bicycle, a bicycle acts as a mass and our legs pushing on the pedals of the bicycle is the force.
Third law of motion:when we jump off from the boat,the boat moves backward.
Hope,it will helpyouu!