Answer:
The time it takes the ball to stop is 0.021 s.
Explanation:
Given;
mass of the softball, m = 720 g = 0.72 kg
velocity of the ball, v = 15.0 m/s
applied force, F = 520 N
Apply Newton's second law of motion, to determine the time it takes the ball to stop;

Therefore, the time it takes the ball to stop is 0.021 s.
I belive it could be 6.5 but I could be wrong
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
Answer:
The magnitude of the hiker’s displacement is 2.96 km
Explanation:
Let the initial displacement of the hiker, = x = 2km
the final displacement of the hiker, = y = 1.4 km
The resultant of the two vectors, According to Pythagorean theorem is the vector sum of the two vectors.
R' = x' + y'
Check the image uploaded for solution;
In Newton's Cradle experiment we know that all cradles of same mass and identical to each other
Now we know that when two identical objects collide elastically then they interchange their velocity
So here we have same illustration
When Newton pulls up a cradle and release it will move hit another cradle which is at rest
Due to elastic collision between them first cradle comes to rest and another cradle will move ahead with same speed this process remains the same and one by one all cradle hit another.
So at the last the cradle at the end will move off with the same speed as the first cradle will hit with the speed.
So in this experiment the cradle at the last end will move off at same distance away from the right end as that of left end we pull the cradle.
So here we can say that in horizontal direction when all cradles are colliding each other there is no external force on the system so momentum is conserved and they all will move off with same speed and hence we observe the above condition.