Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11
Answer:
True
Explanation:
In this particular case, the area of the graph represents the impulse.
In fact, impulse is defined as the change in momentum of an object:
Moreover, impulse is also defined as the product between the magnitude of the force acting on an object and the duration of the collision:
If we plot a graph of the force versus the time, if the force is constant then this graph will have a rectangular shape, and the area under the graph will simply be the product
which corresponds to the definition of impulse.
Answer:
The K.E is maximum when the child is at the vertical position and the P.E is maximum at the extreme deviated position from the vertical.
Explanation:
- A child is swinging on swing up and down has both kinetic and potential energy.
- The total mechanical energy of the system is conserved throughout the system. At any instant the total mechanical energy is given by,
E = K.E + P.E
- The K.E is maximum when the child is at the vertical position.
- The P.E is maximum at the extreme deviated position from the vertical.
- And when K.E is maximum P.E becomes minimum and vice versa as per the law of conservation of energy.
Fission and fusion are two processes that release very little amounts of energy.