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:
pressure, stress pascal N/m2
energy, work, quantity of heat joule N·m
power, radiant flux watt J/s
electric charge, quantity of electricity coulomb -
Final velocity = 5km/sec = 5000 m/sec
Initial velocity = 0, mass = 5kg and force = 500,000 N
F = m(Vf-Vi)/time
Time = m(Vf-Vi)/force = 5*5000/500000 = 0.05 seconds
So, it takes 0.05 sec to arrive at the speed of 5km/s.
Rotational speed of an object around an axis is the numbers of turns of the object divided by time. <span />