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:
1 / 2 m v^2 = L m g (1 - cos θ)
This is the KE due to the pendulum falling from a 25 deg displacement
v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2
v = 1.92 m/s this is the speed due to an initial displacement of 25 deg
Its speed at the bottom would then be
1.92 + 1.2 = 3.12 m/s since it gains 1.92 m/s from its initial displacement
But what are the choices??
Based on the options given, the most likely answer to this query is B) The temperature must be converted to Kelvin. Meaning, the temperature should be in SI unit.
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