Distance and time, distance because that's how far and time because that's how long
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:
Approximately
(assuming that external forces on the cannon are negligible.)
Explanation:
If an object of mass
is moving at a velocity of
, the momentum
of that object would be
.
Momentum of the t-shirt:
.
If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if
denote the momentum of this cannon:
.
.
Rewrite
to obtain
. Since the mass of this cannon is
, the velocity of this cannon would be:
.
The speed increases, because as the angle increases (the wing slants up more steeply), the air has to go farther to get over the wing.