If I knew the answer I would help but I don’t know sorry
D=at²
441m=(5*9.81m/s²)(t²)
t²=441/(5*9.81)
t≈√8.99
t≈3 sec
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:
S=48.29 m
Explanation:
Given that the height of the hill h = 2.9 m
Coefficient of kinetic friction between his sled and the snow μ = 0.08
Let u be the speed of the skier at the bottom of the hill.
By applying conservation of energy at the top and bottom of the inclined plane we get.
Potential Energy=kinetic Energy
mgh = (1/2) mu²
u² = 2gh
u²=2(9.81)(2.9)
=56.89
u=7.54 m/s
a = - f / m
a = - μ*m*g / m
a = - μg
From equation of motion
v²- u² = 2 -μ g S
v=0 m/s
-(7.54)²=-2(0.06)(9.81)S
S=48.29 m