By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m
K.E = 1/2 x 0.17 x 
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/
To calculate the final velocity, let us use third equation of motion.
=
+ 2as
=
+ 2 x 0.3 x 61
= 36 + 36
= 72
V = 
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: brainly.com/question/402617
66° N and 90° N
the area of the artic circle in the northern hemisphere
Answer:
Hmm
Explanation:
I don't know sorry forgive me.
Answer:
The car stops after 32.58 m.
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 20 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -6 m/s²
Time taken by the car to stop

Total Time taken by the car to stop is 0.5+3.33 = 3.83 s

The car stops after 32.58 m.
Distance between car and obstacle is 50-32.58 = 17.42 m
False its atmosphere, lithosphere, hydrosphere, and boisphere