Answer:
velocity of combined objects is 6 m/s
Explanation:
As per law of conservation of momentum
total momentum before collision = total momentum after collision
since the two objects sticks together after collision so we will have
Now plug in all values in it
so the missing value in the table is velocity which is 6 m/s
Answer:
the force will decrease to 3/4 of its original value.
Explanation:
The initial electric force between the two charges is:
where
k is the Coulomb's constant
q is the magnitude of each charge
r is their separation
Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of
while the other charge will be
So, the new force will be
So, the force will decrease to 3/4 of its original value.
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