Suppose you heat a liquid and then bubbles are produced. With no other evidence, can you tell whether a physical or chemical is
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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
Explanation:
It is given that,
Spring constant of the spring, k = 15 N/m
Amplitude of the oscillation, A = 7.5 cm = 0.075 m
Number of oscillations, N = 31
Time, t = 15 s
(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :
Total number of oscillation per unit time is called frequency of oscillation. Here,
m = 0.0895 kg
or
m = 89 g
(b) The maximum speed of the ball that is given by :
Hence, this is the required solution.