When you take your 1400 kg car out for a spin, you go around a corner of radius 55 m with a speed of 18 m/s. the coefficient of
1 answer:
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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
Answer:
The pilot is 2214.22 miles from her starting position
Explanation:
Since the pilot is traveling at a constant speed of 635 mph, the total distance traveled can be easily found as follows:
There was a 10 degrees deviation, so the angle between the trajectory of both legs is 170 degrees.
The distance we need to find is that from the start of the first leg to the end of the second leg, those three distances form a triangle and since the side we're interested in is opposite to the 170 degrees angle, we can determine its length by the law of cosines:
The pilot is 2214.22 miles from her starting position