The position vector can be
transcribed as:
A<span> = 6 i + y j
</span>
i <span>points in the x-direction and j points
in the y-direction.</span>
The magnitude of the
vector is its dot product with itself:
<span>|A|2 = A·A</span>
<span>102 = (6 i +
y j)•(6 i+ y j)
Note that i•j = 0, and i•i = j•j =
1 </span>
<span>100 = 36 + y2
</span>
<span>64 = y2</span>
<span>get the square root of 64 = 8</span>
<span>The vertical component of the vector is 8 cm.</span>
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:
19.2 m/s
Explanation:
The train is moving at 18 m/s and you are walking in the same direction (east) so the speeds are added
18 + 1.2 = 19.2
If you were walking backwards (west) your velocity with respect to the ground would be
18 - 1.2 = 16.8
A horizontal line means the object is. not changing its position it is not moving, it is at rest.