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>
Answer:
The resultant velocity of the plane relative to the ground is;
150 kh/h north
Explanation:
The flight speed of the plane = 210 km/h
The direction of flight of the plane = North
The speed at which the wind is blowing = 60 km/h
The direction of the wind = South
Therefore, representing the speed of the plane and the wind in vector format, we have;
The velocity vector of the plane = 210.
The velocity vector of the wind = -60.
Where, North is taken as the positive y or direction
The resultant velocity vector is found by summation of the two vectors as follows;
Resultant velocity vector = The velocity vector of the plane + The velocity vector of the wind
Resultant velocity vector = 210. + (-60.) = 210. - 60. = 150.
The resultant velocity vector = 150.
Therefore, the resultant velocity of the plane relative to the ground = 150 kh/h north.
Explanation:
7) Given:
v₀ = 2.0 m/s
v = 0 m/s
t = 3.00 s
Find: Δx
Acceleration isn't included in the problem, so use a kinematic equation that doesn't involve a.
Δx = ½ (v + v₀) t
Δx = ½ (0 m/s + 2.0 m/s) (3.00 s)
Δx = 3.0 m
8) Given:
v₀ = 0 m/s
v = 5 m/s
t = 4 s
Find: a
Displacement isn't included in the problem, so use a kinematic equation that doesn't involve Δx.
v = at + v₀
5 m/s = a (4 s) + 0 m/s
a = 1.25 m/s²
9) Given:
v_avg = Δx / t
0.5 m/s = 8 m / t
t = 16 s
Hi!
This is representing constant positive acceleration. We can figure this out by its increasing speed, not a speed that is constant.
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