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draw the following vector quantity Using the coordinate system.
a. 190 newton east
b. 120km/hr, 250 north of east
c. 60 meters southwest​


Sagot :

The given vectors quantities can be described by their properties of both

magnitude and direction.

  • a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.
  • b. The velocity vector extending from  (0, 0) to (108.76, 50.714) on the coordinate plane is attached.
  • c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.

Reasons:

a. The magnitude of the vector = 190 N

The direction in which the vector acts = East

Therefore, in vector form, we have;

[tex]\vec{F}[/tex] = 190 × cos(0)·i + 190 × sin(0)·j = 190·i

The vector can be represented by an horizontal line, 190 units long

Coordinate points on the vector = (0, 0) and (190, 0)

The drawing of the vector with the above points using MS Excel is attached.

b. Magnitude of the velocity vector = 120 km/hr. 25° North of east

Solution;

The vector form of the velocity is; [tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j, which gives;

[tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j

[tex]\vec{v}[/tex] ≈ 108.76·i + 50.714·j

Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)

The drawing of the vector is attached

c. The magnitude of the vector = 60 m

The direction of the vector is southwest = West 45° south

The vector form of the displacement is [tex]\vec{d}[/tex] = 60 × cos(45°)·i + 60 × sin(45°)·j

Which gives;

[tex]\vec{d}[/tex] = 30·√2·i + 30·√2·j

Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)

The drawing of the vector is attached

Learn more about vectors here:

https://brainly.com/question/10409036

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