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Ahat [919]
2 years ago
12

Question is down below need rply fast

Physics
1 answer:
My name is Ann [436]2 years ago
8 0

The X and Y components are as follows;

1. X = 35 * cos 57 = 19. 1m/s; Y = 35 * sin 57 = 29.4 m/s

2. X = 12 * -cos 34  = -10 m/s; Y = 12 * -sin 34 = -6.7 m/s

3. X = 8 * -cos 90  = 0 m/s; Y = 12 -sin 90 = -8 m/s

4. X = 20 * cos 75 = 5. 2m/s; Y = 20 * (-sin 75) =  -19.3 m/s

<h3>What are the horizontal and vertical components of the vectors?</h3>

The horizontal and vertical components of the velocities are given as follows:

  • Horizontal component, X = x cos θ
  • Vertical component, Y = y sin θ

1. 35 m/s at  57° from x-axis

X = 35 * cos 57 = 19. 1m/s

Y = 35 * sin 57 = 29.4 m/s

2. 12m/s at 34° S of W

X = 12 * -cos 34  = -10 m/s

Y = 12 * -sin 34 = -6.7 m/s

3. 8 m/s at South

X = 8 * -cos 90  = 0 m/s

Y = 12 -sin 90 = -8 m/s

4. 20 m/s at  275° from x-axis

X = 20 * cos 75 = 5. 2m/s

Y = 20 * (-sin 75) =  -19.3 m/s

In conclusion, the X and Y components are found by taking cosines and sine of the angles.

Learn more about horizontal and vertical components at: brainly.com/question/26446720

#SPJ1

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Draw the following vector quantity Using the coordinate system.
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The given vectors quantities can be described by their properties of both

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  • 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;

\vec{F} = 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; \vec{v} = 120 × cos(25)·i + 120×sin(25)·j, which gives;

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

\vec{v} ≈ 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 \vec{d} = 60 × cos(45°)·i + 60 × sin(45°)·j

Which gives;

\vec{d} = 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:

brainly.com/question/10409036

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