1.a slanted vector has a magnitude of 41 n and is at an angle of 23 degrees north of east. what are the magnitude and direction of the horizontal and vertical components of this vector?
2.a horizontal component vector has a magnitude of 11 m. if the resultant vector forms and angle with this component of 19 degrees, what is the magnitude of the resultant vector?

1. The horizontal component is 37.74 N due to east

The vertical component is 16.02 N due to north

2. The magnitude of the resultant vector is 11.6338 m

Explanation:

The resultant vector is the vector sum of two or more vectors

It has magnitude and direction

Its magnitude = , where x is its component

in the direction of x-axis and y is its component in the direction of y-axis

Its direction = with respect to x-axis

So the components of the resultant vector of magnitude R and direction

Ф with the positive part of x-axis are:

Horizontal component = R cosФ

Vertical component = R sinФ

1.

Slanted vector has a magnitude of 41 N and is at an angle of 23°

north of east

We need to find the horizontal and vertical components with their

directions

North of east means 23° between east and the ray of the vector

If we consider that east is the positive part of the x-axis

Then R = 41 N and Ф = 23°

The horizontal component = (41) cos(23) = 37.74 N due to east

The vertical component = (41) sin(23) = 16.02 N due to north

The horizontal component is 37.74 N due to east

The vertical component is 16.02 N due to north

2.

A horizontal component vector has a magnitude of 11 m

If the resultant vector forms an angle with this component of 19°

The horizontal component = R cosФ

11 = R cos(19)

Divide both sides by cos(19)

11.63 = R

The magnitude of the resultant vector is 11.6338 m

1/0.14nm = 10^10/1.4 a lot