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What is the resultant displacement of the two vectors below?
Vector 1: 2.0 cm, E
Vector 2: 3.0 cm, N

A 5.8 cm, 59 degrees East of the North axis
B 5.4 cm, 34 degrees East of the North axis
C 3.6 cm, 56 degrees North of the East axis
D 3.6 cm, 34 degrees North of the East axis

Respuesta :

Answer:

C 3.6 cm, 56 degrees North of the East axis

Explanation:

The two vectors are perpendicular to each other, so we can find the magnitude of their resultant simply by using the Pythagorean theorem:

[tex]R=\sqrt{A^2+B^2}[/tex]

where

A = 2.0 cm is the magnitude of the first vector

B = 3.0 cm is the magnitude of the second vector

Substituting,

[tex]R=\sqrt{2^2+3^2}=3.6 cm[/tex]

Now we have to find the angle. If we measure the angle as North of East, the tangent of the angle is equal to the ratio between the component along North and the component along East. Therefore, in this case:

[tex]tan \theta = \frac{B}{A}=\frac{3}{2}=1.5\\\theta = tan^{-1}(1.5)=56.3 \sim 56^{\circ}[/tex]

So, 56 degrees North of East.

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