Question

A vector has components Ax = 52.0 m and Ay = 41.0 m. Find: (a) The length of the vector A.

Answer 1

Answer:

Part a)

A = 66.2 m

Part b)

Angle = 38.35 degree

Explanation:

Part a)

Length of the vector is the magnitude of the vector

here we know that

$A_x = 52.0 m$

$A_y = 41.0 m$

now we have

$A = \sqrt{A_x^2 + A_y^2}$

$A = \sqrt{52^2 + 41^2}$

$A = 66.2 m$

Part b)

Angle made by the vector is given as

$tan\theta = \frac{A_y}{A_x}$

$tan\theta = \frac{41}{52}$

$\theta = 38.25 degree$