Answer to Question #96254 in Mechanics | Relativity for muhammad sheraz khan

Question

Vector A is in the direction 34.0* clockwise from the
Y axis. The x-component of A is Ax= -16.0m. (a) What is the Y-component Of A? (b) What is the magnitude of A?

Answer 1

The components of vector $\bold a$ are as follows: $a \sin \theta = - a_x$, $a \cos \theta = a_y$, where $a = |\bold a|$, $\theta = 34^{\circ}$, $a_x = 16$.

From the first equation, $a = \frac{-a_x}{\sin \theta}$, and plugging it into the second equation, obtain $a_y = -a_x \ctg \theta \approx 23.72$.

The magnitude of $\bold a$ is $a = \sqrt{a_x^2 + a_y^2} \approx 28.61$.