Answer:
x-component=-9.3 m
Magnitude of A=17.7m
Explanation:
We are given that
[tex]A_y=+15 m[/tex]
[tex]\theta=32^{\circ}[/tex]
We have to find the x-component of A and magnitude of A.
According to question
[tex]A_y=\mid A\mid cos\theta[/tex]
Substitute the values then we get
[tex]15=\mid A\mid cos32[/tex]
[tex]\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}[/tex]
[tex]\mid A\mid=17.7[/tex]m
[tex]tan\theta=\frac{perpendicular\;side}{Base}[/tex]
[tex]tan32=\frac{A_x}{A_y}=\frac{A_x}{15}[/tex]
[tex]0.62\times 15=A_x[/tex]
[tex]A_x=9.3[/tex]
The value of x-component of A is negative because the vector A lie in second quadrant.
Hence, the x- component of A=-9.3 m
Answer:
The x component is 24.0 m and the magnitude of vector A is 28.30 m.
Explanation:
Given that,
Vertical component [tex]A_{y}=+15.0 m[/tex]
Angle = 32.0
We need to calculate the x component of A
Using formula of angle
[tex]\tan\theta=\dfrac{A_{y}}{A_{x}}[/tex]
[tex]A_{x}=\dfrac{A_{y}}{\tan\theta}[/tex]
Where, [tex]A_{y}[/tex] = vertical component
[tex]A_{x}[/tex] = horizontal component
Put the value into the formula
[tex]A_{x}=\dfrac{15.0}{\tan32}[/tex]
[tex]A_{x}=24.0\ m[/tex]
We need to calculate the magnitude of A
Using formula of vector magnitude
[tex]|A|=\sqrt{(15.0)^2+(24.0)^2}[/tex]
[tex]|A|=28.30\ m[/tex]
Hence, The x component is 24.0 m and the magnitude of vector A is 28.30 m.
