Respuesta :
Answer:
[tex]\overline{V_x} = 6\sqrt{2}\\\overline{V_y} = -6\sqrt{2}[/tex]
components can be written as:
[tex]\overline{V} = 6\sqrt{2}\,\hat{i}-6\sqrt{2}\,\hat{j}[/tex]
Step-by-step explanation:
The components of a velocity vector at any direction can be thought of separate vectors in 'i' and 'j' directions respectively. These vectors can be added to make the resultant velocity vector.
In our case the magnitude is provided:
[tex]|\overline{V}|= 12[/tex]
the direction is provided to be southeast:
[tex]\theta = 45[/tex]
[tex]\theta[/tex] is measured from positive x-axis clockwise towards the 4th quadrant. it is important to know that quadrant since the signs of the components will be where x is positive and y is negative as the direction is south east (+x,-y).
the components of the velocity will be:
[tex]\overline{V_x} = 12\cos{(45)} = 12\dfrac{\sqrt{2}}{2} = 6\sqrt{2}\\\overline{V_y} = -12\sin{(45)} = -12\dfrac{\sqrt{2}}{2} = -6\sqrt{2}\\[/tex]
So that:
[tex]\overline{V} = \overline{V_x}\hat{i}+\overline{V_x}\hat{j}[/tex]
[tex]\overline{V} = 6\sqrt{2}\,\hat{i}-6\sqrt{2}\,\hat{j}[/tex]
This can be checked using since we know that magnitude of the above vector should be 12.
[tex]|\overline{V}| = \sqrt{(12\cos{\theta})^2+(12\sin{\theta})^2}[/tex]
[tex]|\overline{V}| = \sqrt{(6\sqrt{2})^2+(-6\sqrt{2})^2}\\|\overline{V}| = \sqrt{36(2)+36(2)}\\|\overline{V}| = \sqrt{144} = 12\\[/tex]
The velocity vector of the wind is given by the vector sum of the components of the velocity vector.
The components of the velocity vector of the wind are;
- x-direction component of the wind's velocity, [tex]\vec{v}_x[/tex] = 6·√2
- y-direction component of the wind's velocity, [tex]\vec{v}_y[/tex] = -6·√2
Reasons:
The given parameter are;
The speed of the wind = 12 km/hr
Direction of the wind = Southeast direction
Assumption: The positive y-direction = North
Required:
The components of the velocity vector of the wind
Solution:
A velocity vector, [tex]\vec{v}[/tex] is given by the magnitude and direction, θ, of the vector as follows;
- [tex]\vec{v}[/tex] = |v| × cos(θ)·i + |v| × sin(θ)·j
The magnitude of the velocity of the wind, |v| = 12 km/hr
The direction of the wind, θ = South east = -45°
Therefore;
[tex]\vec{v}[/tex] = 12 × cos(-45°)·i + 12 × sin(-45°)·j
[tex]\displaystyle cos \left(-45^{\circ} \right) = \mathbf{\frac{\sqrt{2} }{2}}[/tex]
[tex]\displaystyle sin\left(-45^{\circ} \right) = -\frac{\sqrt{2} }{2}[/tex]
[tex]\vec{v} = 12 \times \displaystyle \frac{\sqrt{2} }{2} \cdot \mathbf{i} - 12 \times \displaystyle \frac{\sqrt{2} }{2} \cdot \mathbf{j} = 6 \cdot {\sqrt{2} } \cdot \mathbf{i} -6 \cdot {\sqrt{2} } \cdot \mathbf{j}[/tex]
[tex]\vec{v}[/tex] = 6·√2·i - 6·√2·j
The components of the velocity vector of the wind are;
Component in the x-direction, vₓ = 6·√2
Component in the y-direction, [tex]v_y[/tex] = -6·√2
Learn more here:
https://brainly.com/question/18090230
