Resolution of a Vector into Rectangular Components Questions in English

(B) The initial vector is $\vec{A} = 4\hat{i} + 10\hat{j}$.
The magnitude of the vector is $|\vec{A}| = \sqrt{(4)^2 + (10)^2} = \sqrt{16 + 100} = \sqrt{116} \,m$.
When the vector is rotated, its magnitude remains constant.
Let the new vector be $\vec{A}' = 8\hat{i} + n\hat{j}$, where the $x$-component is doubled $(4 \times 2 = 8)$.
Since the magnitude is conserved, $|\vec{A}'| = |\vec{A}|$.
$\sqrt{8^2 + n^2} = \sqrt{116}$.
Squaring both sides, we get $64 + n^2 = 116$.
$n^2 = 116 - 64 = 52$.
$n = \sqrt{52} \approx 7.21 \,m$.
Thus, the new $y$-component is approximately $7.2 \,m$.

(C) Let $\overrightarrow{B} = \hat{i} - \hat{j}$.
To find the component of vector $\overrightarrow{A}$ along the direction of $\overrightarrow{B}$,we calculate the scalar projection of $\overrightarrow{A}$ onto the unit vector of $\overrightarrow{B}$.
The unit vector along $\overrightarrow{B}$ is $\hat{b} = \frac{\overrightarrow{B}}{|\overrightarrow{B}|} = \frac{\hat{i} - \hat{j}}{\sqrt{1^2 + (-1)^2}} = \frac{\hat{i} - \hat{j}}{\sqrt{2}}$.
The component of $\overrightarrow{A}$ along $\overrightarrow{B}$ is given by $\overrightarrow{A} \cdot \hat{b}$.
$\overrightarrow{A} \cdot \hat{b} = (a_x \hat{i} + a_y \hat{j} + a_z \hat{k}) \cdot \left( \frac{\hat{i} - \hat{j}}{\sqrt{2}} \right)$.
$= \frac{a_x(1) + a_y(-1) + a_z(0)}{\sqrt{2}} = \frac{a_x - a_y}{\sqrt{2}}$.

(A) The magnitude of a vector $\overrightarrow{p}$ is given by the formula: $|\overrightarrow{p}| = \sqrt{p_x^2 + p_y^2}$.
Given, $|\overrightarrow{p}| = 25 \text{ units}$ and $p_y = 7 \text{ units}$.
Substituting the values into the formula: $25 = \sqrt{p_x^2 + 7^2}$.
Squaring both sides: $25^2 = p_x^2 + 7^2$.
$625 = p_x^2 + 49$.
$p_x^2 = 625 - 49 = 576$.
Taking the square root: $p_x = \sqrt{576} = 24 \text{ units}$.
Therefore, the $x$-component is $24 \text{ units}$.

(A) Let the magnitude of vector $\vec{A}$ be $A$.
Given that the $y$-component of vector $\vec{A}$ is $A_y = 3.0 \ m$.
The angle $\theta = 30^{\circ}$ is measured counterclockwise from the positive $y$-axis.
Using the definition of the component of a vector,the $y$-component is given by $A_y = A \cos(\theta)$.
Substituting the given values: $3.0 = A \cos(30^{\circ})$.
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $3.0 = A \times \frac{\sqrt{3}}{2}$.
Solving for $A$: $A = \frac{3.0 \times 2}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \ m$.

$(A)$ To find the resultant force, we resolve each force into its $X$ and $Y$ components.
$X$-components:
$\Sigma F_x = 19 + 15 \cos 60^{\circ} - 16 \cos 45^{\circ} - 11 \cos 30^{\circ}$
$\Sigma F_x = 19 + 15(0.5) - 16(0.707) - 11(0.866)$
$\Sigma F_x = 19 + 7.5 - 11.312 - 9.526 = 5.662 \,N$
$Y$-components:
$\Sigma F_y = 15 \sin 60^{\circ} + 16 \sin 45^{\circ} - 11 \sin 30^{\circ} - 22$
$\Sigma F_y = 15(0.866) + 16(0.707) - 11(0.5) - 22$
$\Sigma F_y = 12.99 + 11.312 - 5.5 - 22 = -3.198 \,N$
Resultant force $R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} = \sqrt{(5.662)^2 + (-3.198)^2} \approx \sqrt{32.06 + 10.23} \approx \sqrt{42.29} \approx 6.5 \,N$.
Direction $\theta = \tan^{-1}\left(\frac{\Sigma F_y}{\Sigma F_x}\right) = \tan^{-1}\left(\frac{-3.198}{5.662}\right) \approx \tan^{-1}(-0.565) \approx -29.5^{\circ} \approx 330.5^{\circ}$.
Thus, the resultant is approximately $6.5 \,N$ at $330^{\circ}$.

(D) The velocity vector $\vec{v}$ can be resolved into its rectangular components along the $X$ and $Y$ axes.
Given magnitude $v = 10 \ m/s$ and angle $\theta = 60^{\circ}$.
The components are given by:
$v_x = v \cos \theta = 10 \cos 60^{\circ} = 10 \times \frac{1}{2} = 5 \ m/s$
$v_y = v \sin \theta = 10 \sin 60^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} \ m/s$
Thus,the velocity vector is $\vec{v} = v_x \hat{i} + v_y \hat{j} = 5 \hat{i} + 5 \sqrt{3} \hat{j} \ m/s$.