Let the two forces have equal magnitude A. If | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The magnitude of the resultant $R$ of two vectors of equal magnitude $A$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{A

Vedclass · Accepted Answer

$\cos^{-1}\left(-\frac{7}{9}\right)$

Vedclass · Answer

(B) The magnitude of the resultant $R$ of two vectors of equal magnitude $A$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{A^2 + A^2 + 2A^2 \cos 	heta}$.Given $R = \frac{2A}{3}$,we substitute this into the equation:$\frac{2A}{3} = \sqrt{2A^2 + 2A^2 \cos 	heta}$.Squaring both sides:$\frac{4A^2}{9} = 2A^2(1 + \cos 	heta)$.Dividing both sides by $2A^2$:$\frac{2}{9} = 1 + \cos 	heta$.Solving for $\cos 	heta$:$\cos 	heta = \frac{2}{9} - 1 = \frac{2-9}{9} = -\frac{7}{9}$.Therefore,the angle $	heta = \cos^{-1}\left(-\frac{7}{9}ight)$.

Let the two forces have equal magnitude $A$. If the magnitude of the resultant is $\frac{2A}{3}$,then the angle between those two forces is:

Similar Questions

What displacement must be added to the displacement $25\hat{i} - 6\hat{j} \text{ m}$ to give a resultant displacement of $7.0 \text{ m}$ pointing in the $X$-direction?

Is it possible that $\vec{A} + \vec{B} = \vec{A} - \vec{B}$?

Two forces of $12 \, N$ and $8 \, N$ act upon a body. The resultant force on the body has a maximum value of ........ $N$.

Among the given pairs of vectors,the resultant of two vectors can never be $3$ units. The vectors are

The sum of two forces acting at a point is $16\, N.$ If the resultant force is $8\, N$ and its direction is perpendicular to the minimum force,then the forces are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module