A particle is moving on a circular path of ra | vedclass

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

Question

(C) The velocity of the particle at point $P$ is $\vec{v}_P$ and at point $Q$ is $\vec{v}_Q$. Both have the same magnitude $v$.The change in velocity

Vedclass · Accepted Answer

$2v\sin 20^\circ$

Vedclass · Answer

(C) The velocity of the particle at point $P$ is $\vec{v}_P$ and at point $Q$ is $\vec{v}_Q$. Both have the same magnitude $v$.The change in velocity is given by $\Delta \vec{v} = \vec{v}_Q - \vec{v}_P$.The magnitude of the change in velocity for a particle moving in a circle with uniform speed is given by the formula $|\Delta \vec{v}| = 2v \sin(	heta/2)$, where $	heta$ is the angle subtended at the center.Given $	heta = 40^\circ$, we have:$|\Delta \vec{v}| = 2v \sin(40^\circ / 2) = 2v \sin 20^\circ$.Thus, the correct option is $C$.

$A$ particle is moving on a circular path of radius $r$ with uniform speed $v$. The change in velocity when the particle moves from $P$ to $Q$ is $(\angle POQ = 40^\circ)$.

Similar Questions

Two spheres of the same material have radii $1\ m$ and $4\ m$ and temperatures $4000\ K$ and $2000\ K$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

Among the following,the strongest base is

In a series $LCR$ circuit,what will be the nature of the circuit for frequencies higher than the resonance frequency?

The bonds present in ${N_2}{O_5}$ are ........

In the given figure, the magnetic field at the centre will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module