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Harman [31]
3 years ago
15

A particle is moving in a plane with constant radial velocity \dot{r} = 2.68 ​r ​˙ ​​ =2.68 m/s, having started at the origin. I

t also has a constant angular velocity \dot{\theta} = 2.45 ​θ ​˙ ​​ =2.45 rad/s. When the particle is 3.58 m from the origin, what is the magnitude of its velocity in m/s?
Physics
1 answer:
Aloiza [94]3 years ago
5 0

Answer:

9.2m/s

Explanation:

We are given that

Radial velocity=\dot r=2.68m/s

Angular velocity=\dot \theta=2.45rad/s

Displacement=r=3.58 m

We have to find the magnitude of velocity of particle in m/s.

The magnitude of velocity of particle in polar coordinates is given by

\mid v\mid=\sqrt{(\dot r)^2+r^2(\dot \theta)^2}

Using the formula

The magnitude of velocity of particle

\mid v\mid=\sqrt{(2.68)^2+(3.58)^2(2.45)^2}

\mid v\mid=9.2m/s

Hence, the magnitude of velocity of particle=9.2m/s

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