Physics, asked by sreedhums2606, 11 months ago

A particle is moving with constant speed over circle x2 + y2 = 50, with speed vums
V10 m/s.
Acceleration of the particle, when it is at point (5.5) is (in m/s)​

Answers

Answered by behuman1234567890
2

Explanation:

it contains only centripetal acceleration

Attachments:
Answered by sushmadhkl
0

Given:

Speed(v)= 10 m/s (constant)

Equation of a circle is, x²+y²=50⇒ r² = 50⇒ r = 5√2

Acceleration(a)= ?

To find the acceleration of the particle, when it is at point (5.5)

Solution:

Acceleration (a) is calculated as,

a= \frac{v^{2} }{r}

      =10²/5√2

      =100/5√2

      =20/√2 m/s

      =14.14 m/s

Hence, acceleration of the particle, when it is at point (5.5) is 14.14 m/s².​

 

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