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MrMuchimi
2 years ago
12

A bicycle tire has a radius of 0.23 m. the tire spins at a constant speed with a centripetal acceleration of 42 m/s2. what is th

e tangential speed of the tire? 3.1 m/s 6.5 m/s 9.7 m/s 14 m/s
Physics
2 answers:
swat322 years ago
7 0

Answer:

(a)

Explanation:

3.1 m/s

vredina [299]2 years ago
3 0

Answer:

a = v^2/r      centripetal acceleration

v = (a * r)^1/2 = (42 * .23)^1/2 = 3.1 m/s

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        TM  =  \frac{1}{2}  *  k  *  A^2

Here  k is the spring constant  ,  A is the total displacement of the  the mass  from maximum  compression to maximum extension of the spring

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Here the potential  energy of the mass is mathematically represented as

     PE   = \frac{1}{ 2}  *  k *  [ x ]^2

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is  

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So

     PE   = \frac{1}{ 2}  *  k *  [ \frac{A}{2}  ]^2

So

      KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{2}  *  k  *  [\frac{A}{2} ]^2

=>    KE =  \frac{1}{2}  *  k  *  A^2 - \frac{1}{8}  *  k  *  A ^2

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       \frac{KE}{TM} =  \frac{0.375  k A^2 }{0.5 k A^2}

=>    \frac{KE}{TM} = 0.75

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