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A spinning top rotates so that its edge turns at a constant speed of 0.63 m/s. If the top has a radius of 0.10 m, what is the
centripetal acceleration of the edge of the top?
O 0.016 m/s2
0.73 m/s2
4.0 m/s2
6.3 m/s2

Respuesta :

nandhini123

Answer:

The  centripetal acceleration of the edge of the top [tex]6.3m/s^2[/tex]

Explanation:

Given:

Speed with which the top rotates =  0.63 m/s

Radius of the top = 0.10 m

To Find:

Centripetal acceleration of the edge of the top =?

Solution:

The centripetal acceleration is the rate at which the tangential velocity changes. Also its direction is always inwards along the radius vector of the circular motion.

We know that centripetal acceleration is

[tex]a_c = \frac{v^2}{r}[/tex]

Where

v = velocity

r = radius

Substituting the values

[tex]a_c = \frac{0.63}{0.10}[/tex]

[tex]a_c = 6.3[/tex][tex]m/s^2[/tex]

whitneytr12

The centripetal acceleration of the edge of the top of a spinning that has a radius of 0.10 m and that rotates at a constant speed of 0.63 m/s is 4.0 m/s², so the correct option is c.  

The centripetal acceleration is given by:

[tex] a_{c} = \frac{v^{2}}{r} [/tex]

Where:

v: is the tangential velocity = 0.63 m/s

r: is the radius = 0.10 m

Hence, the centripetal acceleration is:

[tex] a_{c} = \frac{v^{2}}{r} = \frac{(0.63 m/s)^{2}}{0.10 m} = 4.0 m/s^{2} [/tex]

Therefore, the centripetal acceleration of the edge of the top is 4.0 m/s², so the correct option is c.

Learn more about centripetal acceleration here:

  • https://brainly.com/question/10023961?referrer=searchResults
  • https://brainly.com/question/79801?referrer=searchResults

I hope it helps you!

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