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Sonbull [250]
4 years ago
13

Hello jigxjufogonskdcxx ggnkctbntjc​

Physics
2 answers:
igor_vitrenko [27]4 years ago
7 0

Answer:

svbsshsjjsjs Hello :)

Ede4ka [16]4 years ago
3 0

Answer:

Hiiiiiiiiiiiii!!!!!!!!!!!!!!............ gshsujfhiwopqeyegsbdk

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A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block
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Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  v  = 11.76  \ m/ s

Explanation:

From the question we are told that

   The  total distance traveled is  d =  1.2  \ m

    The mass of the block is  m_b  =  0.3 \ kg

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substituting values

     PE  =   3 *  9.8  *  0.60

      PE  =  17.64 \  J

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          KE  =  \frac{1}{2}  *  m v^2

So

      \frac{1}{2}  *  m* v ^2  =  PE

substituting values  

  =>    \frac{1}{2}  *  3 * v ^2  = 17.64

=>       v  = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }

=>    v  = 11.76  \ m/ s

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