Physics, asked by piyushnehra2006, 1 month ago


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Answered by UtsavPlayz
2

Initial Velocity(u )  =5 \times {10}^{14} \: m{s}^{-1}

Final Velocity( v)  = 2u

Acceleration( a)  = {10}^{4} \: m{s}^{-2}

(i) Using Kinematical Equation,

v = u + at

2u = u + at

u = at

5  \times  {10}^{14}  = ( {10}^{4} )t

  \implies  \boxed{t = 5 \times  {10}^{10}  \: s}

(ii) Using Kinematical Equation,

 {v}^{2}  -  {u}^{2}  = 2as

 {(2u)}^{2}  -  {u}^{2}  = 2as

3 {u}^{2}  = 2as

3 {(5 \times  {10}^{14}) }^{2}  = 2( {10}^{4} )s

 \dfrac{3}{2}  \times  \dfrac{25 \times  {10}^{28} }{ {10}^{4} }  = s

 \implies  \boxed{s = 3.75 \times  {10}^{25} \: m}

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