anonymous
  • anonymous
Kinetic energy KE=1/2mv^2 if mass is increased by 4% And velocity reduced by 3% determine the approximate change in K.E... Apparently i need to use binomial theorem (a=x)^n but how do i do it?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Its really simple. New mass = 1+4/100 i.e\[Mass_n = 1 + \frac{ 4 }{ 100 }\]
anonymous
  • anonymous
For velocity subtract \[Velocity_n = 1-0.3 = 0.97\]
anonymous
  • anonymous
Square .97 now

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anonymous
  • anonymous
Tell me you get .
anonymous
  • anonymous
For distinction I need to use binomial.... Thank you. I think i have done something like that.
anonymous
  • anonymous
The Binomial theorem example: (x+y)^2 = x^2 + y^2 + 2xy
anonymous
  • anonymous
How can we possible apply the Binomial theorem here..??
anonymous
  • anonymous
I don't know. Will have to check with my teacher... Thank you for your help.
Michele_Laino
  • Michele_Laino
If I call with \(v_1,m_1\) the new values of the speed and mass, and with \(v_0,m_0\) the corresponding initial value, then I can write this: \[\large \begin{gathered} {m_1} = 1.04{m_0} \hfill \\ {v_1} = 0.97{v_0} \hfill \\ \hfill \\ \frac{{\frac{1}{2}{m_0}v_0^2 - \frac{1}{2}{m_1}v_1^2}}{{\frac{1}{2}{m_0}v_0^2}} = \frac{{{m_0}v_0^2 - 1.04{m_0} \cdot {{\left( {0.97} \right)}^2}v_0^2}}{{{m_0}v_0^2}} = \hfill \\ \hfill \\ = \frac{{{m_0}v_0^2\left\{ {1 - 1.01 \cdot {{\left( {0.97} \right)}^2}} \right\}}}{{{m_0}v_0^2}} = ...? \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
values*
Michele_Laino
  • Michele_Laino
oops.. I have made a typo: \[\large \begin{gathered} \frac{{\frac{1}{2}{m_0}v_0^2 - \frac{1}{2}{m_1}v_1^2}}{{\frac{1}{2}{m_0}v_0^2}} = \frac{{{m_0}v_0^2 - 1.04{m_0} \cdot {{\left( {0.97} \right)}^2}v_0^2}}{{{m_0}v_0^2}} = \hfill \\ \hfill \\ = \frac{{{m_0}v_0^2\left\{ {1 - 1.04 \cdot {{\left( {0.97} \right)}^2}} \right\}}}{{{m_0}v_0^2}} = ...? \hfill \\ \end{gathered} \]
alekos
  • alekos
i see what your doing this is like finding ΔΕ/Ε0 correct ?
anonymous
  • anonymous
Thank you. .
Michele_Laino
  • Michele_Laino
that's right!! @alekos

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