Respuesta :

fieryanswererft

6.21 cm

  • use the pendulum formula : [tex]\sf \bold{\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}}[/tex]

where

  • T is time or period
  • π is pie = 22/7
  • L is pendulum length
  • g is acceleration due to gravity

Given:

  • T = 0.500 s
  • g = 9.8 m/s²

solving step-wise:

[tex]\dashrightarrow \mathrm{T}=2 \pi \sqrt{\dfrac{\mathrm{L}}{\mathrm{g}}}[/tex]

[tex]\sf \dashrightarrow \mathrm{0.5}=2 \pi \sqrt{\dfrac{\mathrm{L}}{\mathrm{9.8}}}[/tex]

[tex]\dashrightarrow \mathrm{\dfrac{0.5}{2 \pi } }=\sqrt{\dfrac{\mathrm{L}}{\mathrm{9.8}}}[/tex]

[tex]\dashrightarrow\sqrt{\dfrac{\mathrm{L}}{\mathrm{9.8}}}= \mathrm{\dfrac{0.5}{2 \pi } }[/tex]

[tex]\sf \dashrightarrow{\dfrac{\mathrm{L}}{\mathrm{9.8}}}= (\mathrm{\dfrac{0.5}{2 \pi } })^2[/tex]

[tex]\sf \dashrightarrow{{\mathrm{L}}= (\mathrm{\dfrac{0.5}{2 \pi } })^2*9.8[/tex]

[tex]\sf \dashrightarrow{{\mathrm{L}}=0.06205922 \ m[/tex]

  • 1 m → 100 cm

[tex]\sf \dashrightarrow{{\mathrm{L}}=6.2059\ cm[/tex]

[tex]\sf \dashrightarrow{{\mathrm{L}}=6.21\ cm[/tex]             { rounded to nearest hundredth }

Let's see

[tex]\\ \rm\rightarrowtail T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

[tex]\\ \rm\rightarrowtail 0.5=2\pi \sqrt{\dfrac{l}{9.8}}[/tex]

[tex]\\ \rm\rightarrowtail 0.783=\pi \sqrt{l}[/tex]

[tex]\\ \rm\rightarrowtail 0.2494=\sqrt{l}[/tex]

[tex]\\ \rm\rightarrowtail \ell=0.0622m[/tex]

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