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Sagot :
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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