answersLogoWhite

0


Best Answer

I would solve this problem by starting from the equation for a pendulum's period:



T ≈ 2π√( L / g )



Rearranging for g, and substituting the length in Cambridge:



g ≈ ( 4π² * L ) / T²


. . = ( 4π * 0.9942 m ) / ( 1.000 s + 1.000 s )²


. . = 9.812 m/s²



And for Tokyo:



g ≈ ( 4π² * L ) / T²


. . = ( 4π * 0.9927 m ) / ( 1.000 s + 1.000 s )²


. . = 9.798 m/s²

User Avatar

Wiki User

11y ago
This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: In Cambridge England a seconds pendulum is 0.9942 m long what is the free-fall acceleration in Cambridge?
Write your answer...
Submit
Still have questions?
magnify glass
imp