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Arada [10]
3 years ago
9

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

2 m and its period is measured to be 2 s. The value of g obtained in this investigation is most nearly __________.A. 1 m/s²B. 2 m/s²C. 5 m/s²D. 10 m/s²E. 20 m/s²
Physics
1 answer:
SVEN [57.7K]3 years ago
8 0

Answer:

Acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer

Explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

T=2\pi \sqrt{\frac{l}{g}}

2=2\times 3.14 \sqrt{\frac{2}{g}}

0.3184= \sqrt{\frac{2}{g}}

Squaring both side

0.1014= {\frac{2}{g}}

g=19.71=20m/sec^2

So acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer.

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WITCHER [35]

Displacement is d  


Vf² = Vi² + 2 g d  


(-20²) = (+10²) + 2 (-9.8) d  


-19.6 d = 300  


d = -15.3 m  


negative means lower


time is t  


d = Vi t + 1/2 g t²



-15.3 = 10 t + (-4.9) t²



4.9 t² - 10 t -15.3 = 0  


t = 3.06 s

Hope this helps -John

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3 years ago
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