The length of a displaced pendulum ball which p... - JAMB Physics 1995 Question
The length of a displaced pendulum ball which passes its lowest point twice every seconds is [g = 10m-2]
A
0.25 m
B
0.45 m
C
0.58m
D
1.00m
correct option: a
T = 2\(\pi\) \(\sqrt\frac{L}{g}\)
T = 2 x 3.142\(\sqrt\frac{L}{10}\)
L= 0.25m
T = 2 x 3.142\(\sqrt\frac{L}{10}\)
L= 0.25m
Please share this, thanks:
Add your answer
No responses