Question

If the periodic time of a pendulum whose length is 36.9 cm equals 1.22 seconds, how much gravitational acceleration is in the pendulum's region of existence?

Answer 1

Answer:

t=2pie√l/√g

sobs

t²=4pie²l/g

t=1.22 approximately t=1

length (L) =36.9 approximately l= 37

1²=4(10)(37)/(g)

1=40(37)/g

g=40(37)

g=1480

g=14.8