After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm. The explorer finds that the pendulum completes 96.0 full swing cycles in a time of 126 s.
What is the value of the acceleration of gravity on this planet?
Express your answer in meters per second per second.
2006-11-18 13:49:19 · 3 answers · asked by Theresa C 2 in Science & Mathematics ➔ Physics
never mind i got the question i had the right formula but i missed calculated it...heheheh
2006-11-18 14:47:51 · update #1
T = 2pi sqrt(l/g)
g = l/(T/2/pi)^2 = 0.530 / ((126 / 96.0)/2/pi)^2 = 12.1 m/s/s
2006-11-18 13:53:11 · answer #1 · answered by jacinablackbox 4 · 0⤊ 0⤋
The period of a simple pendulum is 2Ïâ[L/g], where L is the length of the pendulum. The period of your pendulum is 126/96 sec.
2006-11-18 21:53:41 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋
The formulae for this question appear earlier in the same chapter. Please reread this chapter plus the following two and answer all questions for 8:00 am Monday morning.
2006-11-18 21:54:27 · answer #3 · answered by St N 7 · 0⤊ 0⤋