Gravity on Another Planet
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 105 full swing cycles in a time of 144 s.
What is the value of the acceleration of gravity on this planet?
Express your answer in meters per second per second. m/s^2
2006-12-13 15:58:07 · 4 answers · asked by MattS 1 in Science & Mathematics ➔ Physics
Where is this planet??! :))
jokes apart, let's shoot the problem.
1. the type of motion is a simple harmonic motion executed by the pendulum.
hence, the formula used will be :
T=2 x pi x Sqrt (L/g)
where,
T = tme period of ONE complete oscillation
pi = const. = 3.142
L = Length of the string = 0.53 m
g = gravitational constant = ??
T = 105/144 = 0.73 s (apprx)
hence substituting the values and calculating, we get
g = 12.53 m/s2.
simple... ain't it??!! :))
2006-12-13 16:26:31 · answer #1 · answered by Vinay V 2 · 0⤊ 0⤋
Question 1) The acceleration of gravity is unchanged by the air resistant.
2016-05-23 23:35:06 · answer #2 · answered by Yolanda 4 · 0⤊ 0⤋
g=Gm/r²= determiantion acceleration due to gravity
thats the equation...or either
Tp= 2∏√l/g -cycle of pendulum
im not very good at physics but i've already learnt this and i've forgotten- hope it helps
2006-12-13 16:17:55 · answer #3 · answered by pamplemousse 6 · 0⤊ 0⤋
use equation
2*pi* T = sqrt(g/L) where T= period, L= length, g= gravitational acceleration.
2006-12-13 16:03:48 · answer #4 · answered by John H 4 · 0⤊ 0⤋