The spaceship Intergalactic landed on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.44 m. She sets the pendulum swinging, and her collaborators carefully count 1.10×102 complete cycles of oscillation during 2.00×102 s. What is the result

18808.7 m/s^2

Explanation:

Given

Length of the pendulum L = 1.44 m

Number of complete cycles of oscillation n = 1.10 x 10^2

total time of oscillation t = 2.00 x 10^2 s

The period of the T = n/t

T = (1.10 x 10^2)/(2.00 x 10^2) = 0.55 ^-s

The period of a pendulum is gotten as

T =

where g is the acceleration due to gravity

substituting values, we have

0.55 =

0.0875 =

squaring both sides of the equation, we have

7.656 x 10^-3 = 144/g

g = 144/(7.656 x 10^-3) = 18808.7 m/s^2

answer; how far out of place which the object finished relative to the starting location;

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.