On an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates with a period of 2 seconds. What is the value for the acceleration due to gravity (g) on this planet? (Outcome 1)

Make them both to where they have the denominator of 18that makes 3/2 equal 27/18 and 1/9 equal 2/18so now you have  27/18x+5=2/18subtract 5 from the other sideso now you have 27/18x+2/18-5then subtract 2/18 from 27/18 to get 25/18x=-5then divide 25/18 by -5 and your answer will be  -5/18

Ithink it's  c. i'm sure 

M= 700 kgvalue of g( gravity) is constant = 9.8 m/s² or 10 m/s² w = mg = 700*9.8 = 6860n or 6.680*10³ n

Tr