Respuesta :
Answer:
the correct answers is a    W' = 6.2 N
Explanation:
In this exercise they give the description of a simple pendulum, its angular velocity
     w = [tex]\sqrt { \frac{g}{l} }[/tex]
angular velocity is related to frequency and period
     w = 2π f = 2π / T
we substitute
     T = 2Ï€ \sqrt { \frac{l}{g} } Â
     T² = 4π²  [tex]\frac{ l}{g}[/tex]
With the period on Earth we can find the length of the pendulum
     l = [tex]\frac{ T^{2} g}{4 \pi ^2 }[/tex]
     l = [tex]\frac{ 1^2 9.8}{ 4 \pi ^2}[/tex]
     l = 0.25 m
this longitude is maintained on planet X, so we can find the value of the acceleration of gravity (g ’) on that planet
     g’ = [tex]\frac{4 \pi ^2 l}{T'^2}[/tex]
     g’ = [tex]\frac{4 \pi ^2 0.25}{ 1.8^2}[/tex]
     g’ = 3.05 m / s²
therefore the weight of the body on this planet is
      W ’= m g’
the mass is invariable in all systems
      W ’= 2.0 3.05
      W ‘= 6.1 N
When examining the correct answers is a
