In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be balanced? The answer is 13.4 In the mobile what is the value for m3 to the nearest hundredth of a kilogram?

Complete Question

The complete question is shown on the first uploaded image

Explanation:

From he question we are told that

    The first mass is  

      The second mass is  

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where are hanged ) is attached is zero  

   Therefore we can say that

Making x the subject of the formula  

Looking at the diagram we can see that the tension T  on the string holding the bar where are hanged  is as a result of the masses ()

     Also at equilibrium the moment about the point where the string holding the bar (where ()  and   are hanged ) is attached is  zero

   So basically

 Making subject

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