Tutor there's no missing information this is a lab from phy. Please help me on this

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Date ROTATIONAL EQUILIBRIUM SIMULATION ASSIGNMENT DIRECTIONS: For this activity, you will be using the pHet website http://phet.colorado.edu/sims/html/balancing-act/latest/balancing-act_en.html (linked on my website). When you open it, select the \"Balance Lab\" tab and select \"Mass Labels\Date Move the same 5 kg brick to the 1.0 m position. How does the position of the seesaw compare to its position when the brick was at the 0.5 m position? Why is this the case? Put the supports back in place and remove one of the 5 kg bricks. Place the other at the 0.25 m mark and then add a 10 kg brick to the 0.25 m mark on the other side. Remove the supports. Does the seesaw move? Why/why not? Replace the 10 kg brick with a 20 kg brick. How is the motion of the seesaw affected? Why does this happen? What does it mean to be in a state of rotational equilibrium? PART B. SIMPLE ROTATIONAL EQUILIBRIUM. In this part of the activity, you will be placing objects at various positions to balance the seesaw. You will be told a starting mass (brick or person) and position and then will determine where a second mass should be placed in order to balance it out. To do this, you will be required to do a torque summation. The first one has been done as an example. Check your answer by placing the objects on the seesaw and removing the supports. Date, Mass 1 (kg) | Position 1 (m) Tright = Tlet M(T(sgs) 2 | position 2 (m) 20 0.5 (20)(9.8)(.5) = (5)(9.8)x . 20 x=20 20 1.0 0 80 0.25 2 30 0.5 10 10 1.0 56 5 2.0 % 15 1.0 6o PART C. COMPLEX ROTATIONAL EQUILIBRIUM. For this part you will be working with situations with multiple objects on the same side of the fulcrum. For example: 80 kg 1w a6 17 2 Date Date Known osition Mass of Mystery Position (m) Mass 1 Position 1 Mass 2 Position 2 Mass 3 Position Mass 1 1 (m) TRight = TLeft Package Package Right (kg) (m) (kg) (m) TRight = TLeft (kg) 3 (m) (kg) Left (kg) (60) (9.8) (0.5) = (x)(9.8)(1.5) 60 0.5 A 1.5 20 X = 20 (20)(9.8)(1.5) + (30)(9.8)(0.5) = 0.75 m 20 1.5 m Left 30 0.50 m Left 50 (60)(9.8)x Right 30 1.C C 2.0 X = 0.75 30 0.25 D .75 20 1.0 m Left 60 2.0 m Left 80 60 0.25 H 2.0 Known Position Known Mass of Position Mystery Position Mass 1 Mass 2 TRight = TLeft Package 1 (m) 2 (m) Package (m) (kg) (kg) (kg) 80 1.5 m Left 30 0.5 m Right 60 1.75 m 0.5 m 1.5 m 20 80 Left Left Right 0.5 m 1.75 m 15 E 1.25 m 60 Left Right Right 0.25 m 50 1.0 m Left 80 20 Right 0.25 m 1.0 m B 2.0 m 20 5 Left Left Right If you are trying to balance a seesaw and you have a higher mass on the left, should the mass on PART D. MYSTERY PACKAGES. the right be placed at a greater distance or lower distance to balance it? Why? For this part, you will determine the masses of the mystery packages based on known objects and distances