l T-Mobile LTE 100% 7:40 AM webassign.net A rod of length 0.60 m has a pivot (sh
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l T-Mobile LTE 100% 7:40 AM webassign.net A rod of length 0.60 m has a pivot (shown as a circle with a dot inside) at its mid-point. Two forces are applied to the rod. The force F has a magnitude of 2.70 N, and is applied at the left end of the rod. The force F2 has a magnitude of 7.90 N and is applied to the rod a distance of d 0.0600 m from the right end of the rod, as show in the figure below A. In which direction is the torque on the rod due to F1? O clockwise (negative) B. In which direction is the torque on the rod due to F2? clockwise (negative) C. What is the magnitude of the torque on the rod due to F2, the force applied on the right side of the rod?AT4 x N m D. What is the magnitude of the net (or total) torque on the rod? N m Bob has a mass of 48.0 kg, and Nora has a mass of 39.3 kg. The two partners in physics approach a see-saw. The board of the see-saw has a uniform mass density, and a length of 3.8 m Bob and Nora sit down on opposite sides of the see-saw, each 1.60 m away from the fulcrum, which is at the center of the see-saw board. To start with, Bob and Nora both have their feet on the ground. The figure below shows a whimsical 1950's version of our friends Bob and Nora A. When Bob and Nora lift their feet up, what will happen? (Select One) The see-saw will remain level, .e. horizontal Nora's end of the see-saw will hit the ground, and Bob's end will tilt up in the ai. Bob's end of the see-saw will hit the ground, and Nora's end will tilt upExplanation / Answer
c) T = r F
F1 = 2.70, r1 = 0.3 m
T1 = 0.3 * 2.7 = 0.81 N.m
r2 =0.3 - 0.06 = 0.24 m, F2 = 7.90
T2 = 0.24 * 7.90
T2 = 1.896 N.m
c) Magnitude of net torque = 0.81 + (-1.896)
= 1.086 N.m
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