Please answer this. I will surely upvote!!! Torque due to Unequal ForcesTorqueHanging LoadPosition onMomentMass (g) Weight (N) | Meterstick (cm) | Arm (cm) | Magnitude (N·m) | Direction (CW/CCW)CCWm₁(A)T1(B)(C)(E)(G)m₂(D)(F)x2Percent Difference0.00%Center of Gravity: 50 cmRefer to the table above. What mass m₂ should be placed at x₂=91.6 cm in order to balance a torque of 0.1776 N·m with counterclockwise rotation due to m₁?O 43.6 gO 216gO 19.8 gO 427 g

Torque is the cross-product of position vector and force vector. τ→=r→×F→

Torque due to Unequal Forces Torque Position on Moment Hanging Load Mass (g) Weight (N) Meterstick (cm) | Arm (cm) | Magnitude (N-m) | Direction (CW/CCW) CCW T₁ m₁ (A) (B) (C) (D) (E) (G) m₂ x2 (F) Percent Difference 0.00% Center of Gravity: 50 cm Refer to the table above. What mass m₂ should be placed at x₂=91.6 cm in order to balance a torque of 0.1776 N-m with counterclockwise rotation due to m₁? O 43.6 g O 216g O 19.8 g O 427 g

Torque due to Unequal Forces Torque Position on Moment Hanging Load Mass (g) Weight (N) Meterstick (cm) | Arm (cm) | Magnitude (N-m) | Direction (CW/CCW) CCW T₁ m₁ (A) (B) (C) (D) (E) (G) m₂ x2 (F) Percent Difference 0.00% Center of Gravity: 50 cm Refer to the table above. What mass m₂ should be placed at x₂=91.6 cm in order to balance a torque of 0.1776 N-m with counterclockwise rotation due to m₁? O 43.6 g O 216g O 19.8 g O 427 g

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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