1) Force F =(-8.00 N){+(6.00 N) j acts on a particle with position vector r =...
Force F = (-5.3 N) ? + (3.2 N); acts on a particle with position vector 7 = (2.4 m) î + (3.9 m)). what are (a) the magnitude of the torque on the particle about the origin and (b) the angle between the directions of 7 and 7 (a) Number Units (b) Number Units
A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector r Overscript right-arrow EndScripts = (2.00 m) i Overscript ̂ EndScripts - (3.00 m) j Overscript ̂ EndScripts + (2.00 m) k Overscript ̂ EndScripts, the force is Upper F Overscript right-arrow EndScripts = Fx i Overscript ̂ EndScripts + (7.00 N) j Overscript ̂ EndScripts - (6.00 N) k Overscript ̂ EndScripts, and the corresponding torque about...
A particle moves through an xyz coordinate system while a force acts on the particle. When the particle has the position vector r = (2.00 m) - (3.00 m) + (2.00 m) , the force is F = Fx + (7.00 N) - (6.00 N) , and the corresponding torque about the origin is T = (4.00 N·m) + (10.0 N·m) + (11.0 N·m) . Determine Fx. We were unable to transcribe this imageWe were unable to transcribe this imageWe...
Force F = 1.99 N i -3.18 N k acts on a pebble with position vector T = 0.48 m J -2.20 m k relative to the origin What is the resulting torque on the pebble about the origin? Enter i, j, and k component. What is the resulting torque on the pebble about the point (2.12 m, 0, -3.32 m)? Enter i, j, and k component.
Force = 2.11 N i -3.19 N k acts on a pebble with position vector = 0.51 m j -2.12 m k relative to the origin. What is the resulting torque on the pebble about the origin? Enter i, j, and k component. What is the resulting torque on the pebble about the point (2.02 m, 0, -3.25 m)? Enter i, j, and k component.
At one instant, force F 4.7i N acts on a 0.374 kg object that has position vector (7.80-7.80f) m and velocity vector V - (-1.11+1.11K) m/s. About the origin and in unit- vector notation, what are (a) the object's angular momentum and (b) the torque acting on the object?
Check my work pleas. Last submission +Question Details | Previous Answers For Show Question Details (5.0 N) acts on a particle with position vector r (-1.0 m) (5.0 m) k. (a) What is the torque on the particle about the origin? -25 N m i+ 5 N m (b) What is the angle between the directions of r and F? (If there is no torque, enter o.) 101.1 (c) What is the angle between the directions of force and torque?...
A particle is located at the vector position r = (2¡ +4j) m and the force acting on it is F = 4¡ + 3j N. What is the torque about the origin ? O (-10.0 Nm)ť (-11.0 Nm)k (-10.0 Nm) i (7.00 Nm) (10.0 Nm)k
6) At one instant, force F 4.0j N acts on a 0.25 kg object that has position vector (2.0-2.0k) m and velocity vector (-50i+5.0k) m/s. About the origin and in unit vector notation, what are (a) the object's angular momentum and (b) the torque acting on the object?
A particle is located at the vector position r = (1¡ + 4ſ) m and the force acting on it is f = 4ị + 2ị N. What is the torque about the origin ? (-14.0 Nm) (14.0 Nm) (-14.0 Nm), (-7.00 Nm)k (-14.0 Nm)i