C. A 0-kg particle moves in the horizontal xy-plane motion under the action of a net...
A 1.30-kg particle moves in the xy plane with a velocity of V - (3.90 i 3.20j) m/s. Determine the angular momentum of the particle about the origin when its position vector is r-(1.50i2.20 j) m. R)kg m2/s
A 1.80-kg particle moves in the xy plane with a velocity of V (4.10 1 - 3.10 j) m/s. Determine the angular momentum of the particle about the origin when its position vector is r = (1.50 + 2.20 j) m. i + j +
A 1.30-kg particle moves in the xy plane with a velocity of = (4.10 î − 3.80 ĵ) m/s. Determine the angular momentum of the particle about the origin when its position vector is = (1.50 î + 2.20 ĵ) m.
A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point with coordinates (x, y) with a velocity of vi and an acceleration of -aj. In these expressions, v and a are the speed and the magnitude of acceleration, respectively, and thus are positive quantities. In terms of the variables given, what are the coordinates (a) xo and (b) Yo of the center of the circular path? (a) Xo =...
A particle whose mass is 2.0 kg moves in the xy plane with a constant speed of 3.0 m/s along the direction. What is its angular momentum (in kg/m 2 /s) relative to the point (0, 5.0) meters?
A force acting on a particle in the xy plane at coordinates (x, y) is given by vector F = (F_0/r) (y hat i - x hat j), where F_0 is a positive constant and r is the distance of the particle from the origin. Show that the magnitude of this force is F_0. Show that the direction of vector F is perpendicular to vector r = x hat i + y hat j.
The position of a particle of mass m = 0.80 kg as a function of time is given by ⃗r = xˆi + yˆj = (Rsinωt)ˆi + (Rcosωt)ˆj, where R = 4.0 m and ω = 2πs−1. (a) Show that the path of this particle is a circle of radius R, with its center at the origin of the xy plane. (b) Compute the velocity vector. Show that vx/vy = −y/x. (c) Compute the acceleration vector and show that it...
A particle of mass 2 kg is moving in the xy plane at a constant speed of 0.80 m/s in the +x direction along the line y=4m. As the particle travels from ,x= -3m to x= +3m, the magnitude of its angular momentum with respect to the origin is A] not constant B] 0 C} 4.8 kg m2/s D} 8.4 kg m2/s E} 8 kg m2/s
A particle moves in the xy plane with constant acceleration. At time t=0 s, the position vector for the particle is r=9.70mx^+4.30my^. The acceleration is given by the vector a=8.00m/s^2x^+3.90m/s^2y^. The velocity vector at time t=o s is v=2.80m/sx^ - 7.00m/sy^. What is the magnitude of the position vector at time t= 2.10 s? What is the angle between the position vector and the positive x-axis at time t= 2.10 s?
A Particle P In Fig. 12-24, a particle P with mass 1.3 kg has position vector r of magnitude 1.3 m and velocity v of magnitude 4.5 m/s. Figure 12-24 A force F of magnitude 1.5 N acts on the particle. All three vectors lie in the xy plane oriented as shown. (a) About the origin, what is the rotational momentum of the particle? _____kg·m2/s k (b) About the origin, what is the torque acting on the particle? _____N·m k...