The vector position of a 3.45 g particle moving in the xy plane varies in time according to r₁ = (3î + 3ĵ)t + 2ĵt² where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.65 g particle varies as r₂ = 3î − 2ît² – 6ĵt.(a) Determine the vector position (in cm) of the center of mass of the system at t = 2.90 s.7cm(b) Determine the linear momentum (in g. cm/s) of the system at t = 2.90 s.p =cm(c) Determine the velocity (in cm/s) of the center of mass at t = 2.90 s.cm=cmg. cm/s=(d) Determine the acceleration (in cm/s2) of the center of mass at t = 2.90 s.cm/s²cm/s(e) Determine the net force (in µN) exerted on the two-particle system at t = 2.90 s.FnetμN

Name: The vector position of a 3.45 g particle moving in the xy plane varie
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: The vector position of a 3.45 g particle moving in the xy plane varies in time according to r₁ = (3î + 3ĵ)t + 2ĵt² where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.65 g particle varies as r₂ = 3î − 2ît² – 6

According to bartelby guidelines only the first three sub-parts are solved. Kindly repost other…

Science

Physics

ector position of a 3.45 g particle moving in the xy plane varies in time according to r₁ = (3î + 3ĵ)t + 2ĵt² where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.65 g particle varies as r₂ = 3î – 2ît² – 6jt. Determine the vector position (in cm) of the center of mass of the system at t = 2.90 s. cm Determine the linear momentum (in g - cm/s) of the system at t = 2.90 s. 9. cm/s 2. p= Determine the velocity (in cm/s) of the center of mass at t = 2.90 s. 7 Vcm = cm/s

ector position of a 3.45 g particle moving in the xy plane varies in time according to r₁ = (3î + 3ĵ)t + 2ĵt² where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.65 g particle varies as r₂ = 3î – 2ît² – 6jt. Determine the vector position (in cm) of the center of mass of the system at t = 2.90 s. cm Determine the linear momentum (in g - cm/s) of the system at t = 2.90 s. 9. cm/s 2. p= Determine the velocity (in cm/s) of the center of mass at t = 2.90 s. 7 Vcm = cm/s

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)...

See similar textbooks

Similar questions

A 45.0 kg girl is standing on a 137 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.41 m/s relative to the plank. (Let the direction the girl is moving in be positive. Indicate the direction with the sign of your answer.) (a) What is her velocity relative to the surface of ice? m/s(b) What is the velocity of the plank relative to the surface of ice? m/s
One day while out on a space walk to repair your ship, you notice that you have forgotten to tie off your tether! You and the heavy wrench you are carrying are drifting slowly away from the ship with a velocity of V1 = -2.87 m/s (this is negative since you are moving left in the figure). Your mass (including your suit) is M = 106 kg and you are carrying a wrench of mass m = 9.23 kg. You are one smart astronaut and and you throw the wrench in the direction away from your ship as hard as you can. The result is that afterward, your velocity is V2 = 0.64 m/s (this is positive since you are now moving right in the figure). Determine all the following:a) The velocity of the wrench at moment 2: v2 = m/sb) The total kinetic energy of you and the wrench before the throw: KE1 = Joulesc) The total kinetic energy of you and the wrench after the throw: KE2 = Joulesd) Where did all the extra energy come from at moment #2? (multiple choice) 1. Lowering the gravitational potential energy of the…
A 1.90-kg particle has a velocity (2.20 1 - 4.00 ĵ) m/s, and a 4.00-kg particle has a velocity (1.00 +8.00 ĵ) m/s. (a) Find the velocity of the center of mass. Î + m/s (b) Find the total momentum of the system. Î + J kg m/s
A system of four particles moves along one dimension. The center of mass of the system is at rest, and the particles do not interact with any objects outside of the system. Find the velocity v4 of particle 4 at t1 = 2.99 s given the details for the motion of particles 1, 2, and 3. particle 1: m¡ = 1.21 kg, v1 (t) = (5.87 m/s) + (0.255 m/s?) × t particle 2: m2 = 2.69 kg, v2 (t) = (9.59 m/s) + (0.609 m/s² ) x t particle 3: m3 = 4.37 kg, V3 (t) = (7.87 m/s) + (0.153 m/s²) × t particle 4: m4 = 4.91 kg Ua = m/s
A 45.0-kg girl is standing on a 139-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.47 m/s to the right relative to the plank. (Let the direction the girl is moving in be positive. Indicate the direction with the sign of your answer.) (a) What is her velocity relative to the surface of ice? = ____ m/s (b) What is the velocity of the plank relative to the surface of ice? = _____ m/s
Mass MAMAM_A = 33 kg and massMB= 30 kg, They have velocities (in m/s) ,v→A = 12 i^ - 20 j^ and v→B = - 20 i^ + 13 j^. Part A Determine the velocity of the center of mass of the system.
In the figure, two particles are launched from the origin of the coordinate system at time t - O. Particle 1 of mass m₁ - 5.20 g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed of 10.7 m/s. Particle 2 of mass m₂ -4.50 g is shot with a velocity of magnitude 15.6 m/s, at an upward angle such that it always stays directly above particle 1 during its flight. (a) What is the maximum height Hmax reached by the com of the two-particle system? In unit-vector notation, what are the (b) velocity and (c) acceleration of the com when the com reaches Hmax? (a) Number 3.05 (b) Number 10.7 (c) Number 0 Units i+ 0 m 3.3 j Units m/s j Units m/s^2
In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocity of +3.77 m/s, grabs him. The tackler holds onto the receiver, and the two move off together with a velocity of +1.82 m/s. The mass of the tackler is 94.1 kg. Assuming that momentum is conserved, find the mass of the receiver. Number i Units Before collision After collision
A system composed of the following particles: a 0.500 kg particle at the origin; a 1.25 kgkg particle at x = 0.150 mm, y = 0.200 mm; and a 0.750 kg particle at x = 0.200 mm, y = -0.800 mm. a. Find the x-coordinate of the center of mass of the system. b. Find the y-coordinate of the center of mass of the system.
Please Asap
A 0.346 kg puck, initially at rest on a hor- izontal, frictionless surface, is struck by a 0.275 kg puck moving initially along the r axis with a speed of 2.01 m/s. After the colli- sion, the 0.275 kg puck has a speed of 1.01 m/s at an angle of 37° to the positive x axis. Determine the magnitude of the velocity of the 0.346 kg puck after the collision. Answer in units of m/s. Answer in units of m/s. 1. 1.3422 2. 1.21901 3. 0.347028 4. 1.14264 5. 1.07153 6. 0.637653 7. 0.328271 8. 0.591588 9. 0.551494 10. 1.00165
A system of four particles moves along one dimension. The center of mass of the system is at rest, and the particles do not interact with any objects outside of the system. Find the velocity v4 of particle 4 at t1 = 2.75 s given the details for the motion of particles 1, 2, and 3. particle 1: mj = 1.33 kg, vi (t) = (7.41 m/s) + (0.277 m/s²) × t particle 2: m2 = 3.05 kg, v2 (t) = (8.05 m/s) + (0.567 m/s²) × t particle 3: m3 = 4.25 kg, v3 (t) = (6.33 m/s) + (0.195 m/s²) × t particle 4: m4 = 5.27 kg