The center of mass of Faith\'s atmosphere is a little less than halfway between
ID: 1414206 • Letter: T
Question
The center of mass of Faith's atmosphere is a little less than halfway between the Faith's surface and the boundary of the atmosphere near the surface of the Faith near the outer boundary of the atmosphere near the center of the Faith none of the above A 640-N hunter gets a rope around a 3200-N polar bear They are stationary 20 m apart, on frictionless level ice. When the hunter pulls the polar bear to him. the polar bear will move: 1.0 m 3.3 m 10 m 12 m 17 m The center of mass of a system of particles has a constant velocity if the forces exerted by the particles on each other sum to zero the external forces acting on particles of the system sum to zero the velocity of the center of mass is initially zero the particles are distributed symmetrically around the center of mass the center of mass is at the geometric center of the system A light rope passes over a light frictionless pulley attached to the ceiling An object with a large mass is tied to one end and an object with a smaller mass is tied to the other end Starting from rest, the object moves downward, and the lighter object moves upward with an acceleration of the same magnitude 'Much of the following statements is true for the system consisting of the two objects' The center of mass remains at rest The net external force is zero The velocity of the center of mass is a constant The acceleration of the center of mass is g. downward None of the above statements are true The position of the center of mass of a system of particles moves as x = 4.5t + 2.4 t^2 + 1.1 t^3, where r b in meters If the system starts from rest at t = 0. What is its velocity at t = 3.0 s? 10 m/s 21 m/s 49 m/s 63 m/sExplanation / Answer
Answer 1) D Near the center of earth
Answer 2) B
Given m1 = 640N , m2 =3200N , X1 = 0 , X2 =20m
Hunter polar bear represent an isolated system and initially they are at rest their center of mass will be at rest.
X = (m1x1 + m2x2) / m1 + m2 = (m1g1x +m2g2x) / m1g +m2g
= (640*0 + 3200*20) / (640+ 3200) =16.7m
so now The 3200 N polar bear moves 20-16.7= 3.3m
Answer 3) B The external forces acting on particles of the system sum to zero
Answer 4) E None of the above statements are true N
Answer 5) C
The velocity is dx/dt = 4.5 +4.8t+3.3t2
The maximum occurs when dv/dt = 4.8+6.6t =0 , from which t = 0.727sec
Velocity at t= 3.0s so , [v = 4.5 + 4.8t+3.3t2 ] t=3.0sec = 48.6 m/s = 49m/s
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