Question
PLEASE HELP!!! GUARANTEE THE POINTS
Need solutions, answers are provided
1) A particle can only move in the xy plane. At t = 0, it starts at the origin with an initial velocity given by the vector ?3m/s , 4 m/s . The particle accelerates with the components 0 , 2 m/s .
(A) What is the particle
1) A particle can only move in the xy plane. At t = 0, it starts at the origin with an initial velocity given by the vector ?3m/s , 4 m/s . The particle accelerates with the components 0 , 2 m/s . (A) What is the particle???s distance from the origin at t = 2 s? Answer must be 13.4 m. (B) What is the particle???s speed at t = 2 s? Answer must be 8.5 m/s 2) A 2.5-kg block travels in a horizontal circle of radius 0.80 m along a tabletop. It is attached to a light horizontal string fixed at the center of the circle. The block???s speed is initially 6.0 m/s. After completing one full revolution around the circle, the block???s speed is 3.0 m/s. (A) What is the energy lost to friction after the block has moved one revolution? Answer must be -33.75 J (B) What is the coefficient of kinetic friction between the block and the tabletop? Answer must be 0.27 (C) How many revolutions does the block make in coming to rest given its initial speed above? Answer must be 1/3 3) A 4.0-kg bomb is being tested by a demolition company. The bomb hangs initially at rest by a light string from a very tall tower. At 12 noon, it is detonated. The bomb fragments into three pieces that fly in three different directions (see diagram). Pieces A and B are each 1.0 kg; piece C is 2.0 kg. Ignore gravity. (A) What is the magnitude of piece C???s velocity? Answer is 46.4 m/s (B) What is the direction of piece C???s velocity? Express your answer as an angle to piece B???s velocity. Answer is 98?degree
Explanation / Answer
1) x = ut + at^2 / 2
x = ( -3 x2 +0) , (4x2 +2x2^2/2) = -6 , 12 m
d = sqrt(6^2 + 12^2) = 13.42 m
b) v = u + at
v = (-3 + 0) , ( 4 + 2x2) = -3 , 8 m/s
speed = sqrt(3^2 + 8^2) = 8.54 m/s
2) energy lost = change in K.E. = 2.5 x (3^2 - 6^2) /2
= - 33.75 J
b) energy lost = work done bt friction
33.75 = u.mg x 2pir
u = 33.75 / (2.5 x 9.8 x 2 x pi x 0.80) = 0.27
c) initial total energy = 2.5 x 6^2 /2 = 45 J
n x 33.75 = 45
n = 1.33 rev
3.
momentum conservation,
1 x 90i + 1 x (120cos130i + 120sin130j) + 2(vcosQi - vsinQ j) = 0
91.93 = 2vcosQ
12.87 = 2vsinQ
Q = 90 + tan-1(12.87/91.93) = 98 degree
V = 46.4 m/s
