Answer:
The fraction of kinetic energy lost in the collision is 0.25 .
Explanation:
We know , momentum will be conserved along x direction .
Therefore ,
Now , fraction lost in kinetic energy is :
Therefore , the fraction of kinetic energy lost in the collision is 0.25 .
Answer:
The angular momentum of the system is 0.18kgm²/s
Explanation:
L = angular momentum of the particle 1 + angular momentum of the particle 2
L = ml²ω + ml²ω
L = 2ml²ω
Given that,
m = 500g are connected by a massless cord
=0.500kg
l = 30cm
= 0.30m
angular speed ω = 2.0 rev/s
angular momentum of the system =
L = 2ml²ω
L = 2 × 0.500 × (0.30)² × 2
L = 2 × 0.500 × 0.09 × 2
L = 0.18kgm²/s
Thus, the angular momentum of the system is 0.18kgm²/s
Spiral Galaxies, Irregular Galaxies and Elliptical Galaxies?
The velocity of the object at the end of 3.00 seconds is 29.4 m/s.
Find the initial velocity of an object by dividing the time it took the object to travel the specified distance by the total distance. An object at rest has zero velocity and remains so. Such an object does not change its state of motion unless an unbalanced force acts on it. The velocity of a body at rest must be zero, but the acceleration of a body at rest obviously cannot be zero.
Particles thrown vertically upwards will momentarily stop at the highest point of their movement. In free fall the object is affected only by gravitational acceleration. Acceleration is a non-zero value, so the velocity changes and increases as the object falls. This is because acceleration is defined as the rate of change of velocity with respect to time.
Learn more about The velocity here:-brainly.com/question/25749514
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<span>Answer:
First we need to find the acceleration.
torque on cylinder Ď„ = T * r where T is the string tension;
T = m(g - a) where a is the acceleration of the cylinder. Then
Ď„ = m(g - a)r
But also τ = Iα. For a solid cylinder, I = ½mr²,
and if the string doesn't slip, then α = a / r, so
τ = ½mr² * a/r = ½mra.
Since Ď„ = Ď„, we have
m(g - a)r = ½mra → m, r cancel, leaving
g - a = ½a
g = 3a/2
a = 2g/3 where g, of course, is gravitational acceleration.
We know that v(t) = a*t, so for our cylinder
v(t) = 2gt / 3 â—„ linear velocity
and ω = v(t) / r = 2gt / 3r ◄ angular velocity</span>