Answer:
Kinetic energy of the system = 2547.41 Joules.
Explanation:
Given:
Disk:
Mass of the disk (m) = kg
Radius of the disk (r) = cm = m
Cylinder:
Mass of the annular cylinder (M) = kg
Inner radius of the cylinder = m
Outer radius of the cylinder = m
The angular speed of the system = rev/s
Angular speed in in terms of Rad/sec = rad/sec
Formula to be used:
Rotational Kinetic energy, =
So, before that we have to work with the moment of inertia (MOI) of the system.
⇒ MOI of the system = MOI of the disk + MOI of the cylinder
⇒ MOI (system) =
⇒ MOI (system) =
⇒ MOI (system) = kg.m^2
Now
The rotational Kinetic energy.
⇒
Plugging the values.
⇒
⇒ Joules
Then
The kinetic energy of the rotational system is 2547.41 J.
Because it also can be apart of the milky way galaxy
Answer:
Work done in all the three cases will be the same.
Explanation:
1) The free falling body has only one force acting on it, the gravitational force. The work done on the body = mgH (Gravitational potential energy)
2) There are two forces acting on the body going down on a frictionless inclined plane - gravity and the normal force. The gravitational potential energy will be the same. The work done due to the normal force is zero, since the direction of the force is perpendicular to the displacement. Hence, total work done on the body = mgH
3) In the case of the body swinging on the end of a string, the change in gravitational potential enrgy will once again be the same since difference in height is H. The additional force on the body is the tension due to the string. But the work done due to this force is <em>zero, </em>since the displacement of the body is perpendicular to the tension. Therefore, the total work done on the body is once again mgH.
Answer: 23 electrons.
Explanation: an atom have equal number of protons and electrons.
Answer:
Explanation:
Let the slit width be a , slit separation be d and screen distance be D . Let the wavelength of light used be λ .
For diffraction through single slit
position of first minima from central fringe
x = λ (D / a)
For interference pattern
position of 5 th maxima from central fringe
x = 5 λ (D / d )
Since these positions are equal
λ (D / a) = 5 λ (D / d )
5 a = d
a / d = 1 / 5 = . 2