Answer:
(a) 
(b) 
Explanation:
Given data
The angular velocity of two cylinders ω=257 rad/s
The mass of the two cylinders m=2.88 kg
The radius of small cylinder r₁=0.319 m
The radius of larger cylinder r₂=0.605 m
For Part (a)
The rotational kinetic energy of the cylinder is given by:

Where I is rotational of inertia of solid cylinder about its central axis.
So
 
Substitute the given values
So

For Part (b)

Substitute the given values

 
        
             
        
        
        
Answer:
Part a)

Part b)

Explanation:
As we know that by parallel axis theorem we will have

Part a)
here we know that for a stick the moment of inertia for an axis passing through its COM is given as

now if we need to find the inertia from its end then we will have



Part b)
here we know that for a cube the moment of inertia for an axis passing through its COM is given as

now if we need to find the inertia about an axis passing through its edge



 
        
             
        
        
        
Answer:
<h2>154.73N</h2>
Explanation:
The question is incomplete. Here is the complete question.
Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.
Check the diagram related to the question in the attachment below for better understanding.
The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.
The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).
Ty = 15sin31°
Ty = 7.73N
W = mass * acceleration due to gravity
W = 15.0*9.8
W = 147N
The normal force is therefore expressed as;
N = Ty + W
N = 7.73 + 147
N = 154.73N
 
        
             
        
        
        
Really long we’ll not long but far in distance
        
             
        
        
        
Answer:
Explanation:
wave length of light λ = 623 x 10⁻⁹ m . 
Distance of screen D = 76.5 x 10⁻² m 
width of slit        =      d 
Distance on the screen between the second order minimum and the central maximum       =  2  λ D / d
1.11 x 10⁻²  = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d
d =  ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²
= 85872.97 x  10⁻⁹
=  85.87297 x  10⁻⁶
= 85.87 μm
width a of the slit is = 85.87 μm