Answer:
1/2mv² = ke²
Explanation:
Let's suppose the material in question is a spring with spring constant k, mass m and position k, the kinetic energy possessed by the string will be;
K.E = 1/2mass×velocity² i.e 1/2mv²
Its elastic potential energy will be the work done on the spring when stretched which is equal to 1/2kx²
E.P = 1/2kx²
The equation describing the case where the kinetic energy is twice the elastic potential energy will be;
K.E = 2EP... 1)
Substituting the KE and EP formula into (1), we have;
1/2mv² = 2(1/2ke²)
1/2mv² = ke² which gives the required equation
Answer:
New moment of inertia will be
Explanation:
It is given initially angular velocity 
Moment of inertia 
Angular momentum is equal to 
Now angular velocity is decreases to 
As we know that angular momentum is conserved
So 

So new moment of inertia will be 
Answer:
Explanation:
For entry of light into tube of unknown refractive index
sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube
r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.
sin ( 90 - 25 ) / sin( 90 - C) = μ
sin65 / cos C = μ
sinC = 1.33 / μ , where 1.33 is the refractive index of body liquid.
From these equations
sin65 / cos C = 1.33 / sinC
TanC = 1.33 / sin65
TanC = 1.33 / .9063
TanC = 1.4675
C= 56°
sinC = 1.33 / μ
μ = 1.33 / sinC
= 1.33 / sin56
= 1.33 / .829
μ = 1.6 Ans