Answer:
ni sin theta i = nr sin theta r where theta i is the angle of incidence and theta r the angle of refraction
For nr greater than ni (air to water) theta r must be smaller and the image would be bent towards the normal
<span>The solution for this
problem is:</span>
density = mass / volume <span>
7860 = 1 / ((4/3) pi r^3)
r^3 = 1 / (7860 * 4/3*pi)
r = (1 / (7860 * 4/3*pi))^(1/3)</span>
= 0.067 m <span>
Inertia = (2/5)mr^2</span>
= (2/5) x 1 x 0.067^2
= 0.0017956 kg-m^2 <span>
1/2.3 = 0.4348 rev/s
0.4348 x 2pi = 2.732 rad/s
Angular momentum = Inertia x rad/s
0.0017956 x 2.732 = 0.00490557 kg m^2/s</span>
speed of point A is given as
here r = distance from the axis
so here we have
now the distance of point A and B is 0.71 m
while the distance of axis from point A is 0.91 m
so distance of axis from point B will be given as
now for the speed of point b is given as
so end point B will move with speed 37.26 m/s
Try passive assistance if that will work, if not, <span> the definition would be a pilometric contraction.</span>
