Answer:
a)R= sqrt( wt³/12wt)
b)R=sqrt(tw³/12wt)
c)R= sqrt ( wt³/12xcos45xwt)
Explanation:
Thickness = t
Width = w
Length od diagonal =sqrt (t² +w²)
Area of raectangle = A= tW
Radius of gyration= r= sqrt( I/A)
a)
Moment of inertia in the direction of thickness I = w t³/12
R= sqrt( wt³/12wt)
b)
Moment of inertia in the direction of width I = t w³/12
R=sqrt(tw³/12wt)
c)
Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)
R= sqrt ( wt³/12xcos45xwt)
Answer
given,
a = 2 t - 10
velocity function
we know,


integrating both side

v = t² - 10 t + C
at t = 0 v = 3
so, 3 = 0 - 0 + C
C = 3
Velocity function is equal to v = t² - 10 t + 3
Again we know,


integrating both side


now, at t= 0 s = -4

C = -4
So,

Position function is equal to 
Find the solution in the attachments
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