Answer:
Ix = Iy =
Radius of gyration x = y = 
Step-by-step explanation:
Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.
Mass of disk = ρπR2
Moment of inertia about its perpendicular axis is
. Moment of inertia of quarter disk about its perpendicular is
.
Now using perpendicular axis theorem, Ix = Iy =
=
.
For Radius of gyration K, equate MK2 = MR2/16, K= R/4.
Answer: 13
Step-by-step explanation:
Answer:
that one is A but im not sure about the other ones
Step-by-step explanation:
Answer:
Step-by-step explanation:
Step-by-step explanation:
sol;
x+1=y...(1)
3y-7=2x....(2)
or, 3(x+1)-72x [from (1)]
or, 3x+3-7=2x
or, 3x-2x=7-3
x=4
now,
putting the value of x in (1)
y=x+1
=4+1
=5
PR and SQ are the diagonal of PQRS.