Answer:
Step-by-step explanation:
Given
Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height
Suppose they have same mass and radius
time Period is given by
,where h=height of release
a=acceleration
Where I=moment of inertia
a for hoop
a for Uniform solid cylinder
a for spherical shell
a for Uniform Solid
time taken will be inversely proportional to the square root of acceleration
thus first one to reach is Solid Sphere
second is Uniform solid cylinder
third is Spherical Shell
Fourth is hoop
The straight edge of the semicircle is 4 units, as this is this the diameter.
The circumference of the full circle is
C = pi*d = pi*4 = 4pi
which is the distance around the full circle (aka circle's perimeter)
So half of this is 4pi/2 = 2pi
Add this onto the straight edge length to get 4+2pi as the exact distance around the entire semicircle. This includes both straight and curved portions.
If you use the approximation pi = 3.14, then 4+2*pi = 4+2*3.14 = 10.28 is the approximate answer. To get a more accurate answer, use more decimal digits in pi.
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In summary,
exact answer in terms of pi is 4+2pi units
approximate answer is roughly 10.28 units (using pi = 3.14)
Substitute
12(3) + (52)
12(3) = 36
36 + 52 = 88
