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Diano4ka-milaya [45]
3 years ago
5

The figure below shows a circle with segment AB as it's diameter. The center of the circle is NOT known. A square needs to be in

scribed in the circle with A and B as a pair of opposite vertices.
Which step will be used in the construction? PLEASE HELP

A. constructing four arcs on the circles circumference with the compass width as the radius
B. constructing a segment that is tangent to the circle at point A
C. constructing a perpendicular to a line segment through a point not lying on it.
D. constructing a perpendicular bisector of a line segment.​

Mathematics
1 answer:
Firlakuza [10]3 years ago
3 0
It’s B because it is the correct answer
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A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
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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

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

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t_2=k\sqrt{\frac{3}{2}}=1.224k

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second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
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