Moment of inertia of single particle rotating in circle is I1 = 1/2 (m*r^2)
The value of the moment of inertia when the person is on the edge of the merry-go-round is I2=1/3 (m*L^2)
Moment of Inertia refers to:
- the quantity expressed by the body resisting angular acceleration.
- It the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.
The moment of inertia of single particle rotating in a circle I1 = 1/2 (m*r^2)
here We note that the,
In the formula, r being the distance from the point particle to the axis of rotation and m being the mass of disk.
The value of the moment of inertia when the person is on the edge of the merry-go-round is determined with parallel-axis theorem:
I(edge) = I (center of mass) + md^2
d be the distance from an axis through the object’s center of mass to a new axis.
I2(edge) = 1/3 (m*L^2)
learn more about moment of Inertia here:
<u>brainly.com/question/14226368</u>
#SPJ4
Sorry!
This cannot be answered. We don't have weight, height, etc.
Physical changes: melting, evaporating, and condensation. This is a physical change.
The formula is:
Work = Force · Displacement
F = m · g
F = 16 kg · 9.8 m/s² = 156.8 N
and we know that:
d = 0.8 m
W = 156.8 N · 0.8 m = 125.44 J
Answer:
W = 125.44 J.
Answer:
m = 14*26 = 364
Explanation:
overall magnification is given as m
mo magnification of objective lens
me magnification of EYE lens
where mo is given as
and me as
d is distant of distinct vision = 25.0 cm for normal eye
fe = focal length of eye piece
focal length of objective lense is 0.140 cm
we know that
m = 14*26 = 364
