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)
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The maximum height attained is 460 m.
<h3>What is the maximum height?</h3>
We know that the final velocity of a body is 0 m/s at the maximum height which is the greatest height that is attained by the body. We now use the formula;
v^2 = u^2 -2gh
Given that v = 0 m/s
u^2 = 2gh
h = u^2/2g
v = final velocity
u = initial velocity
h = maximin height
g = acceleration due to gravity
h = (95)^2/2 * 9.8
h = 460 m
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Answer:
1/i + 1/o = 1/f thin lens equation
i = 33 * 8.9 / (33 - 8.9) = 12.2 cm to right of first lens
27 - 12.2 = 14.8 cm to left of second lens
i = 14.8 * 8.9 / (14.8 - 8.9) = 22,3 cm to right of second lens