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
Answer:
C. The decrease in speed as the wave approaches shore.
Explanation:
The waves break when approaching the shore because the depth decreases. Thus, the wave travels more slowly and increases its height. There comes a time when the part of the wave on the surface travels faster than the one that travels under water, the ridge destabilizes and falls against the ground.
Answer:
20.7 s
Explanation:
The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:
v = a*t + v₀
Solve for t:
t = (v - v₀)/a
Wave speed = (wavelength) x (frequency)
= (1.5 m) x (500 / sec)
= 750 m/s .
