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:
Explanation:
Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,
Given that the uncertainty in x is 0.1 mm.
Therefore,
Therefore, uncertainty in the transverse momentum of photon is
Answer: Friction
Explanation:
Friction and the normal force would be the two initial forces to overcome.
Answer:
17.565 kgm/s
Explanation:
Momentum = mass × velocity
I = mv..................... Equation 1
But we can calculate the value of v using the equation of motion under gravity.
v² = u²+2gs............. Equation 2
Where u = initial velocity, s = maximum heigth, g = acceleration due to gravity.
Given: u = 0 m/s (at the maximum heigth), s = 7.0 m.
Constant: g = 9.8 m/s²
Substitute these values into equation 2
v² = 0²+ 2×7×9.8
v² = 137.2
v = √137.2
v = 11.71 m/s.
Also given: m = 1.50 kg
substitute these values into equation 1
Therefore,
I = 1.5×11.71
I = 17.565 kgm/s
Answer:
b
Explanation:
because a quark have many kinds of flavor
