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:
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First, we would need to know the decaying isotope.
Next, we use the decay formula
A = Ao e^(-kt)
After determining the remaining amount after two hours, the decay reaction can be used to determine the number of gamma rays released. If the given is in terms of mole, then the total energy is
E = 140n KeV where n is the number of moles of gamma rays released
Answer:
691.13 nm
Explanation:
d = width of the slit = 0.11 x 10⁻³ m
θ = angle of diffraction pattern = 0.72° degree
λ = wavelength of the light = ?
m = order = 2 (since second minimum)
for the second minimum diffraction pattern we use the equation
d Sinθ = m λ
Inserting the values
(0.11 x 10⁻³) Sin0.72 = (2) λ
λ = 691.13 x 10⁻⁹ m
λ = 691.13 nm
What is the longest the bolt can be and still be acceptable
