Answer:
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
Explanation:
We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is
w² = mg d / I
In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow
d = L / 2
The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated
I = ¼ m r2 + ⅓ m L2
I = m (¼ r2 + ⅓ L2)
now let's use the concept of density to calculate the mass of the system
ρ = m / V
m = ρ V
the volume of a cylinder is
V = π r² L
m = ρ π r² L
let's substitute
w² = m g (L / 2) / m (¼ r² + ⅓ L²)
w² = g L / (½ r² + 2/3 L²)
L >> r
w = √[g /L (½ r²/L2 + 2/3 ) ]
When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE
<u>Answer:</u> The Young's modulus for the wire is 
<u>Explanation:</u>
Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.
The equation representing Young's Modulus is:
where,
Y = Young's Modulus
F = force exerted by the weight = 
m = mass of the ball = 10 kg
g = acceleration due to gravity = 
l = length of wire = 2.6 m
A = area of cross section = 
r = radius of the wire = 
(Conversion factor: 1 m = 1000 mm)
= change in length = 1.99 mm = 
Putting values in above equation, we get:
Hence, the Young's modulus for the wire is
Answer:
D. a cation that has a smaller radius than the atom.
Explanation:
When electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than the atom.
