The time period resulting in oscillations will be 1.986 seconds.
<h3>What is the period of oscillation?</h3>
The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.
The time period of the oscillation is;

Hence the time period resulting oscillations will be 1.986 seconds.
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A: particles are more spread out in gas
Explanation:
According to the law of conservation of energy
,
Potential energy = kinetic energy
I =
mgh =
v = 7.4 m/s
thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y =
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m