Answer:
the answer is 1, oil and water
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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Answer:D.no of the above
Explanation:get right with Christ
Answer:
a.
b. 
Explanation:
<u>Given:</u>
- Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .
<h2>
(a):</h2>
The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

At time t = 3 seconds,

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>
<h2>
(b):</h2>
The velocity of the particle at some is defined as the rate of change of the position of the particle.

For the time interval of 2 seconds,

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

It is the displacement of the particle in 2 seconds.
Answer:
The center of mass of the two-ball system is 7.05 m above ground.
Explanation:
<u>Motion of 0.50 kg ball:</u>
Initial speed, u = 0 m/s
Time = 2 s
Acceleration = 9.81 m/s²
Initial height = 25 m
Substituting in equation s = ut + 0.5 at²
s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m
Height above ground = 25 - 19.62 = 5.38 m
<u>Motion of 0.25 kg ball:</u>
Initial speed, u = 15 m/s
Time = 2 s
Acceleration = -9.81 m/s²
Substituting in equation s = ut + 0.5 at²
s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m
Height above ground = 10.38 m
We have equation for center of gravity

m₁ = 0.50 kg
x₁ = 5.38 m
m₂ = 0.25 kg
x₂ = 10.38 m
Substituting

The center of mass of the two-ball system is 7.05 m above ground.