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Answer:
= ( ρ_fluid g A) y
Explanation:
This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force
for the first part, let's write Newton's equilibrium equation
B₀ - W = 0
B₀ = W
ρ_fluid g V_fluid = W
the volume of the fluid is the area of the cube times the height it is submerged
V_fluid = A y
For the second part, the body introduces a quantity and below this equilibrium point, the equation is
B - W = m a
ρ_fluid g A (y₀ + y) - W = m a
ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a
ρ_fluid g A y + (B₀-W) = ma
the part in parentheses is zero since it is the force when it is in equilibrium
ρ_fluid g A y = m a
this equation the net force is
= ( ρ_fluid g A) y
we can see that this force varies linearly the distance and measured from the equilibrium position
A)acceleration is in the direction of motion
The period of the block's mass is changed by a factor of √2 when the mass of the block was doubled.
The time period T of the block with mass M attached to a spring of spring constant K is given by,
T = 2π(√M/K).
Let us say that, when we increased the mass to 2M, the time periods of the block became T', the spring constant is not changed, so, we can write,
T' = 2π(√2M/K)
Putting T = 2π(√M/K) above,
T' =√2T
So, here we can see, if the mass is doubled from it's initial value. The time period of the mass will be changed by a factor of √2.
To know more about time period of mass, visit,
brainly.com/question/20629494
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The horizontal change between two points on a graph is called the 'run'.
The vertical change between two points is called the 'rise'.