Answer:
    = ( ρ_fluid g A) y
 = ( 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
 = ( ρ_fluid g A) y
we can see that this force varies linearly the distance and measured from the equilibrium position
 
        
             
        
        
        
Answer:
Explanation:
To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum 
I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω
I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .
I₁ = 1/2 mr²
= .5 x 175 x 2.13² 
= 396.97 kgm²
I₂ = m r²
= 55.4 x 2.13²
= 251.34 mgm²
ω₁ = .651 rev /s 
= .651 x 2π rad /s 
ω₂ = tangential velocity of man / radius of disc
= 3.51 / 2.13
= 1.65 rad/s 
I₁ ω₁ + I₂ ω₂ = ( I₁  + I₂ ) ω
396.97 x  .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω
 ω = 3.14 rad /s 
kinetic energy = 1/2 I ω²
= 3196 J
 
        
             
        
        
        
Answer:
B electrons protons and neutrons
hope i helped...
Explanation:
 
        
                    
             
        
        
        
Answer:
A,B,D,E,F
Explanation:
I took the test for yall.
 
        
             
        
        
        
Answer:
70 cm
Explanation:
0.5 kg at 20 cm
0.3 kg at 60 cm
x = Distance of the third 0.6 kg mass
Meter stick hanging at 50 cm
Torque about the support point is given by (torque is conserved)

The position of the third mass of 0.6 kg is at 20+50 = 70 cm