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mixas84 [53]
3 years ago
15

As a rock sinks deeper and deeper into water of constant density, what happens to the buoyant force on it? As a rock sinks deepe

r and deeper into water of constant density, what happens to the buoyant force on it? It increases. It may increase or decrease, depending on the shape of the rock. It may increase or decrease, depending on the weight of the rock. It remains constant. It decreases. g
Physics
1 answer:
Sidana [21]3 years ago
3 0

Answer:

It remains constant

Explanation:

As we know that buoyant force on an object given as

Fb = ρ Vd g

ρ= Density of fluid

Vd=Volume displace by body

g=10 m/s²

Fb =buoyant force

So from above we can say that buoyant force does not depends on the depth. It only depends on the fluid density and volume displace by body.

So when rock gets deeper and deeper the buoyant force will remain constant.

It remains constant

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Answer:

Velocity = 4.33[m/s]

Explanation:

The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.

E_{M}=E_{p}  + E_{k} \\E_{p} = potential energy [J]\\E_{k} = kinetic energy [J]\\where:\\E_{p} =m*g*h\\E_{p} = 4*9.81*0.5=19.62[J]\\E_{k}=\frac{1}{2} *m*v^{2}  \\E_{k}=\frac{1}{2} *4*(3)^{2} \\E_{k}=18[J]\\Therefore\\E_{M} =18+19.62\\E_{M}=37.62[J]

All this energy will become kinetic energy and we can find the velocity.

37.62=\frac{1}{2} *m*v^{2} \\v=\sqrt{\frac{37.62*2}{4} } \\v=4.33[m/s]

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Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has
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  h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

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final point. Lower, just before the crash

         Em_f = K = ½ m v_{b}^2

energy is conserved

         Em₀ = Em_f

         m g h = ½ m v²

         v_b = \sqrt{2gh}

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

         p₀ = 2m 0 + m v_b

final instant. Right after the crash

         p_f = (2m + m) v

       

the moment is preserved

         p₀ = p_f

         m v_b = 3m v

         v = v_b / 3

         

          v = ⅓ \sqrt{2gh}

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

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Final point. Higher

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          ½ 3m v² = 3m g h’

           

we substitute

         h’=  \frac{v^2}{2g}

         h’ =  \frac{1}{3^2} \  \frac{ 2gh}{2g}

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