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boyakko [2]
3 years ago
9

Why do you think fixed boundaries ""flip"" waves and loose boundaries do not?

Physics
1 answer:
Vinil7 [7]3 years ago
5 0

Answer:

When the obstacle is fixed, the law of action and reaction, makes the reflected wave is inverted.

When the obstacle is mobile, he mobile point, it moves in the direction of the wave, therefore there is no inversion of it.

Explanation:

Waves when they reach an obstacle behave like a shock, therefore if we use the conservation of momentum the wave must reverse its speed, this explains that the speed changes sign, the wave is reflected.

When the obstacle is fixed, the wave when it reaches the obstacle exerts a force on the point, by the law of action and reaction the point exerts on the wave a force of equal magnitude but in the opposite direction, this reaction force which makes the reflected wave is inverted.

When the obstacle is mobile, this is without friction, when the wave arrives it exerts a force on the mobile point, it moves in the direction of the wave, reaching the maximum amplitude of the incident wave, when it is reflected the point begins to go down along with the wave, therefore there is no inversion of it.

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A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A
Kruka [31]

Answer:

\omega_f = 585.37 \ rev/s

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)=  800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

I \omega_i = I' \omega_f

(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f

M\times 800 = ( M + m )\omega_f

3\times 800 = (3+1.1)\times \omega_f

2400 = (4.1)\times \omega_f

\omega_f = 585.37 \ rev/s

3 0
3 years ago
A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the
Serjik [45]
The elastic potential energy of a spring is given by
U= \frac{1}{2}kx^2
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The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
W=-\Delta U =  \frac{1}{2}kx_i^2 -  \frac{1}{2}kx_f^2

In our problem, initially the spring is uncompressed, so x_i=0. Therefore, the work done by the spring when it is compressed until x_f is
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8 0
3 years ago
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
yuradex [85]

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

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we substitute

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         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

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we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

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