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Mamont248 [21]
2 years ago
8

If you increase the elasticity of a material, what happens to the speed of a mechanical wave going through it?

Physics
1 answer:
Sophie [7]2 years ago
5 0

Answer:

At the molecular level, materials are held together by bonds, which act like springs for small displacements from the equilibrium spacing between neighboring atoms. Push the atoms close, the bond pushes back to keep them apart. Pull them apart, the bond pulls the atoms closer. For those small displacements, it acts like a spring

The speed of the wave will be related to the stiffness of of those springs - you compress the material - how quickly do all of those little springs rebound and push their neighboring atoms away, sending that wave of compression through the material.

Explanation:

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denis-greek [22]

Potential and kinetic energy both decrease with the acorn's falling potential and kinetic energy.

The acorn's potential energy is at its peak when it reaches the top of the tree, yet its kinetic energy is zero (i.e., it is not accelerating).

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The Doppler effect occurs when a source of sound moves
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The Doppler effect occurs when a source of sound or light
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Oksanka [162]

Answer:

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6 0
2 years ago
A three branch parallel circuit has resistors of 27 W, 56 W, and 15 W. What is the total resistance?
Mekhanik [1.2K]

Answer:8.2 Ω

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6 0
2 years ago
Water (rhoH20 = 1000.0 kg/m3 ) flows through a garden hose that goes up a step 20.0 cm high. The cross-sectional area of the hos
Soloha48 [4]

Answer:

 P₂ = 138.88 10³ Pa

Explanation:

This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation

Let's start by looking for the top speed

        Q = A₁ v₁ = A₂ v₂

We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)

         v₂ = A₁ / A₂ v₁

They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)

         v₂ = 2  1.20

         v₂ = 2.40 m / s

Now let's write the Bernoulli equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂

Let's clear the pressure at point 2

       P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)

we put our reference system at the lowest point

        y₁ - y₂ = -20 cm

Let's calculate

       P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)

       P₂ = 143 103 - 2,160 103 - 1,960 103

       P₂ = 138.88 10³ Pa

3 0
3 years ago
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