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daser333 [38]
4 years ago
12

Explain what happens to the motion of a particle as a wave passes through a medium. How is the motion of the particles like the

motion of a mass on a spring
Physics
1 answer:
bezimeni [28]4 years ago
3 0
  1. As they vibrate, they pass the disturbance of energy to the particles next to them, which passes the energy to the particles next to them, and so on.
  2. Energy travels down the spring and reflected like the motion of a mass on a spring.  

<u>Explanation</u>:

  • Particle motion is the superposition of a large component due to the fluid drag toward the filter barrier and a random component due to Brownian motion.
  • The particles of the medium just vibrate in place. As they vibrate, they pass the unsettling influence of energy to the particles beside them, which passes the energy to the particles by them, etc.  
  • Energy goes down the spring and reflected like the movement of a mass on a spring.
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Given :

Height from which ball is dropped , h = 40 m .

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Velocity when ball covered 20 m and velocity when it hit the ground .

Solution :

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By equation of motion :

v^2=u^2+2gh\\\\v=\sqrt{2\times 10\times 20}\ m/s\\\\v=20\ m/s

Now , distance covered when body reaches ground is , 40 m .

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v=\sqrt{2\times 10\times 40}\ m/s\\\\v=20\sqrt{2}\ m/s

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A woman can lift barrels a vertical distance of 1 meter or can roll them up a 2-meter long ramp to the same elevation. If she us
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travels along a perfectly flat (no banking) circular track of radius 532 m. The car increases its speed at uniform rate of at ≡
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Answer:

<em>The coefficient of static friction = 0.392</em>

Explanation:

When the tire starts to skirt, The net force on the car is equal to the maximum friction.

F = μR................ Equation 1

R = mg.............. Equation 2

F = ma ............ Equation 3

Substituting equation 2 and 3 into Equation 1

ma = μmg.

a = μg

μ = a/g................................... Equation 4

Where F = frictional Force, μ = Coefficient of frictional force, R = normal reaction. a = total acceleration of the car. g = acceleration due to gravity.

<em>Note: The tangential acceleration  is perpendicular to the radius. The total acceleration is</em>

a = √(a₁² + a₂²)........................ Equation 4

Where a₁ = tangential acceleration of the tire, a₂ = centripetal acceleration of the car.

a₁ = 2.96 m/s²

a₂ = v²/R..................... Equation 5

Where v = speed of the car = 36.1 m/s, R = radius of the circular part traveled by the car = 532 m

Substituting into equation 5

a₂ = 36.1²/532

a₂ = 1303.21/532

a₂ = 2.45 m/s².

Therefore,

a = √(2.45²+2.96²)

a = √(6.0025 + 8.7612)

a = √14.764

a = 3.84 m/s².

Substituting into equation 4

μ = 3.84/9.8

<em>μ = 0.392.</em>

<em>Thus the coefficient of static friction = 0.392</em>

<em></em>

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