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Elanso [62]
3 years ago
14

Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from a

bout 20.0 Hz to about 20.0 kHz.a)If you were to mark the beginning of each complete wave pattern with a red dot for the long-wavelength sound, how far apart would the red dots be?b)If you were to mark the beginning of each complete wave pattern with a blue dot for the short-wavelength sound, how far apart would the blue dots be?c)In reality would adjacent red dots be far enough apart for you to easily measure their separation with a meterstick?d)In reality would adjacent blue dots be far enough apart for you to easily measure their separation with a meterstick?e)Suppose you repeated part A in water, where sound travels at 1480 {\rm{ m/s}}. How far apart would the red dots be ?f)Suppose you repeated part A in water, where sound travels at 1480 {\rm{ m/s}}. How far apart would the blue dots be ?g)Could you readily measure their separation with a meterstick?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
8 0

Answer:

Check the explanation

Explanation:

The beat frequency is

df = f2 - f1

the wavelength is

lamda1 = (v/f1)

and lamda2 = (v/f2)

where v = 340 m/s,f1 = 25.0 kHz and f2 = 20.0 kHz

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one-half of A'

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An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object's position from the top of the
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Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec =  78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

<em>u=0m/s   t=1 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

<em>u=0m/s   t=2 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

<em>u=0m/s   t=3 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

<em>u=0m/s   t=4 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

<em>u=0m/s   t=5 s  a=9.8m/s²</em>

s = ut + (1/2)at²

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Linear and rotational Kinetic Energy + Gravitational potential energy

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The ball rolls off a tall roof and starts falling.

Let us first consider the potential energy or more specifically gravitational potential energy (mgh; m = mass of the ball, g = acceleration due to gravity, h = height of the roof). This energy comes because someone or something had to do work to take the ball to the top of the roof against the force of gravity. The potential energy is naturally maximum at the top and minimum when the ball finally reaches the ground.

Now, the ball starts to roll and falls off the roof. It shall continue rotating because of inertia (Newton's first law). This contributes to the rotational kinetic energy (\frac{1}{2}I\omega^2; I=moment of inertia of the ball & \omega = angular velocity).

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