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AURORKA [14]
3 years ago
9

A tuning fork is sounded above a resonating tube (one end closed), which resonates at a length of 0.20 m and again at 0.60 m. If

the tube length were extended further, at what point will the tuning fork again create a resonance condition?

Physics
1 answer:
Nina [5.8K]3 years ago
6 0

To solve this problem we will apply the concepts related to resonance. The velocity with which sound travels in any medium may be determined if the frequency and the wavelength are known. Since the pipe is closed at one end, that produces frequencies ratio 1:3:5, then

n_1 = \frac{V}{4L_1}

Third resonance occurs at 5n_2

n_2 = \frac{V}{4L_2}

At resonance 5n_2=n_1

5\frac{V}{4L_2} = \frac{V}{4L_1}

L_2 = 5L_1

L_2 = 5*0.2

L_2 = 1m

Therefore at 1m will the tuning fork again create a resonance condition

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<h3>What is law conservation of energy?</h3>

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An object traveling at 1.5 rad
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The object's final velocity, given the data is 10.5 rad/s

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This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

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  • a is the acceleration
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  • u is the initial velocity
  • t is the time

<h3>How to determine the final velocity</h3>

The following data were obtained from the question

  • Initial velocity (u) = 1.5 rad/s
  • Acceleration (a) = 0.75 rad/s²
  • Time (t) = 12 s
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The final velocity can be obtained as follow:

a = (v – u) / t

0.75 = (v – 1.5) / 12

Cross multiply

v – 1.5 = 0.75 × 12

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1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
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Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

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When the projectile hits the 50m mark, y=0; therefore,

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solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

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\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

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the vertical component of the velocity is

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v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

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3 years ago
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