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inna [77]
3 years ago
5

A plumber is trying to fix a clog in a vertical pipe, but does not know how far down in the pipe the clog and the water level is

. He is a good whistler, so he whistles into the open end of the pipe and finds that a frequency of 197hz (cycles per second) returns a loud resonant sound. He knows the speed of sound is 340 m/s. Calculate the depth of the air in the pipe (in meters) above the clog.
Physics
1 answer:
PIT_PIT [208]3 years ago
5 0

Answer:

the shortest distance to the obstruction is 0.431 m

Explanation:

We can see this system as an air column, where the plumber is open and where the water is closed, in the case when he hears the sound there is a phenomenon of resonance and superposition of waves with constructive interference.

For the lowest resonance we must have a node where the water is and a maximum where the plumber is a quarter of the wavelength

       λ = ¼ L

If we are in a major resonance specifically the following resonance. We have a full wavelength plus a quarter of the wavelength

    λ = 4L / 3

The general formula is

    λ = 4L / n            n = 1, 3, 5, 7,…

In addition the wave speed is the product of the frequency by the wavelength

    v = λ f

Let's replace

    v = (4L / n) f

    L = v n / (4 f)

Now we can calculate the depth or length of the air column

If we have the first standing wave n = 1

    L = 340 1 / (4 197)

    L = 0.431 m

If it is the second resonance n = 3

    L = 340 3 / (4 197)

    L = 1.29 m

We can see the shortest distance to the obstruction is 0.431 m

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3 years ago
1. A 1.32 x 104 meter steel railroad track with a coefficient of linear expansion of 12 x 10-6 per degree Celsius changes temper
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Answer:

1) 0.1584 m

2) To allow for expansion without derailment

3) 0.101376 m

4) 213.675 °C

5) 266.67 m

6) 8.33 × 10⁻⁶ /°C

7) The alloy meets the requirement

8) 1.95 × 10⁻³ /°C

9) 32.095 m

10) -12157.72°C

Explanation:

1) Equation for the coefficient of linear expansion = \frac{\Delta L}{L} = \alpha _L \Delta T

Where:

ΔL = Change in length = Required

L = Initial length = 1.32 × 10⁴ m

\alpha _L = Coefficient of linear expansion of steel = 12 × 10⁻⁶ /°C

ΔT = Change in temperature = 37°C - 27°C = 10°C

Plugging the values in the equation for the temperature expansion of steel, we have m;

ΔL = L × \alpha _L ×ΔT = 1.32 × 10⁴ × 12 × 10⁻⁶ × 10  = 0.1584 m

2. Here we have that by segmenting railroad tracks into short pieces, the expansion of the metal tracks with temperature can be absorbed by the gaps between the segment without distorting the shape and direction (pattern) of the tracks

3. Here we have;

\alpha _L = Coefficient of linear expansion of iron = 12 × 10⁻⁶ /°C

ΔT = Temperature change = 27°C - 3°C = 24°C

L = Height of the Eiffel Tower = 352 meters

∴ ΔL = L × \alpha _L ×ΔT = 352 × 12 × 10⁻⁶ × 24 = 0.101376 m

Therefore, the height of the Eiffel Tower changes from 352 m to about 352.101376 m each year, with an average change in height experienced each year = 0.101376 m

4. Here, we have

L = 13.0 ft

ΔL = 1 in.

\alpha _L = 30 × 10⁻⁶ /°C

ΔT = Required temperature change

From  \frac{\Delta L}{L} = \alpha _L \Delta T

\Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{1}{156 \times 30 \times 10^{-6}} = 213.675^{\circ}C

5. Here, we have;

L = \frac{\Delta L}{\alpha _L \Delta T}

∴ L = 1/(150×25 × 10⁻⁶) = 266.67 m

The bars original length = 266.67 m

6. Here we have;

\alpha _L = \frac{\Delta L}{L \times \Delta T}

Where:

ΔL = 3.00 - 3.002 = 0.002 m

L = 3.00 m

ΔT = 110°C - 30°C = 80°C

∴ \alpha _L = 0.002/(3.00 × 80) = 8.33 × 10⁻⁶ /°C

7. Here we have;

ΔL = L × \alpha _L ×ΔT = 3 × 8.33 × 10⁻⁶ × 210 = 0.00525 m

Therefore, final length = 3.00 m + 0.00525 m = 3.00525 m

Since 3.00525 m < 3.017 m hence the alloy meets the requirement.

8. Here, we have

L = 3.2 m

ΔL = 0.5 m

ΔT = 84°C - 24°C = 60°C

∴ \alpha _L = 0.5/(3.2 × 60) = 1.95 × 10⁻³ /°C

The coefficient of linear expansion of the material from which the rod is made = 1.95 × 10⁻³ /°C

9. Here, we have

Length of steel girder, L = 32.10 m

ΔT = 8°C - 22°C = -14°C

\alpha _L = 12 × 10⁻⁶ /°C

ΔL = L × \alpha _L ×ΔT

Hence ΔL = 32.1 × 12 × 10⁻⁶× -14 = -0.0054 m

New length = 32.1 - 0.0054 = 32.095 m

10. Here we have;

ΔL = 92.6 cm - 123 cm  = -30.4 cm

\alpha _L = 2.0 × 10⁻⁵ /°C

L = 123 cm

∴ \Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{-30.4}{123 \times 2.0 \times 10^{-5}} = -12357.724^{\circ}C

Therefore, the temperature will be 200 - 12357.724 = -12157.72°C.

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A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with ra
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