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3 years ago

Que a state properties of Sounds ] 1 laws of replactions of light 2 2

Engineering

1 answer:

Vanyuwa [196]3 years ago

5 0

Answer:

A sound is an energy source, much like electricity, heat or light. That makes a loud ringing noise when you hit a bell. Now put your finger on the bell after you’ve hit it, rather than just listening to the bell. You can feel it shaking and this motion or moving, i.e., the body’s to and fro motion is termed as vibration.

Sound is a vibration that passes through the medium in the form of longitudinal waves. It implies that sound waves are waves wherein the particles of the medium vibrate parallel to the direction of wave propagation. Sound forms are known as mechanical waves because they require a propagating medium. The medium may be

Solids

Liquids

Gases

Properties of sound

Frequency or pitch

Frequency is the number of cycles of periodic compression and rarefaction which occur every second as the wave propagates via the medium.

The human ear’s interpretation of the sound level within the range of human hearing is called the pitch.

The higher the sound frequency, the higher the pitch is and a lower frequency means a lower pitch.

Speed

The speed at which the sound waves travel via the medium is called sound speed. The speed of sound for different mediums is different. Sound moves in solids faster, as the atoms in a solid are packed tightly.

Amplitude or Loudness

The loudness determines the amplitude of the sound waves.

The sound amplitude is a measure of the magnitude of the overall sound disturbance.

The amplitude is a measure of the vibrational energy.

More energetic vibration is responsible for a greater amplitude.

Timbre

Timbre is the property used to distinguish between sounds of the same frequency.

Timbre based on what substance the sound is created from.

Reflection of sound

As sound waves encounter a solid or light surface, it bounces back into the same medium. It is termed as sound reflection. Sound waves obey the laws of reflection, like light waves.

Explanation:

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#keep learning!!

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Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

3 years ago

Mnsdcbjksdhkjhvdskjbvfdfkjbcv hjb dfkjbkjfvvfebjkhbvefgjdf

Julli [10]

Answer:

ik true eedcggftggggfffff

8 0

3 years ago

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Pls help me it’s due today

hichkok12 [17]

Answer:

C. 14.55

Explanation:

12 x 10 = 120

120 divded by 10 is 12

so now we do the left side

7 x 3 = 21 divded by 10 is 2

so now we have 14

and the remaning area is 0.55

so 14.55

6 0

3 years ago

Given below are the measured streamflows in cfs from a storm of 6-hour duration on a stream having a drainage area of 185 mi^2.

sertanlavr [38]

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

Derive the unit hydrograph using the inverse procedure

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh * duration

= 29700 * 6 hours ( 216000 secs )

= 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000 / 185 mi^2

= 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh / volume of runoff in equivalent depth

= 50 ft^3 / 1.49 in = 33.56 ft^3/sec.in

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3 years ago

Anything that is made to meet a need or desire is?

slavikrds [6]

Answer:

I think it is process or technology

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2 years ago

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Que a state properties of Sounds ] 1 laws of replactions of light 2 2​

Que a state properties of Sounds ] 1 laws of replactions of light 2 2